The University of Hong Kong Department of Physics

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1 The University of Hong Kong Department of Physics PHYS2250 Introductory Mechanics Teaching notes September 2017

2 Course title Introductory Mechanics Aim This course covers the foundation of mechanics in one semester. It serves as a core course for students who are planning to take physics, astronomy, or mathematics/physics as major. It also serves students who intend to take physics as minor. Both conceptual ideas and mathematical treatment in mechanics are emphasized. Contents Kinematics, Newton s Laws of Motion and Their Applications, Linear Momentum and its Conservation, Variable Mass Problems, System of Particles and Centre of Mass, Torque and Rotation, Angular Momentum and its Conservation, Work, Energy and its Conservation, Gravitation, Simple Harmonic Motions, Fluid Static and Pressure, Archimedes Principle and Buoyancy, Bernoulli s Equation, Surface Tension and Capillary Tube. Length One semester (The course is offered in the first semester and it is repeated in the second semester.) Assessment 50% from a two-hour written examination and 50% continuous assessment (including laboratory works (15%), assignments (10%) and a 90-minute quiz (25%)). Textbooks P. A. Tipler and G. Mosca: Physics for Scientists and Engineers (Freeman, 6 th edition). References R. D. Knight: Physics for Scientists and Engineers (Pearson, 2008). D. Kleppner and Robert J. Kolenkow: An Introduction to Mechanics (Mc- Graw Hill, 1978, International Edition). R. Resnick, D. Halliday and K. Krane: Physics Volume 1 (John Wiley and Sons, 2002). R. Serway and J. W. Jewett: Physics for Scientists and Engineers (Thomson, 2014, 9th edition). T. Duncan: Advanced Physics (John Murray Publishers, 2000, 5th edition). Atam P. Arya, Introduction to Classical Mechanics (Prentice Hall).

3 Contents Table of Contents i 1 Dimensional Analysis 1 2 Motion in One Dimension Vector Position, velocity and acceleration vectors One dimensional motion Further example: Non-constant acceleration Newton s Law of Motion Newton s first law Reference frame and relative motion Newton s second law Newton s third law Application of Newton s law in one dimension Pushing a packing crate Mass hanging in a lift Motion in a Plane Projectile motion Shooting a falling object Uniform circular motion Polar coordinate system Application of Newton s Law s with tension and normal force Frictional force i

4 5.2.1 Static friction Kinetic friction Some examples concerning uniform circular motion Impulse and Momentum Definition of linear momentum Collision between bodies Impulse and momentum Conservation of linear momentum Collision of particles in two dimensional space Systems of Particles Center of mass Center of mass of some rigid bodies Momentum of system of particles System of variable mass Rotational Kinematics Rotation with constant angular acceleration Relation between linear and angular variables Rotational Dynamics Torque Rotational inertia and Newton s 2nd law in rotation Parallel axis theorem Rotational inertia of solid bodies Perpendicular axis theorem Equilibrium of rigid body Non-equilibrium situation: pure rotation Non-equilibrium situation: rotational and translational motion Angular Momentum Definition Angular momentum and angular velocity Conservation of angular momentum Stability of spinning object ii

5 11 Work, Kinetic Energy and Potential Energy Work done by a constant force Work done by a variable force One dimensional case Two dimensional case Work-energy theorem Work done and kinetic energy in rotational motion A combination of rotational and translational motions Kinetic energy in collision Conservative force Potential energy Conservation of mechanical energy One dimensional conservative system Conservation of Energy Internal energy in a system of particle Some examples of conservation of energy Gravitation Newton s law of universal gravitation Gravitation near the Earth s surface Effect of Earth s rotation Gravitational force due to an uniform spherical shell Gravitational potential energy Potential energy of many-particle system Energy consideration of satellite motion Simple Harmonic Motions Hooke s law Potential energy and kinetic energy in a spring-mass system Damped oscillation Resonance Fluids Static equilibrium in fluids: Pressure and depth Pascal s principle iii

6 15.3 Archimedes principle and buoyancy Continuity of fluid flow Bernoulli s equation Surface tension Pressure difference across a curved surface Capillary rise formula iv

7 Chapter 1 Dimensional Analysis All the physical quantities used in mechanics are originated from the three basic dimensions: Mass Length Time SI units kg m s For example: acceleration [L T 2 ] force [M L T 2 ] energy [M L 2 T 2 ] momentum [M L T 1 ] For any equation to be held, dimension on the right hand side equals to that of left hand side. Dimensional analysis is a simple and direct method to obtain the physical equation. The assumption of the method is that the relation between quantities obeys the power law. We know that a force is required to keep an object in circular motion. The magnitude of force depends on the object mass, speed and the radius of the motion. Find the empirical formula to relate these quantities. Solution F m a v b r c 1

8 Chapter 1 Dimensional Analysis 2 [F ] = [m a ] [v b ] [r c ] [M L T 2 ] = [M] a [L T 1 ] b [L] c = [M] a [L] b+c [T] b a = 1, b = 2, c = 1, and F mv2 r. The frequency of vibration f of a mass m at the end of a spring that has a stiff constant k is related to m and k by a relation of the form f m a k b. Use dimensional analysis to find a and b. It is known that [f] = [T] 1. Solution Consider the dimension of stiff constant k first. According to the Hooke s law, the magnitude of restoring force of a spring F relates to the extension x by F = kx. dimensional analysis, [F ] = [M] [L] [T] 2 and [x] = [L]. Therefore, [k] = [F ] [x] = [M] [L] [T] 2 [L] = [M] [T] 2 Assume that the frequency f relates m and k by the power law. We have f m a k b [T] 1 = [M] a [MT 2 ] b = [M a+b T 2b ] a + b = 0 and 2b = 1 a = b = 1 2 k Hence, we obtain the relation f m. Using The wave speed on a stretched string is believed to relate the linear density and tension of a string respectively. Find an empirical formula of the wave speed v to relate µ and F. Solution Assume that v µ α F β, where the proportional constant is dimensionless. We consider the dimension of quantities on both sides. [v]=[l] [T] 1 and [µ]=[m] [L] 1, [F ]=[M] [L] [T] 2 We have [µ] α [F ] β =[M] α+β [L] α+β [T] 2β. Therefore, we have α + β = 0, α + β = 1 and F 2β = 1. The solutions are α = 1/2 and β = 1/2. i.e. v µ.

9 Chapter 2 Motion in One Dimension 2.1 Vector a) Definition A vector is a quantity that has both magnitude and direction. It is represented by an arrow. A vector v is sometimes written as v. y v y v θ v x x v v x is called the x-component of vector v and v y is called the y-component of vector v. The magnitude of the vector v is denoted by v or v, where v = v = vx 2 + vy 2 A vector is said to be a constant vector, if its direction does not change and its magnitude is a constant. 3

10 Chapter 2 Motion in One Dimension 4 b) Addition of vectors y b a+b a x c) Multiplication of vector by a scalar y If α is positive, α a has the same direction as a, but its magnitude equals to α times the magnitude of a. If α = 1, α a = a, the direction of it is opposite to a and their magnitudes are the same, i.e. a = a. a αa x d) Component form of a vector î and ĵ are called the unit vector of x and y directions respectively. They have magnitude of unity and fixed directions; they are constant vectors, i.e. î = ĵ = 1. Thus, v = v x î + v y ĵ. With component form, u = u x î + u y ĵ, v = v x î + v y ĵ j y v xi vx v v y j v y u + v = (u x + v x ) î + (u y + v y ) ĵ α u = αu x î + αu y ĵ i x

11 Chapter 2 Motion in One Dimension Position, velocity and acceleration vectors Position vector It is a vector to describe the position of a particle at time t. Generally, it is a vector function of t. It is a constant vector if the particle is stationary. y r r(t 1 ) r(t 2 ) x r(t) = r x (t)î + r y (t)ĵ r = r(t 2 ) r(t 1 ) displacement vector Velocity vector Average velocity in time period t 1 t 2 : Instantaneous velocity at time t: v ave = r t = r(t 2) r(t 1 ) t 2 t 1 If the time interval t is infinitesimal small, i.e. t 0, the instantaneous velocity at time t 1 : r v inst = lim t 0 t = d r(t) dt = r (tangential to curve r(t)) Acceleration vector Likewise, the average acceleration in time period t 1 t 2 : a ave = v t = v(t 2) v(t 1 ) t 2 t 1 v And, the instantaneous acceleration a inst = lim t 0 t = d v(t) dt = d2 r(t) dt 2 = r When we talk about velocity or acceleration, we are usually referring to the instantaneous velocity and instantaneous acceleration.

12 Chapter 2 Motion in One Dimension 6 A car is travelling along a straight road and is decelerating. Does the car s acceleration a necessarily have a negative value? Solution We begin with the meaning of the term decelerating which has nothing to do with whether the acceleration a is positive or negative. The term means only that the acceleration vectors points opposite to the velocity vector and indicates that the moving object is slowing down. When a moving object slows down, its instantaneous speed (the magnitude of the instantaneous velocity) decreases. One possibility is that the velocity vector of the car points to the right, in the positive direction, as shown in figure (a). The term decelerating implies that the acceleration vector points to the left, which is the negative direction. Here, the value of the acceleration a would indeed be negative. However, there is another possibility. The car could be travelling to the left, as shown in figure (b). Now, since the velocity vector points to the left, the acceleration vector would point to the right, according to the meaning of decelerating. But right is the positive direction, so the acceleration a would have a positive value in figure (b). We see, then, that a decelerating object does not necessarily have a negative acceleration. Describe the motion of a car if its velocity satisfies the relation d v dt = 0. How about the case if its velocity satisfies the relation d v = 0? dt Solution Since the acceleration a = d v dt, so d v dt = 0 means the magnitude of the acceleration is zero, i.e. its velocity does not change in both magnitude and direction as time passes. So, if a car moves such that d v dt = 0, then it must move in a straight line with a constant speed.

13 Chapter 2 Motion in One Dimension 7 But if the car now moves under the constraint that d v = 0, then only the magnitude dt of the velocity will keep constant. Although the car moves with constant speed, it may change its direction of motion. The projectile motion: y v 0 v(t 1 ) r(t 1 ) a(t 1 ) θ x a(t) = g ĵ v 0 = v 0 cos θ î + v 0 sin θ ĵ v(t) = (v 0 cos θ) î + [(v 0 sin θ) gt] ĵ r(t) = (v 0 cos θ)t î + [(v 0 sin θ)t 1 2 gt2 ] ĵ A river of width L is streaming due north. Along the middle line of the river, the stream has a constant speed u 0, while at both banks it is stationary. ( The speed at any point in the river varies and has the relation u = u 0 + k x L ) 2, where x is the perpen- 2 dicular distance measured from the west bank and k is a constant. A boat starts from the west bank and it moves due north east with speed v 0. Find the locus of the boat and the location of it when it reaches the east bank. You may consider a coordinate system where the origin is the starting point of the boat and the y-axis is along the west bank. Solution Given that u = u 0 + k(x L/2) 2. As x = 0, u = 0, we have 0 = u 0 + k(l 2 /4). i.e. k = 4u 0 /L 2.

14 Chapter 2 Motion in One Dimension 8 Hence, we have u = u 0 4u ( 0 x L ) 2, which gives u = 4u ( 0x 1 x ). L 2 2 L L Let v be the velocity of the boat relative to the bank. i.e. v = v ( ) 0 v0 î + + u ĵ. 2 2 t L ( ) dt L ( ) vy We have, y = v y dt = v y dx = dx, where v x = v 0 and 0 0 dx 0 v x 2 v y = v 0 + 4u 0x 2 L (1 x ). That is L 2 L ( v0 y = + 4u 0x v L (1 x ) L ) y = x + 2 2u 0 v 0 L x2 4 2u0 3v 0 L 2 x3 y x=l = L ( u 0 v 0 4 Alternatively, we have, x = v x t = v 0 2 t and y = t 0 ) 2u0 3v 0 v y dt = t 0 dx ( ) v0 + u dt. 2 y = = = t 0 t 0 t 0 [ v0 + 4u 0x 2 L [ v u 0v 0 2L t ( 1 L) x ] dt ( 1 v 0 t 2L ] [ v u 0v 0 2L t 2u 0v 2 0 L 2 t 2 )] dt dt When x = L, = v 0 2u0 v 0 t + t 2 2u 0v0 2 2 L 3L 2 = x + 2 2u 0 v 0 L x2 4 2u0 3v 0 L 2 x3 t3 y = L + 2 2u 0 L 4 2u0 L v 0 3v ( 0 y = L u 0 4 ) 2u0 v 0 3v 0

15 Chapter 2 Motion in One Dimension One dimensional motion A stone is thrown from the ground towards the sky with vertical speed v 0. y t=0 way up max. height v=0 way down just before touching ground a=g a=g v<v0 a=g a=g a=g 0 v=v 0 y y max v<v 0 y v=v 0 decelerate accelerate Take upward as positive y direction. a = dv dt = g v = a dt = dv (a = dt = d2 y dy, where v = dt2 dt ) g dt = gt + A, A = constant To determine A, substitute the initial condition at t = 0, v t=0 = A = v 0 Alternatively, one can consider definite integral. v = v 0 gt (2.1) dv dt = g v v 0 dv = t 0 g dt v v 0 = gt v = v 0 gt v = dy dt = v 0 gt (2.2) y = (v 0 gt) dt = v o t 1 2 gt2 + B, B = constant To determine B, substitute t = 0 again, y(t = 0) = B = 0 y = v 0 t 1 2 gt2 (2.3)

16 Chapter 2 Motion in One Dimension 10 At maximum height, dy dt = 0, i.e. v 0 gt max = 0 t max = v 0 t=tmax g. ( ) v0 y max = v 0 1 ( ) 2 g 2 g v0 = v2 0 g 2g Time for the stone to hit the ground, say t 0, y = v 0 t gt2 0 = 0 t 0 = 0 or t }{{} 0 = 2v 0 g initial time a g t v v 0 t v 0 v0 2 2g y v t 1 gt v g 0 2v 0 g t A screw at the ceiling of an elevator falls down while the elevator is moving upward with an acceleration a. The height of the elevator is h and the gravitational acceleration is g. Find the time when the screw just touches the floor of the elevator. Evaluate the time when a = 1.22 m/s 2 and h = 2.74 m.

17 Chapter 2 Motion in One Dimension 11 Solution The screw: y h = ut 1 2 gt2 The elevator: y 0 = ut at2 Eliminate y, we have When a = 1.22 ms 2, t = ut at2 = h + ut 1 2 gt2 1 2 (a + g)t2 = h 2h t = a + g 2(2.74 m) = s. ( ) ms 2 Mass B has a downward velocity in meters per second given by v B = t 2 /2 + t 3 /6, where t is in seconds. Calculate the acceleration of A in terms of t.

18 Chapter 2 Motion in One Dimension 12 Solution Let the total length of the string be l. We have 4 l l 2 + constant = l, which gives 4 l l 2 = 0 or l 1 = l 2 2, i.e. v A = v B 2. v A = 1 ( ) t t3 t24 t3 = 6 12 The velocity of A is negative as the length l 1 decreases with time when A is moving up. a A = t 2 t2 4. Here, a A is negative, because mass A is accelerating upward. The magnitude of it is a A = t 2 + t2 4. A boy of height h is standing under a light pole of height H. He moves away from the pole along a horizontal and straight path with a constant speed v. Find the rate of change of the length of his shadow. Solution Let the coordinates of boy and one end of his shadow be x and x respectively. Since x /H = (x x)/h, we have x = (xh)/(h h). Note that l = x x, which gives dl dt = dx dt dx dt Therefore, the rate of change of length of shadow dl dt where v = dx/dt. = d ( ) xh dt H h Hv = H h v hv = H h dx dt

19 Chapter 2 Motion in One Dimension 13 A rocket at rest is launched to move upwards. At the first 30 seconds, the rocket moves with an upward acceleration of 18 ms 2. Then the rocket s engine shuts down. As a result, the rocket keeps on moving upward for a while and then falls back to the ground. (a) Calculate the maximum height that the rocket can reach. (b) Calculate the total time of flight of the rocket. Solution After the lift-off, the motion of the rocket can be divided into the two parts: 1) moving with upward acceleration 18 ms 2 during the first 30 seconds, 2) free-fall to the ground with downward acceleration g = 9.8 ms 2 after its engine is shut down. (Note that we have taken upward as positive direction and neglect the air resistance.) (a) Suppose the rocket reach the maximum height H after shutting down its engine for time t 2. Then the height H is given by H = 1 2 at2 1 + v 1 t gt2 2 where a = 18 ms 2, t 1 = 30 s is the time for the upward acceleration and the velocity of the rocket is v 1 (at the instant that the engine shuts down). Besides, the velocity v 1 is equal to v 1 = at 1 = (18 ms 2 ) (30 s) = 540 ms 1 Obviously, the velocity of the rocket should be zero at the maximum height H. So we can find the time t 2 as follows: 0 = v 1 gt 2 t 2 = v 1 g Substituting the result of t back into eq.(5.6), we obtain H = 1 ( ) v1 2 at2 1 + v 1 1 ( ) 2 g 2 g v1 g = 1 2 at2 1 + v2 1 2g = 1 2 (18 ms 2 ) (30 s) 2 + (540 ms 1 ) ms 2 = m

20 Chapter 2 Motion in One Dimension 14 (b) The total time of flight is given by where t 3 t = t 1 + t 2 + t 3, is time taken for the rocket in free-fall to reach the ground form the maximum height H. For the flight from the height H to ground, H = 1 2H 2 gt2 3 t 3 = g Therefore, t = t 1 + t 2 + t 3 = t 1 + v 1 g + 2H g 540 ms 1 = 30 s ms + 2 = s m 9.8 ms 2 A man near the cliff is pulling a boat in the river (see figure). The cliff is of height h. The speed and acceleration of the man are v and a respectively (not necessary constants), when the rope makes an angle φ to the horizontal. At the same instant the horizontal speed and the horizontal acceleration of the boat are v and a respectively. The rope keeps taut during the motion, find v if the rope keeps taut during the motion. Find also the acceleration a. Express your answer in terms of the given parameters. Solution There are two approaches to find the speed v of the boat. The first one bases on conceptual idea and the second one relies on mathematical treatment.

21 Chapter 2 Motion in One Dimension 15 Method I: Observe that v is not a component of v along the horizontal. But, instead v is a component of v along the rope because v is the final velocity of the boat and the rope is always tight. Method II: v cos φ = v v = v cos φ Notice the right-angled triangle in the figure, we have l 2 = h 2 + x 2. The lengths l and x vary with time while h is a constant. Differentiate on both sides with time, we obtain 2l l = 2 xẋ 2 lv = 2xv v = lv x v v = cos φ where l = dl dt with time. = v and ẋ = dx dt = v. The negative sign appears as l and x decreases To find the acceleration a, we differentiate the expression v with time. Since v = we have a = dv dt = d ( ) v = a cos φ + v φ sin φ dt cos φ cos 2 φ where a = dv dt. The problem is solved if φ is known. v cos φ,

22 Chapter 2 Motion in One Dimension 16 Let s consider the right-angled triangle again. h = l sin φ 0 = l φ cos φ + l sin φ 0 = l φ cos φ v sin φ v tan φ φ = l Plugging the above result to Eqn. (2.4), we obtain where h = l sin φ. a = a cos φ + v ( v tan φ l cos 2 φ a = a cos φ + v2 tan 2 φ l cos φ a = a cos φ + v2 tan 3 φ h ) sin φ 2.4 Further example: Non-constant acceleration Dropping a stone at a height of h with the air resistance not ignorable. Given that the air drag force per unit mass = kv, where k is a positive constant and v is the instantaneous velocity of the stone. y 0 y(0)=h a= g kv g is positive k is positive v is negative as it points downward a = g kv dv = g kv dt dv Separation of variable and integrating on both sides, we have g + kv = dt. 1 ln(g + kv) = t + A, where A = constant and g + kv > 0 k g + kv = e k( t+a) = Be kt, where B = e ka v = 1 k (Be kt g)

23 Chapter 2 Motion in One Dimension 17 At t = 0, v(0) = 0 B g = 0 B = g. v = 1 k (ge kt g) = g k (1 e kt ), i.e. dy dt = g k (1 e kt ) g k v t g y = k (1 e kt )dt + C, = g k t g k 2 e kt + C where C = constant. At t = 0, y = h h = g k 2 + C C = h + g k 2. Therefore, y = g k t g k 2 e kt + h + g k 2 = h g k t + g k 2 (1 e kt )

24 Chapter 3 Newton s Law of Motion 3.1 Newton s first law If there is no net force acting on a body, the body will preserve its state of motion, i.e. if the body is at rest, it remains at rest; if the body moves with a velocity, it will keep on moving with that constant velocity. 3.2 Reference frame and relative motion Reference frame: Where do we observe the motion of an object? S : Earth s frame r ms r ms S S : Car s frame moving with v S S with respect to S rs S v S S S where r S S - Position vector of the car observed from the earth, r ms - Position vector of the object observed from the earth, r ms - Position vector of the object observed from the moving car. 18

25 Chapter 3 Newton s Law of Motion 19 Notice that: d r S S = v S dt S From the vector diagram: where r ms = r S S + r ms d r ms = d r S S + d r ms dt dt dt or v ms = v S S + v ms or v ms = v ms v S S v S S - velocity of the car observed from the earth, v ms - velocity of the object measured from the earth, v ms - velocity of the object measured from the moving car. From v ms = v S S + v ms, we obtain a ms = a S S + a ms. If v S S is a constant, i.e. a S S = 0, the observers are inertial to each other. Hence, a ms = a ms. The observers in S and S have the same observation on the acceleration of a moving object and they write down the same equation of motion. A truck has a h = 2 m rear door. When it stops on the road with the rear door opened, the rain drops can fall into the truck up to d = 1 m from the rear door (as shown in figure). The rain drops no longer can fall into the truck once the truck increases its speed to 15 km h 1 while moving on a level road. Find the velocity of the rain drops (relative to the ground). Solution Because of the relativeness of the velocity, we have v rain to car = v rain to ground v car to ground (3.1)

26 Chapter 3 Newton s Law of Motion 20 Since the rain cannot fall into the car when the car increase its velocity to 15 km hr 1 relative to the ground, v rain to car should point vertically. Then, the three relative velocities will have the relation as shown in the below figure. And the angle α is given by α = tan 1 ( h d ) ( ) 2 m = tan 1 = m Using eq. (3.1), we obtain the speed of the rain drops v rain to ground = v car to ground cos α = 15 km h 1 cos 63.4 = 33.5 km h 1 A ship A is 5 km due north of a ship B. A is streaming due west at 15 km/hr and B is streaming due north-west at 10 km/hr. Find the distance and the time of their nearest approach to each other. Solution

27 Chapter 3 Newton s Law of Motion 21 v AB = v A v B = ( 15 î) ( 10 cos 45 o î + 10 sin 45 o ĵ) = ( 15 î) ( 10 î + 10 ĵ) 2 2 = ( ) î 10 2 ĵ v AB = 7.93 î 7.07 ĵ Therefore, tan θ = v AB,x v AB,y = , i.e. θ = 48.28o. From the right figure, we observe that the shortest distance between A and B is d, where d = 5 sin o = 3.73 km. Hence, 5 cos 48.28o t = = 0.31 hr, where v AB = 7.93 v AB km/hr = km/hr. John is running at constant speed v 0 = 0.7 m/s along a straight path. His father Peter, at a normal distance 10 m from the path and a distance 20 m from John, observes Johns approaching. Suppose that Peter starts to run at constant speed, along a straight course, immediately when he observes John, what is the minimum speed of Peter such that they can meet? Solution Subtract the velocity vector of John from both John and Peter. John becomes stationary. Now, Peter can meet John if the relative velocity vector of Peter to John is along the line joining them. The required velocity vector, i.e. the shortest vector v min, is obtained if it is normal to this line. From the figure, we notice that sin θ = 10 m = 0.5, and thus 20 m v min = (0.7 m/s) sin θ = (0.7 m/s) (0.5) = 0.35 m/s.

28 Chapter 3 Newton s Law of Motion Newton s second law If there is a net force F acting on an object m, then F equals to the rate of change of the object momentum, i.e. F d p = dt. OR in a simple statement: if the mass of the object is a constant m, we have F = m a, as F = d p dt = d(m v) ( ) d v = m = m a, where a is the object acceleration. dt dt Unit of force: Newton = kg m s 2 ([MLT 2 ]) One Newton of force is the magnitude of force acting on a 1 kg object that will accelerate the object with acceleration of 1 ms 2. A ball bearing is released from rest and drops through a viscous medium. The retarding force acting on the ball bearing has magnitude kv, where k is a constant depending on the radius of the ball and the viscosity of the medium, and v is the bearing s velocity. Find the terminal velocity acquired by the ball bearing and the time it takes to reach a speed of half the terminal velocity. Solution As the ball falls through the medium, it is accelerated by gravity and the viscous force (which has direction opposite to that of the gravitational force). To find the acceleration of the bearing, we use Newton s second law i F i = m a to relate the net force on the ball to its acceleration. Taking downward as positive direction, mg kv = ma where k becomes positive and a is the acceleration of the ball at any time. The initial value of a is g, since at the moment of release v = 0. As the value of v increases, the acceleration decreases until a = 0 when v = v 0, the terminal velocity. Therefore, mg kv 0 = 0 v 0 = (m/k)g

29 Chapter 3 Newton s Law of Motion 23 In order to find out at which time v = v 0, we must calculate t as a function of v (or vice versa). At any time t, it will be found that thus mg kv = ma = m dv dt, (mg kv) dt = m dv dt = m dv mg kv = dv g (k/m)v t v dv dt = g (k/m)v 0 where, in the integration limits, v = 0 at t = 0 and v = v at t = t. Hence, t = v Let u = g (k/m) v, then du = (k/m) dv. 0 0 dv g (k/m)v To find the new integration limits, we realized that when v = 0, u = g, and when v = v, u = g (k/m) v. Therefore, g (k/m)v (m/k) du t = g u [ m ] g (k/m) v = k ln u = m k g [ln g (k/m)v ln g ] = m k ln 1 (k/mg) v The time to acquire half the terminal velocity, T, is found by inserting v = v 0 /2 = mg/2k in the above equation T = m k ln 1 ( ) k mg ( mg 2k ) = m k ln 1 2 = 0.69 m/k

30 Chapter 3 Newton s Law of Motion Newton s third law As a body exerts a force on another body, the second body also exerts a force on the first. The two forces are equal in magnitude but opposite in direction. F F 3.5 Application of Newton s law in one dimension Pushing a packing crate A worker w is pushing a packing crate of mass m 1 = 4.2 kg. In front of the crate is a second crate of mass m 2 = 1.4 kg. Both crates slide across the floor without friction. The worker pushes on crate 1 with a force F = 3.0 N. Find the accelerations of the crafts and the force exerted by crate 1 on crate 2. a a R 1 R 3 m 1 m m 1 2 F R m 2 2 R2 m g 1 m 2 g where R 1 - reaction from the ground on m 1, R 2 - action reaction pair between m 1 and m 2, R 3 - reaction from the ground on m 2. Note that R 1 = m 1 g, R 3 = m 2 g and no vertical motion (i.e. a = 0).

31 Chapter 3 Newton s Law of Motion 25 Taking right side as positive. Using Newton s 2 nd law, we find the equation of motions: F R 2 = m 1 a (3.2) R 2 = m 2 a (3.3) Substitute (3.3) into (3.2), we get: F m 2 a = m 1 a F a = m 1 + m 2 R 2 = m 2 a = m 2F m 1 + m 2 Remark: It is interesting to compare the reaction force if the two crates are interchanged. When we move two crates along a straight line, the crate of smaller mass should be put in front of the heavier one in order to reduce damage Mass hanging in a lift A mass m is hanging by a massless string which is attached to the ceiling of a lift. The mass of lift is M. Case 1: The lift has no acceleration but moves upward with constant speed v (see figure below). T 2 T 2 T T 1 T 1 mg T 1 T 1 mg Mg Mg Taking upward as positive. Since a = 0, the equation of motions are given by T 1 mg = 0 T 1 = mg T 2 T 1 Mg = 0 T 2 = (m + M)g

32 Chapter 3 Newton s Law of Motion 26 N.B. Results are the same for the following cases: 1) a = 0 and v > 0, 2) a = 0 and v < 0, 3) a = 0 and v = 0 Case 2: Lift accelerates upward with acceleration a. Note that a must be positive for this case. Therefore, the equation of motions are now given by T 1 mg = ma T 1 = m(g + a) T 2 T 1 Mg = Ma T 2 = (m + M)(g + a) Note that T 1 and T 2 in this case are larger than that in the previous case (a = 0). Compute the least acceleration with which a 45-kg woman can slide down a rope if the rope can withstand a tension of only 300 N. Solution According to Newton s second law, F i = m a. i Therefore, taking down as positive direction, the tension of the rope T and the acceleration of the woman a are thus related by: mg T = ma where m is the mass of the woman. Now it is given that the maximum tension of the rope is T max = 300 N. Hence, the minimum acceleration of the woman a min is given by: a min = mg T max m = [(45 kg)(9.8 ms 2 ) 300 N] 45 kg = 3.13 ms 2

33 Chapter 3 Newton s Law of Motion 27 What acceleration a must the cart Y have in order that block B does not slide down (it is at rest relative to the cart)? The inclined surface of the cart is smooth. Solution When there is no relative motion between the block and the cart, the block moves together with the cart, i.e. an acceleration a along the horizontal. Obviously, the block is at equilibrium along the vertical. Consider the free-body diagram of the block. { N sin θ = ma N cos θ = mg The above equations give tan θ = a/g or a = g tan θ. A movable table has rough surface of frictional coefficient µ and a set up with it (see figure). The set up contains one frictionless and massless pulley, one massless cord and two blocks m and M, where m < M. Assume that the cord keeps tight during the motion of the system. By considering the relative motion of the blocks with respect to the table, find the acceleration of m relative to the table and the tension of the cord if the table is accelerating downward with a T.

34 Chapter 3 Newton s Law of Motion 28 Solution Consider the motion of m: mg N = ma T (3.4) T f = ma, (3.5) where f = µn and a is the acceleration of m relative to the table. Eqn. (3.5) becomes T µn = ma (3.6) The motion of M along the vertical: Mg T = M (a + a T ) (3.7) Use Eqns. (3.6) and (3.7), we have Mg µn ma = M (a + a T ) (3.8) Use Eqns. (3.4) and (3.8), we have Mg + µ (ma T mg) ma = Ma + Ma T a = M (g a T ) mµ (g a T ) M + m

35 Chapter 3 Newton s Law of Motion 29 Hence, we obtain a = (g a T ) (M mµ) M + m (3.9) From Eqns. (3.7) and (3.9), we obtain the tension Therefore, T = M (g a a T ) = M (g (g a T ) (M mµ) a T ) ( M + m ) gm + gm Mg + MaT + gmµ mµa T Ma T ma T = M M + m M = M + m (gm + gmµ mµa T ma T ) M = M + m (gm (µ + 1) ma T (µ + 1)) T = mm (µ + 1) (g a T ) M + m

36 Chapter 4 Motion in a Plane 4.1 Projectile motion y v 0 θ x At t = 0, { vx = v 0 cos θ a x = 0, { vy = v 0 sin θ a y = g. Let us study the horizontal and vertical motions of the object in turn. I. Horizontal motion d 2 x dt = 0 2 dx dt = v x = A, At t = 0, A = constant dx dt = v x = v 0 cos θ v x (t) = v 0 cos θ x(t) = v 0 cos θ dt + B = (v 0 cos θ) t + B, B = constant 30

37 Chapter 4 Motion in a Plane 31 At t = 0, x = 0 B = 0 x(t) = (v 0 cos θ) t II. Vertical motion d 2 y dt = g 2 dy dt = v y = gt + C, At t = 0, dy dt = v y = v 0 sin θ C = constant C = v 0 sin θ v y (t) = gt + v 0 sin θ y(t) = 1 2 gt2 + v 0 sin θ dt + D = 1 2 gt2 + (v 0 sin θ) t + D, D = constant At t = 0, y = 0 D = 0 y(t) = (v 0 sin θ) t 1 2 gt2 In conclusion: or: a x (t) = 0 v x (t) = v 0 cos θ x(t) = (v 0 cos θ) t a(t) = gĵ a y (t) = g v y (t) = v 0 sin θ gt y(t) = (v 0 sin θ) t 1 2 gt2 Let t 0 be the time of flight, then v(t) = v 0 cos θ î + (v 0 sin θ gt) ĵ r(t) = (v 0 cos θ) t î + [(v 0 sin θ) t 1 2 gt2 ] ĵ Let R be the horizontal range, then 0 = (v 0 sin θ) t gt2 0 t 0 = 0 (neglected as it is the initial status) or t 0 = 2v 0 sin θ g R = x(t 0 ) = (v 0 cos θ) t 0 = 2v2 0 sin θ cos θ g Maximum range occurs at θ = 45. = v2 0 sin 2θ g

38 Chapter 4 Motion in a Plane Shooting a falling object y R m v φ x At time t = 0, the mass m is released and a bullet is fired with a velocity v 0. Interesting to notice that no matter what the value of v 0 is, the bullet hits the mass. Let r 1 be the position vector of the bullet and r 2 be the position vector of the mass. d 2 r 2 dt 2 = g ĵ d r 2 dt = gt ĵ + A, A = constant vector At t = 0, v 2 = 0 A = 0 At t = 0, r 2 = R B = R d r 2 dt = gtĵ r 2 (t) = 1 2 gt2 ĵ + B r 2 (t) = R 1 2 gt2 ĵ d 2 r 1 dt 2 = g ĵ d r 1 dt = gt ĵ + C, C = constant vector At t = 0, v 1 = v 0 C = v 0 d r 1 dt = gt ĵ + v 0 r 1 (t) = v 0 t 1 2 gt2 ĵ + D

39 Chapter 4 Motion in a Plane 33 At t = 0, r 1 = 0 D = 0 r 1 (t) = v 0 t 1 2 gt2 ĵ Rewriting r 1, r 2 into î, ĵ components, R = R cos φ î + R sin φ ĵ v 0 = v 0 cos φ î + v 0 sin φ ĵ where R = R. r 1 (t) = (v 0 cos φ) t î + [(v 0 sin φ) t 1 2 gt2 ] ĵ To check if they will collide, we can check: r 2 (t) = R cos φ î + [R sin φ 1 }{{} 2 gt2 ] ĵ const. As the bullet passes through x = R cos φ, are the y position component of the bullet and the mass the same? Let t 0 be the time that the bullet passes x = R cos φ. (v 0 cos φ)t 0 = R cos φ t 0 = R v 0 ( ) R y-component of the bullet, y 1 (t 0 ) = v 0 sin φ 12 ( ) 2 g Rv0 y-component of the mass, y 2 (t 0 ) = R sin φ 1 ( ) 2 R 2 g = y 1 (t 0 ) v 0 Thus the bullet will hit the mass. v 0 The following figure shows a ball being thrown upward from a point on the side of a hill which slopes upward uniformly at an angle of 28. Initial velocity of the ball: v 0 = 33 ms 1, at an angle of θ 0 = 65 (with respect to the horizontal). At what distance up the slope does the ball strike and in what time?

40 Chapter 4 Motion in a Plane 34 Solution When the projectile moves, the horizontal component of the its velocity remains constant at the initial value v 0x. Thus the time of flight t is related to v 0x as follows: t = x x = v 0x v 0 cos θ 0 where v 0 is the initial velocity and θ 0 is the angle of projection. Using this relation, we obtain the trajectory equation of the projectile: y = v 0y t 1 2 gt2 = ( ) x (v 0 sin θ 0 ) 12 ( v 0 cos θ x 0 v 0 cos θ 0 = gx 2 x tan θ 0 2v0 2 cos 2 θ 0 Note that v 0y = v 0 sin θ 0 is the vertical component of the initial velocity. Putting the parameters of the ball into the above equation, we find its trajectory equation y = x tan 65 (9.8)x 2 = 2.14x 2(33) 2 cos x2 65 The points on the incline are related by the equation: y = (tan 28 )x = 0.532x. At the value of x for which the ball hits the incline, we have 0.532x = 2.14x x x x 2 = 0 x = 63.8 m (The solution, x = 0, is rejected since it refers to the starting point.) The distance along the incline, S, obeys: x = S cos 28 S = x/ cos 28 = 63.8 m/ cos 28 = 72.3 m And the time to reach this point is: Remark: t = x v 0 cos θ 0 = 63.8 m (33 ms 1 ) cos 65 = 4.57 s One may consider a coordinate system having axes along and normal to the inclined plane, e.g. x and y. The ball strikes on the incline when y = 0. ) 2 In one contest, a spring-loaded plunger launches a ball at a speed of 3.0 m/s from one corner of a smooth, flat board that is tilted up at a 20 o angle. To win, you must make the ball hit a small target at the adjacent corner, 2.50 m away. At what angle θ should you tilt the ball launcher?

41 Chapter 4 Motion in a Plane 35 Solution The problem is the same as the projectile motion with reduced gravitational acceleration a = g sin 20 o. The component g cos 20 o is cancelled by the normal force. So, we have x = (v 0 cos θ) t y = (v 0 sin θ) t 1 (4.1) 2 at2 From the equation set, we have t = That is ( x y = v 0 sin θ v 0 cos θ x, and hence v 0 cos θ ) 1 2 (g sin 20o ) ( ) g sin 20 y = x tan θ x 2 o 2v0 2 cos 2 θ ( ) 2 x. v 0 cos θ (4.2) Plugging the given parameters x = 2.5 m, y = 0 m, and v 0 = 3 m/s into the equation, we obtain tan 2 θ 7.2 g sin 20 o tan θ + 1 = 0 After solving, we have tan θ = or That is θ = 55.7 o or 34.3 o.

42 Chapter 4 Motion in a Plane 36 A wedge of mass M rests on a frictionless horizontal plane. A block of mass m is released from rest on the smooth and inclined surface of the wedge which has an elevation angle θ. The acceleration of the wedge and the acceleration of the block relative to the wedge are a 1 and a M respectively. Write down the equations of motion of the wedge and the block relative to an inertial observer. Hence, find a 1 and a M. Check the limit when M >> m. Solution Before doing this problem, we notice two remarks first. If M is not moving then we have the trajectory sketched in the below figure. If M is moving to the left, then the actual trajectory is sketched as below, where a M is the acceleration of the block relative to the wedge of mass M, and a 1 is the acceleration of the wedge relative to the table. Now, we look at the force diagram of the wedge and the block and write down the equations of motion along the horizontal and vertical respectively (if necessary).

43 Chapter 4 Motion in a Plane 37 For M : N 2 sin θ = Ma 1 (4.3) For m : mg N 2 cos θ = m a M sin θ (4.4) N 2 sin θ = m (a M cos θ a 1 ) (4.5) This is a set of simultaneous equations with 3 unknowns. After solving, we obtain mg sin θ cos θ a 1 = M + m sin 2 θ a M = (M + m) g sin θ M + m sin 2 θ The reaction between the wedge and the block is given by N 2 = mmg cos θ M + m sin 2 θ Consider the equation set (4.6), when M is very large, e.g. M >> m, we have (4.6) (4.7) From eq. (4.7), N 2 mg cos θ. a 1 0 and a M g sin θ. Repeat the above example but a non-inertial observer that is fixed on the wedge. Write down the equations of motion of the block relative to this observer, along and normal to the inclined surface of the wedge. Find a 1 and a M again. Solution

44 Chapter 4 Motion in a Plane 38 Consider the motion of the block relative to the wedge. Along the incline: mg sin θ + ma 1 cos θ = ma M, (4.8) where ma 1 cos θ is the fictitious force acting on the block along the incline. Normal to the incline: N 2 mg cos θ + ma 1 sin θ = 0, (4.9) where ma 1 sin θ is the fictitious force acting on the block normal to the incline. As a reminder, the fictitious force points opposite to the motion of the observer. Using eqs. (4.3), (4.8) and (4.9), we solve the answers of a 1 and a M again. 4.3 Uniform circular motion - On a horizontal frictionless plane v 1 θ y v p P P P 1 2 θ θ r θ v 2 x v 1 = v cos θ î + v sin θ ĵ v 2 = v cos θ î v sin θ ĵ Time required for moving from P 1 to P 2 : Note that v = v 1 = v 2 x-component of average acceleration: t = 2rθ v a av,x = v 2x v 1x t v 2x = v cos θ, v 1x = v cos θ a av,x = 0 (θ is in radian)

45 Chapter 4 Motion in a Plane 39 y-component of average acceleration: a av,y = v 2y v 1y t v 2y = v sin θ, v 1y = v sin θ ( ) ( ) 2v sin θ v 2 sin θ a av,y = = 2rθ/v r θ sin θ lim θ 0 θ Now we can take the limit θ 0 to obtain the instantaneous acceleration at P. [ ( ) ( )] ( ) v 2 sin θ v 2 a y = lim = θ 0 r θ r Notice that with this condition, a x = 0. At P, the instantaneous acceleration is purely y-direction, pointing towards the center = 1 of the circular motion and has a magnitude of v 2 /r. For a uniform circular motion, the centripetal acceleration is a c = v 2 /r and is always pointing towards the center of the circle. y v a r x A mass m slides without friction on the roller coaster track shown in the below figure. The curved sections of the track have radius of curvature R. The mass begins its descent from the height h. At certain value of h, the mass will begin to lose contact with the track. Indicate on the diagram where the mass loses contact with the track and calculate the minimum value of h for which this happens.

46 Chapter 4 Motion in a Plane 40 Solution Just before the inflection point A of the track (i.e. α > 30 o ), the normal reaction of the track on the mass is N = mv2 + mg sin α > 0, R where v is the velocity of the mass and its direction points toward O. The mass should not lose contact before reaching A, because both terms in the RHS of the above equation is positive. The normal force N is decreasing when the mass is climbing up and before reaching A. At the inflection point A, the angle α = β = 30 o since sin α = sin β = R 2R = 1 2. Just after point A (i.e. β > 30 o ), the normal reaction is N = mg sin β mv2 R, which could take zero or negative value. The direction of N is away from O. Here, the force mg sin β increases and mv2 R decreases when the mass is away from A (i.e. β

47 Chapter 4 Motion in a Plane 41 is increasing), and thus N increases when the mass moves away from A. To conclude, The earliest the mass can start to lose contact with the track is at A when N 0, i.e. mv 2 R mg sin 30o. The conservation of mechanical energy mg[h R sin 30 o ] = 1 2 mv2 then requires h 3R 4. Hence the minimum h required is 3R Polar coordinate system In many situations, it is convenient to use plane polar coordinates (r, θ) instead of rectangular coordinates (x, y) to describe the motion of a particle, where r is the length of the position vector. We define two unit vectors in polar coordinates that are perpendicular to each other. ( The unit vectors (î, ĵ) are for rectangular, their directions are fixed; another pair of unit vectors (ê r, ê θ ) are for polar, their directions vary with the position vector.) { êr = cos θ î + sin θ ĵ ê θ = sin θ î + cos θ ĵ dê r dθ = sin θ î + cos θ ĵ = ê θ dê θ dθ = cos θ î sin θ ĵ = ê r (4.10) (4.11) The position vector r in terms of polar coordinates is given by r = rê r. (Note that ê r and ê θ are functions of θ, i.e. ê r = ê r (θ) and ê θ = ê θ (θ). They are unit vectors, but not constant vectors because their directions are varying.)

48 Chapter 4 Motion in a Plane 42 Thus the velocity v is v = d r dt = d dt (rê r) = dr dt êr + r d ê r dt (4.12) ( dê r dt = dê r dθ dθ dt = θê θ ) (4.13) = ṙ ê r + r θ ê θ (4.14) = v r ê r + v θ ê θ (4.15) where v r is radial velocity component of v along ê r and v θ is angular velocity component of v along ê θ. The acceleration of the system is given by a = d v dt = d dt (ṙ ê r + r θ ê θ ) (4.16) = dṙ dt êr + ṙ dê r dθ dθ dt + dr dt θê θ + r d θ dt êθ + r θ dê θ dθ dθ dt = rê r + ṙ(ê θ ) θ + ṙ θê θ + r θê θ + r θ( ê r ) θ = ( r r θ 2 ) ê r + (r θ + 2ṙ θ) ê θ (4.17) where ( r r θ 2 ) = a r is radial acceleration and (r θ + 2ṙ θ) = a θ is the angular component of the acceleration. A radar monitors a flying object with trajectory r = cos θ, the measurement is in meter. At the position θ = 45 o, the radar records the angular velocity of the object to be 0.6 rad/s. Find the speed of the object at the instant. Solution Since r = cos θ, we have ṙ = 0.5 θ sin θ. The velocity vector in the polar coordinate system gives v = ṙ ê r + r θ ê θ = 0.5 θ sin θ ê r + r θ ê θ. Plugging in the conditions, i.e. θ = 45 o, r = ( cos 45 o ) m = m and θ = 0.6 rad/s. We obtain

49 Chapter 4 Motion in a Plane 43 v = ˆr ˆθ. The speed of the object is ( 0.212) 2 + (0.812) 2 = m/s. A particle of mass m is released inside a smooth tube AB when the tube is rotating with constant angular speed ω about A horizontally. The initial position of the particle is r 0 from A and the tube has length l. Find the speed of the particle will it leaves the tube. Solution Since the tube is smooth, there is no force acting on the particle along the tube. The equation of motion of the particle along the tube is given by r ω 2 r = 0 Using the chain rule, we have r = dṙ dt = dṙ dr dr dt = becomes ( ) dṙ ṙ ω 2 r = 0 dr By considering the separation of variables, we obtain Integrating on both sides, vr 0 ṙ dṙ = ω 2 r dr l ṙ dṙ = ω 2 r dr r 0 vr 2 = ω 2 (l 2 r0) 2 ( ) dṙ ṙ. The equation of motion dr

50 Chapter 4 Motion in a Plane 44 The speed of the particle when it leaves the tube is given by v = vr 2 + vθ 2 = ω 2 (l 2 r0) 2 + l 2 ω 2 = ω 2 (2l 2 r0). 2 A smooth rod OA of length 2l rotates in a horizontal plane about its end O with constant angular speed 6g/l. One end of a light elastic string of natural length l and force constant 2mg/l is attached to O and the other end to a small smooth ring P of mass m which passes around the rod. The ring is projected from the midpoint of the rod with speed 3 gl relative to the rod in the direction OA. Show that while the ring is in contact with the rod its distance r from O satisfies the equation r 4gr = 2g. Find also the l boundary conditions of this equation. Solution Since the rod and the ring are smooth, the only force that acts on the ring along the rod is the restoring force due to the elastic string. The magnitude of the force is given by 2mg (r l). The motion of the ring is governed by l On the left side of Eq. 2mg (r l) = m( r r l θ 2 ). (4.18) (4.18), a negative sign appears and it simply represents the direction of the restoring force, pointing from the ring to the rotation center O. The

51 Chapter 4 Motion in a Plane 45 terms on the right side comes from the radial part of Eq. (4.17). Plugging into Eq. (4.18) the angular speed θ = 6g/l, we have 2g (r l) = r r l ( ) 6g l After simplification, we have r 4gr = 2g. There are two boundary conditions associated l with this equation, e.g. r(0) = l and ṙ(0) = 3 gl..

52 Chapter 5 Application of Newton s Law 5.1 s with tension and normal force A block of mass m is at rest on a table. The middle and right figures show the freebody diagrams of the block and the table respectively. As the block is in equilibrium: R W = 0, where W = mg. Thus, the reaction force on the block from the table is R = mg. Notice that R and W are not the action and reaction pair stated in Newton s third law of motion, because the forces are acting on the same object. On the other hands, Newton s three law of motion asserts that the action force on the table by the block has magnitude R = mg and a downward direction, as shown in the right figure. The reaction force on the table from the ground is N = R + W. The mass and the weight of table are m and W respectively. A mass is hanging by a string on a frictionless inclined plane. y x R m T T R m T θ mg θ mg 46

53 Chapter 5 Application of Newton s Law 47 R + T + m g = 0 x-direction: T mg sin θ = 0 T = mg sin θ y-direction: R mg cos θ = 0 R = mg cos θ y x T R m T α R m T α θ mg θ mg x-direction: T cos α mg sin θ = 0 T = mg sin θ sec α y-direction: R + T sin α mg cos θ = 0 R = mg cos θ mg sin θ tan α Under the ceiling is a mass system which has a non-stretchable and massless string, a frictionless axis, a massless pulley, and two masses m 1 and m 2. T 3 T 2 T 1 m T 2 2 m 2 g T 1 m 1 m g 1 m 1 > m 2 As the pulley has no linear motion, we can write T 3 = T 1 +T 2. Here, we do not assert that T 1 = T 2 = T, but instead we try to prove it by contradiction. Assume that T 1 T 2, there is a non-zero torque τ acting on the pulley and the pulley will has angular acceleration. As the mass of the pulley is zero and thus its moment of inertia I is also zero. Then a finite τ = I θ gives τ/i = θ which is impossible. The knowledge of torque will be introduced in latter chapters. Analyzing the force on the two masses, we find that m 1 g T = m 1 a and T m 2 g = m 2 a

54 Chapter 5 Application of Newton s Law 48 Then a and T can be found by solving these two equations. A mass m 2 is hanging over a pulley by a string which has its next end connected to a block m 1. The block can slides on a frictionless inclined surface, as shown in the figure. The pulley is massless and frictionless. R m 1 T T For the block m 1 : x-direction: T m 1 g sin θ = m 1 a θ m 1 g m 2 g m 2 T = m 1 (a + g sin θ) y-direction: R m 1 g cos θ = 0 x R = m 1 g cos θ y R a T T For the block m 2 : θ m 1 g m 1 a m 2 g m 2 m 2 g T = m 2 a T = m 2 (g a) Combining the two expressions of T, we obtain: m 2 (g a) = m 1 (a + g sin θ) a = (m 2 m 1 sin θ)g m 1 + m 2 A massless rod of length l has its two ends connected to two identical particles A and B. The system is held to align with a smooth vertical wall where both particles meet the wall and the lower particle B lies on the smooth horizontal floor. A disturbance is given so that the system slides down with A moving down along the wall and B leaving the wall along the floor. You may consider a reference frame where its origin O locates at the joint of the wall and the floor, OA = y and OB = x, where x and y are positive quantities. Show that cos θ = 2/3 when A leaves the wall, where θ is the acute angle that the rod makes with the wall.

55 Chapter 5 Application of Newton s Law 49 Solution From the figure, we have the relation x 2 + y 2 = l 2. Hence, we obtain xẋ + yẏ = 0 (5.1) and xẍ + ẋ 2 + yÿ + ẏ 2 = 0 (5.2) The conservation of mechanical energy gives mg (l y) = 1 2 mẋ mẏ2 Eq. (5.2) becomes Consider A: Note that T is a compressive force. { R1 T sin θ = 0 2g (l y) = ẋ 2 + ẏ 2 (5.3) xẍ + yÿ + 2g (l y) = 0 (5.4) T cos θ mg = mÿ

56 Chapter 5 Application of Newton s Law 50 Consider B: { T sin θ = mẍ R 2 mg T cos θ = 0 When A leaves the wall, R 1 = 0, we obtain T = 0 and ÿ = g. Also, we have ẍ = 0 when T = 0. From Eq. (5.4), x(0) + y( g) + 2g (l y) = 0, which gives y = 2l/3. Since y = l cos θ, we can write 2l = l cos θ, i.e. cos θ = 2/ Frictional force Static friction N f s F f s m µ s N F = f s mg start to move F If a mass m is placed on a table with friction. We have experience that we must have large enough force to have it moved. Before the mass is moved, F = f s and they have opposite direction. Thus the net force on the mass is zero. There exist a maximum static friction force f s,max such that f s f s,max = µ s N where µ s is the coefficient of static friction.

57 Chapter 5 Application of Newton s Law Kinetic friction As a mass m is moving on a frictional surface, there exist a frictional force, for which its direction is against the motion. Its magnitude is always: f k = µ k N. N.B.: µ k < µ s for the same surface. f f s f k f s F v mgsinθ θ mgsinθ θ f k v = 0 f s A block is projected upward with speed v. Take upward motion as positive: µ k mg cos θ mg sin θ = ma. Hence, we obtain a = µ k g cos θ g sin θ. That means the block decelerates upward with µ k g cos θ + g sin θ. A block is released from rest on an incline plane, its subsequent motion depends on the coefficient of static friction µ s : a) if mg sin θ < f s,max = µ s mg cos θ or µ s > tan θ, then it will continue stay at rest. v b) if mg sin θ > f s,max or µ s < tan θ, it will slide down. mgsinθ θ f k If mg sin θ > f s,max and on its way down, µ k mg cos θ mg sin θ = ma a = µ k g cos θ g sin θ < 0, it accelerates downward.

58 Chapter 5 Application of Newton s Law Some examples concerning uniform circular motion a) Conical pendulum y T θ y T θ T R x x mg mg y-direction: x-direction: T cos θ mg = ma y = 0 T = mg cos θ T sin θ = ma x ( ) sin θ a x = T = m ( no acceleration in y-direction) ( mg cos θ ) ( ) sin θ = g tan θ m N.B. a x is pointing towards the circle center and is also perpendicular to v. It is thus the centripetal acceleration for the uniform circular motion. a x = g tan θ = v2 R v = gr tan θ Period T is given by: T = 2πR v = 2π R cot θ g b) The rotor A hollow cylinder is rotating about its axis with uniform circular motion. A person with his hands back and is against the cylinder wall.

59 Chapter 5 Application of Newton s Law 53 v For the person to stick on the wall without falling down, mg < f s,max = µ s N (5.5) axis of f rotation s N mg radius R cylinder wall The person is also under uniform circular motion and the normal reaction from the wall provides the required centrpetal acceleration. Thus, from (5.5), mg < µ s mv 2 R N = mv2 R or v > gr gr i.e.v min = µ s µ s If the rotor spins too slow, the person will fall down. c) A bicycle moving around a curve on a level road The bicycle is moving with constant speed v. N = mg and the friction provides the required centrpetal N force. f s = mv2 R For the bicycle not slipping, f s = mv2 R < f s,max = µ s N = µ s mg v 2 < µ s Rg or v < µ s Rg O R f s mg i.e. if the bicycle runs too fast, it will slip.

60 Chapter 5 Application of Newton s Law 54 d) A bicycle moving around a curve on a banked road N θ y x O R f s θ θ mg As a reminder, the discussion below regards the bicycle as a point mass. y-direction: N cos θ f s sin θ mg = 0 N = mg + f s sin θ cos θ (5.6) x-direction: N sin θ + f s cos θ = ma x = mv2 R (5.7) Remark: Even if there is no friction between the bicycle and the road, the bicycle can turn around the curve without slipping, i.e. N = mg cos θ v 2 = R m & N sin θ = mv2 R ( mg ) sin θ = Rg tan θ cos θ v = Rg tan θ For a rough road and no slipping, (5.6) and (5.7) give f s mv2 + mg tan θ = cos θ R f s = mv2 R cos θ mg sin θ (5.8)

61 Chapter 5 Application of Newton s Law 55 Case I: f s > 0 or mv2 R cos θ > mg sin θ or v > Rg tan θ mgtanθ mv 2 R f s cos θ Both the friction term and the mg term contributes to the centripetal force, the centripetal force mv2 R is large. Case II: f s < 0 or mv2 R cos θ < mg sin θ or v < Rg tan θ The centripetal force required is smaller than the mg term, because the frictional force has a direction against the mg term. When will slipping occur? Refer to case I: v > Rg tan θ (Frictional force points downward) The magnitude of v is limited when the frictional force has maximum static friction f s,max. From (5.6), N = mg + µ sn sin θ cos θ N = mg cos θ µ s sin θ From (5.7), N sin θ + µ s N cos θ = mv2 max R ( ) mg (sin θ + µ s cos θ) = mv2 max cos θ µ s sin θ R ( ) sin θ + µs cos θ v max = gr cos θ µ s sin θ When v > v max, the bicycle slips outward. To conclude, if Rg tan θ < v downward and there is no slipping. gr ( sin θ + µs cos θ cos θ µ s sin θ ), the frictional force points

62 Chapter 5 Application of Newton s Law 56 Refer to case II: v < Rg tan θ (Frictional force points upward) Note that f s cos θ is in opposite direction to that of mg tan θ, i.e. mg tan θ f s cos θ = mv2 R Note also that mv2 R may be so small such that mg tan θ f s,max cos θ = mv2 min R Clearly, the bicycle slips inward when v < v min. But from (5.6), mg tan θ µ sn cos θ = mv2 min R N = mg µ sn sin θ cos θ N = mg cos θ + µ s sin θ From (5.7), N sin θ µ s N cos θ = mv2 min R ( ) mg (sin θ µ s cos θ) = m v2 min cos θ + µ s sin θ R ( ) sin θ µs cos θ v min = gr cos θ + µ s sin θ When v < v min, the bicycle slips inward. To conclude, if Rg tan θ > v upward and there is no slipping. gr ( ) sin θ µs cos θ, the frictional force points cos θ + µ s sin θ A test tube is filled with a liquid of uniform density ρ such that the height of the liquid is L. The tube is set to rotate by a centrifuge. When the tube is rotating with a constant angular speed ω about a fixed vertical axis, it is almost horizontal. The axis is at a distance d from the liquid surface while the tube remains horizontal. Find the pressure on the bottom of the tube during the motion. The atmospheric pressure is p 0.

63 Chapter 5 Application of Newton s Law 57 Solution The force acting on the liquid element is S dp, where S is the cross sectional area and dp is the pressure difference on the two sides of the liquid element. The mass and acceleration of the element are given as ρ dx S and ω 2 x respectively. By Newton s second law of motion, Integrating both sides, we have S dp = ρ dx S ω 2 x dp = ρ ω 2 x dx p p 0 dp = ρ ω 2 L+h p p 0 = (ρ ω 2 ) x2 2 h x dx L+h p p 0 = ρ ω2 2 [(L + h)2 h 2 ] p = p 0 + ρ Lω2 2 n (L + 2h)

64 Chapter 6 Impulse and Momentum 6.1 Definition of linear momentum m v p = mv Newton s 2nd law stated in momentum: The rate of change of the momentum of a body is equal to the total external force acting on the body. i.e. i F i = d p dt If the mass of the body is unchanged, i F i = m d v dt = m a 6.2 Collision between bodies F 1 F 3 m F 2 In a collision between particles, even though we do not know the force acting on the particles during the process, we can still relate the motions before and after the collision whenever there is no external force acting on the system. For example, the mass and momentum can transfer to each other. m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 58

65 Chapter 6 Impulse and Momentum Impulse and momentum Consider two bodies in collision and pay attention to one of these two bodies. It experiences a force during the collision. F For the small time interval dt, d p = F dt t i t t f t p(tf ) p(t i ) d p = p(t f ) p(t i ) = change in momentum after the collision tf t i tf t i F dt F def dt = J impulse This equation can also be applied generally on a body experiencing a force F (t) for a time interval from t i to t f. Three particles A, B and C, of masses 4, 6 and 8 kg, respectively, lie at rest on a smooth horizontal table. They are connected by taut light inextensible strings AB and BC and ABC = 120 o. An impulse I is applied to C. If the magnitude of I is 88 Ns and it acts in the direction BC, find the initial speeds of A, B and C. Solution Let I 1 and I 2 be the impulsive tensions (impulse) in the strings BC and AB respectively. Since A is acted on only by I 2, its initial speed u will be in the direction of AB. Since

66 Chapter 6 Impulse and Momentum 60 the string AB is taut, B must have a speed u in the direction AB. Also, let B have a speed v in a direction perpendicular to AB [this is necessary because I 1 acts in a different direction to I 2 ]. Finally, since both I and I 1, the impulses acting on C, have the same direction BC, let C have an instantaneous speed V in the direction BC. Since BC is taut, the speeds of B and C in the direction BC are equal. u cos 60 o + v cos 30 o u 3v = V 2 + = V (6.1) 2 Considering the motion of A, Considering the motion of B along and perpendicular to AB, Considering the motion of C, 4u = I 2 (6.2) 6u = I 1 cos 60 o I 2 6u = I 1 2 I 2 (6.3) 6v = I 1 cos 30 o 3I1 6v = (6.4) 2 Eliminating I 2 from equations (6.2) and (6.3) 8V = 88 I 1 (6.5) 6u = I 1 2 4u I 1 = 20u (6.6) Substituting from equation (6.6) into equation (6.4), 6v = 20 3u 2 v = 5u 3 (6.7) Substituting from equation (6.7) into equation (6.1), ( ) u u = V V = 3u (6.8) 3 2 Substituting for V and I 1 [equations (6.6) and (6.8)] into equation (6.5), From (6.7), v = From (6.6), V = 6. Speed of A is 2 m/s. Speed of B is u 2 + v 2 = Speed of C is 6 m/s. 8(3u) = 88 20u u = = 4 3 m/s.

67 Chapter 6 Impulse and Momentum Conservation of linear momentum Rather than looking into the experience of one of the body in the collision process, we look into the total momentum of the two bodies before and after the collision. F e1 Consider two particles having initial momentum p i1 and p i2 before the col- P i1 m 1 F 21 F 12 m 2 P i2 lision. During the collision process, the im- F e2 pulse force acting on m 1 from m 2 is F 12 and that on m 2 from m 1 is F 21. Before collision They are also experiencing external forces of F e1 and F e2. As F 12 and F 21 are action-reaction pair. F 12 = F 21 Before the collision, total momentum of the system: In the collision time interval t, P i = p i1 + p i2 p 1 = ( F 12 + F e1 ) t and p 2 = ( F 21 + F e2 ) t where p j is the change in momentum of mass m j during the collision. After the collision, final momentum of m 1 and m 2 : p f1 = p i1 + p 1 = p i1 + ( F 12 + F e1 ) t p f2 = p i2 + p 2 = p i2 + ( F 21 + F e2 ) t After the collision, total momentum of the system: P f = p f1 + p f2 = p i1 + p i2 + ( F e1 + F e2 ) t + ( F }{{} 12 + F 21) t }{{} = F e,tot =0 Hence, Pf P i = P = F e,tot t P t = F e,tot

68 Chapter 6 Impulse and Momentum 62 F ei To generalize the case, consider a N particles system consisting of masses m 1, m 2,..., m N. Define: P = i m i v i. For each particle m i, it experiences an external force F ei and other forces arising from its neighbors m j, i.e. Fij. m 1 F i1 m 2 d P dt = i F e,i = F e,tot m 3 F i3 m i F in F i2 Only the external forces, but NOT the internal forces, play roles on changing the total momentum of the system. m N Moreover, if F e,tot = 0, then dp dt = 0 i.e. if the total external force acting on a system of particles is zero, the total momentum of the system conserves. 6.5 Collision of particles in two dimensional space u 1 y m 1 m 2 u 2 Before collision v 1 x m 1 m 2 v 2 After collision If there is no external force acting on the system, we can write m 1 u 1x + m 2 u 2x = m 1 v 1x + m 2 v 2x m 1 u 1y + m 2 u 2y = m 1 v 1y + m 2 v 2y

69 Chapter 6 Impulse and Momentum 63 A particle of mass m is released on the smooth inclined face of a wedge of mass 2m and inclination angle α, which is initially at rest and is free to move on a smooth horizontal plane. Find the ratio of the acceleration of the wedge to that of the particle relative to the wedge when the particle is sliding down. Solution Let the velocity of the wedge be V and that of the particle relative to the wedge be u. Since there is no horizontal force acting on the system, by the conservation of horizontal momentum, 2mV + m(v u cos α) = 0. Differentiate both sides with time, we have V = u cos α 3 a = a cos α 3 where the acceleration of the wedge, a = dv and the acceleration of the particle relative dt to the wedge, a = du dt. Therefore, we have a : a = cos α : 3. A small block of mass M is placed on the top of a fixed and smooth sphere which has radius R. A moving particle of mass m and speed v 0 collides with the block horizontally, it sticks on the block and the composite slides down the sphere. (a) Find the angle θ when the composite leaves the sphere, where θ is measured from the vertical to the line joining the composite and the center of sphere. (b) Find the minimum initial speed of the particle such that the composite so formed leaves the sphere immediately when the collision occurs.,

70 Chapter 6 Impulse and Momentum 64 Solution (a) Let V be the instantaneous speed of the composite when the particle is just embedded into the block. Consider the conservation of linear momentum in the horizontal: mv 0 = (m + M)V (6.9) V = mv 0 M + m (6.10) Let v be the instantaneous speed of the composite when it is just leaving the sphere. Consider the conservation of mechanical energy: 1 2 (M + m)v 2 + (M + m)gr(1 cos θ) = 1 2 (M + m)v2 (6.11) The composite performs the circular motion on the sphere: (M + m)g cos θ N = (M + m) v2 R The composite leaves the sphere when N = 0. From Eq. (6.12), (6.12) Rg cos θ = v 2 (6.13) From Eq. (6.11) and Eq. (6.13), we obtain Using Eq.(6.10) and Eq.(6.14) gives 1 2 V 2 + gr(1 cos θ) = 1 Rg cos θ 2 cos θ = V 2 + 2gR 3gR (6.14) cos θ = m 2 v 2 0 3gR(M + m)

71 Chapter 6 Impulse and Momentum 65 (b) For the composite to leave the sphere when it is at the top of sphere: (M + m) g cos θ N = (M + m) v2 R N 0, when θ = 0 V 2 R g m 2 v0 2 gr (M + m) 2 gr (M + m) v 0 m Suppose that a boy stands at one end of a boxcar sitting on a railroad track. Let the mass of the boy and the boxcar be M. He throws a ball of mass m with velocity v 0 (relative to the ground) toward the other end, where it collides elastically with the wall and travels back down the length (L) of the car, striking the other side inelastically and come to rest. If there is no friction in the wheels of the boxcar, describe the motion of the boxcar. Solution Along the horizontal, no external force is acting on the system which consists the boxcar, the boy and the ball. Therefore, if V and v are the velocity of the boxcar (also the boy) and the velocity of ball respectively (both quantities are relative to the ground), the conservation of linear momentum gives MV + m v = 0 V = m v (6.15) M

72 Chapter 6 Impulse and Momentum 66 at all times. Before the first collision, v = v 0 and so V = m M v 0 for 0 < t < L (1 + m/m)v 0 where we have on the right the time expression for the ball to reach the boxcar wall, traveling at speed v 0 + V = v 0 + m M v 0 = with respect to the floor of the boxcar. ( 1 + m ) v 0 M The effect of the first elastic collision will be simply to reverse both velocity vectors in eq. (6.15). Thus, v = v 0 and V = + m M v 0 for L (1 + m/m)v 0 < t < 2L (1 + m/m)v 0 Finally, after the second inelastic collision, the ball and boxcar have common velocity. Hence, V = 0 for t > 2L (1 + m/m)v 0 Notice that a nonzero common velocity will move the center of mass of the system, which is not allowed as there is no external force acting on the system. It is seen that the boxcar first moves to the left a distance ( ( ) m 0) M v L m = L (1 + m/m)v 0 m + M and then moves an equal distance to the right, coming to rest at its starting point. This result that there is no net displacement of the boxcar if the ball return to its initial position within the boxcar holds whether or not the first collision is elastic (because the center of mass of the system must remain at rest).

73 Chapter 7 Systems of Particles In previous chapters, we have deal with problems of point mass or particle. Now we turn our focus to system containing many particles, e.g. rigid body. 7.1 Center of mass y m 3 r 3 r 1 m 1 r 2 m 2 x Consider a system containing N particles m 1, m 2,..., m N. r i (t) : position of m i at time t v i (t) : velocity of m i at time t r N a i (t) : acceleration of m i at time t m N Definition center of mass r CM = m 1 r 1 + m 2 r m N r N m 1 + m m N = i m i r i M where M = i m i. v CM = d r CM dt = 1 M (m 1 v 1 + m 2 v m N v N ) Hence, a CM = d v CM dt = 1 M (m 1 a 1 + m 2 a m N a N ) M a CM = m 1 a 1 + m 2 a m N a N = F 1 + F F N (7.1) 67

74 Chapter 7 Systems of Particles 68 F i is indeed the total force experienced by mass m i. F i = F int,i + F ext,i where F int,i total internal force acting on m i originated from other particles m j i, F ext,i total external force acting on m i. m1 F 1i F i1 m 2 Fint,i = F i1 + F i F in (no F ii ) From (7.1), F ext,i m i F in F Ni F i2 m N F 2i Notice F ij = F ji and thus = M a CM F ext,1 + ( F12 + F F 1,N 1 + F 1N ) + F ext,2 + ( F 21 + F F 2,N 1 + F 2N ) + F ext,3 + ( F 31 + F F 2,N 1 + F 2N ) F ext,n + ( F N1 + F N2 + F N F N,N 1 ) M a CM = F ext,1 + F ext, F ext,n = i F ext,i i.e. to say N-particle system with external forces acting on individual particles and internal forces between each of the particle behaves as if a single point mass at the position r CM experiencing a force of i F ext,i. 1N CM 2N A uniform laminar of 1-kg mass is acting on by some forces. F = 1î + 2î 3ĵ = î 3ĵ = (1) acm y 3N a CM,x = 1 ms 2, a CM,y = 3 ms 2 x (Only motion of center of mass is known, but not the rotation.)

75 Chapter 7 Systems of Particles 69 y m 0 v 0 φ explode at t0 m at + t 2 t 0 m 1 at t 0 + t (x 1 0,y 0 ) projectile motion x A particle of mass m 0 explodes at time t 0 into two masses of m 1 and m 2. A short time t after the explosion, m 1 was found at (x 0, y 0 ). Find the position of m 2 at t 0 + t. Consider m 1 and m 2 are pieces to form the system after explosion and we assume that m 2 does not hit the ground in the discussion. Before the explosion, the CM of the system is just the position of the mass m 0. The trajectory of it is a parabola given by r CM and the gravitational acceleration g points vertically downward. a CM = d2 r CM dt 2 v CM = d r CM dt = g ĵ = (v 0 cos φ 0 ) î + (v 0 sin φ 0 gt) ĵ r CM = (v 0 cos φ 0 )t î + [(v 0 sin φ 0 )t 1 2 gt2 ] ĵ After the explosion, m 0 splits into m 1 and m 2. Therefore, m 1 and m 2 have their own positions r 1 (t) and r 2 (t). But as the explosion only involves internal forces, the CM s position will follow the original parabola. At t = t 0 + t, r 1 (t 0 + t) = x 0 î + y 0 ĵ and r CM = m 1 r 1 + m 2 r 2 m 0 r 2 (t 0 + t) = m 0 r CM m 1 r 1 m 2 m ( 2 m0 = v 0 cos φ 0 t m ) 1 x 0 î + m 2 m 2 [ m0 m 2 (v 0 sin φ 0 t 12 gt2 ) m ] 1 y 0 ĵ m 2

76 Chapter 7 Systems of Particles Center of mass of some rigid bodies Locate the center of mass of a triangular laminar, as shown in the figure. The laminar has uniform mass distribution. Solution Imagine the laminar is a collection of rods which are parallel to BC. The CM of the laminar lies on the line joining the CMs of rods. This line is the median of the triangle. Likewise, imagine again the laminar is a collection of rods which are parallel to AC. Thus, the CM of the laminar lies on the line joining the CMs of these rods. Therefore, the CM of laminar ABC is located at the intersection of the medians. From the knowledge of simple geometry, we know that the medians of ABC meet at a point G, the centroid of the triangle, and AG = 2 3 AE BG = 2 3 BF CG = 2 3 CD Remark: To locate the CM of a semi-circular disk. One may divide the disk to many small sectors which are considered as isosceles triangles having small subtended angle at the center of disk.

77 Chapter 7 Systems of Particles 71 An uniform circular laminar of radius R 2 has a hole in it. The hole is in the form of circle and it has a radius R 1. The circumferences of the hole and the laminar meet tangentially. y y R R Ω 1 x R Ω 2 x y R Ω 3 x Solution Ω 1, Ω 2 and Ω 3 are symmetric about the x-axis, CM of them must lie on x-axis. Center of mass of the required object: x CM = i y CM = 0 m i x i M = x dm M From the above figures, we notice that x CM,Ω3 = 0 But x CM,Ω2 = R 1 R 2 x dm = Ω 3 x dm + Ω 1 x dm Ω 2 Making using the definition of the center of mass, we have M Ω3 x CM,Ω3 = M Ω1 x CM,Ω1 + M Ω2 x CM,Ω2 M Ω3 (0) = (M Ω3 M Ω2 ) x CM,Ω1 + M Ω2 (R 1 R 2 )

78 Chapter 7 Systems of Particles 72 x CM,Ω1 = M Ω 2 (R 2 R 1 ) M Ω3 M Ω2 Let σ be the density of the laminar, then M Ω2 = πr 2 1σ and M Ω3 = πr 2 2σ Hence x CM,Ω1 = πr2 1σ (R 2 R 1 ) πr 2 2σ πr 2 1σ = R 2 1 R 1 + R 2 A uniform wire of radius R is bent into a semi-circle. Locate the center of mass of this wire. Solution The wire is symmetric about the y-axis and thus the CM of it must lie on the y-axis. Consider the small arc segment as shown in the figure. It is at a distance y above the x-axis, where y = R sin φ Mass of the segment: dm = ρr dφ where ρ is the density of the wire. y CM = y dm M = π (R sin φ) (ρr dφ) 0 ρπr = R π π 0 sin φ dφ = 2R π

79 Chapter 7 Systems of Particles 73 g y R 1 R 2 x A solid ball with radius R 1 is placed inside a hollow sphere with radius R 2, as shown in the figure. The ball is then released both the ball and the sphere roll back and forth. What is the final equilibrium position? The masses of the ball and the sphere are both m. At final state y x 0 ground x ground Consider motion in x-direction, as external force is zero, the x component of the CM does not change. Before release, x CM = m 0 + m (R 1 R 2 ) 2m = R 1 R 2 2 After reaching equilibrium, x CM = m x 0 + m x 0 2m = x 0 = R 1 R 2 2 A uniform laminar of mass m has an area bounded by y = x 2 and y = 2. Locate the center of mass of the laminar. Solution

80 Chapter 7 Systems of Particles 74 Let the center of mass be ( x, ȳ) and the surface density be σ. We notice that x = 0, because the laminar has symmetry about the y-axis. By the definition of center of mass, y dm we have ȳ =, where dm = σ(2x) dy, 2 x 0. Rewrite the expression in terms dm of y, we have dm = σ(2 y) dy, 2 y 0. 2 y dm = 2σ y 3 2 dy 0 ( ) y = 2σ = 4σ ( ) dm = 2σ y 1 2 dy 0 ( ) y = 2σ = 4σ ( ) Substitute the answers of both intergrals into the expression of ȳ, we obtain 4σ ( ) ȳ = 5 4σ ( ) = Therefore, the center of mass of this laminar is (0, 6 ). Note also that if m is the mass of 5 the laminar, m = dm = 4σ ( ) and we can relate σ = 3m 3 8 though it is not required 2 in this problem. A rod of length 1 m has a non-uniform mass distribution on its length. The mass per unit length of it varies with position according to ρ = ρ 0 (1 x/2), where x is measured from one end of it along the rod. Locate the center of mass of the rod. Solution From the definition of CM, we have x = we obtain x dm, where dm = ρ dx = ρ 0 (1 x ) dx. Thus, dm 2

81 Chapter 7 Systems of Particles 75 x = = x ρ 0 (1 x 2 ) dx ρ 0 (1 x 2 ) dx (x x2 2 ) dx (1 x 2 ) dx = ( x2 2 x3 6 ) (x x2 1 4 ) = ( ) (1 1 4 ) = 4 9 Therefore, the CM is 4 9 m from the heavier end of the rod. (a) Use direct integration to locate the center of mass of a uniform hemispherical shell which has mass m and radius r. (b) By using the result in (a), locate the center of mass of a uniform solid hemisphere Solution which mass m and radius r. (a) The mass of small ring, as shown in the figure, is given by width of ring {}}{ dm = σ (2πr sin θ) (rdθ) }{{} circumference of ring where σ is the surface density of the hemisphere. = 2πσr 2 sin θ dθ

82 Chapter 7 Systems of Particles 76 Therefore, x = = xdm = dm 2πr 2 σ π 2 r cos θ 0 (2πσr2 dθ) 2πr 2 σ π = r 2 0 r cos θ dm π 2 = r sin θ d sin θ = r sin2 θ 2 0 π 2 0 sin θ cos θ dθ That is, x = r. Due to symmetry of the hemisphere about the x-axis, ȳ = 0, thus, 2 we have ( x, ȳ) = ( r 2, 0). (b) Let the mass of a spherical shell with width dx, as shown in the figure, be dm, = r 2 where dm = ρ (2πx 2 dx) and ρ is the volume density of the hemisphere. }{{} volume of spherical shell

83 Chapter 7 Systems of Particles 77 x = = 3 2r 3 ( x 2 ) dm dm r 0 = r 0 ( x 2 ) (2πρx2 dx) 2 3 πr3 ρ x 3 dx = 3 2r 3 (r4 4 ) = 3r 8 That is, x = 3r. Due to symmetry of the hemisphere about the x-axis, ȳ = 0, thus, 8 we have ( x, ȳ) = ( 3r 8, 0). A pulley has two masses m 1 and m 2 hanging over it by a string, as shown in the figure, where m 2 > m 1. Assume that the pulley and all strings are massless and frictionless, please find the CM acceleration of the pulley system when the masses are released from rest. Solution The pulley system is enclosed in the green box, as shown in the figure.

84 Chapter 7 Systems of Particles 78 Taking downward as positive and consider Newton s second law of motion, we have F net = m a CM, where m = m 1 +m 2 and F net = (m 1 +m 2 )g R. Need to mention that the tensions of the string are internal forces of the system and they do not appear in the net force expression. For the motion of individual masses, we can relate the acceleration of them by Solving the above equations, we otain { T m1 g = m 1 a m 2 g T = m 2 a T = 2m 1m 2 g m 1 + m 2 But, R = 2T because the pulley is massless. Therefore, we have R = 4m 1m 2 g m 1 + m 2 The net force on the pulley system, F net = m a CM becomes (m 1 + m 2 )g 4m 1m 2 g m 1 + m 2 = (m 1 + m 2 ) a CM Remark: a CM = g (m 2 m 1 ) 2 (m 1 + m 2 ) 2 Using the definition of CM, we can write m 1 x 1 + m 2 x 2 = (m 1 + m 2 ) x CM. Hence, (7.2) m 1 ẍ 1 + m 2 ẍ 2 = (m 1 + m 2 ) ẍ CM m 1 a + m 2 a = (m 1 + m 2 ) a ( ) CM m1 + m 2 a = a CM m 2 m 1 ( ) m1 + m 2 g (m2 m 1 ) 2 a = m 2 m 1 (m 1 + m 2 ) 2 a = g (m 2 m 1 ) m 1 + m 2 The last expression is the same as that obtained by solving equation set (7.2). 7.3 Momentum of system of particles Consider a system of N particles m 1, m 2,..., m N having position vectors r 1, r 2,..., r N. Total momentum of the system P = i m i v i = i p i (7.3)

85 Chapter 7 Systems of Particles 79 But since r CM = i m i r i M. M d r CM dt = i m i d r i dt M v CM = i p i (7.4) Combining (7.3) and (7.4) P = M v CM and d P dt = M a CM To find the total momentum, other than adding all p i, we can also get it by finding M v CM. Or the system of N particles behaves as if it is a point mass having mass M, velocity v CM and acceleration a CM. Moreover, from last chapter or the recall at the beginning of this chapter: d P dt = i F ext,i = F ext,tot Therefore, M a CM = i F ext,i If the total external force is zero, we have d P dt = 0, or simply, P = 0. The direct result of this is P = constant. This is referred to as the conservation of linear momentum for system of particles! v CE M m v mc A canon on a frictionless ground fires a cannon ball. The canon ball is fired with speed of v mc relative to the canon. In x-direction, there is no external force. Momentum conserved which implies 0 = Mv ce + mv me

86 Chapter 7 Systems of Particles 80 But v me = v mc + v ce 0 = Mv ce + mv mc + mv ce v ce = mv mc m + M 7.4 System of variable mass Time t At t + t, a small mass was m v system being studied ejected out from the original mass. u is the velocity of the small Time t + t mass relative to the Earth. However, it should be noted u m m+ m v+ v system being studied that the small mass velocity is usually given in relative to the original mass. At time t, total momentum: P(t) = m v At time t + t, total momentum: Hence, we find P(t + t) = (m + m)( v + v) + ( m) u = m v + m v + m v + m v m u = m v + m v + m( v u) + m v P = P(t + t) P(t) = m v + m( v u) + m v F P ext = lim t 0 t = md v dt + dm ( v u) dt or Fext = m d v dt dm dt v rel

87 Chapter 7 Systems of Particles 81 where v rel = u v is the velocity of the small particle relative to the original mass. Sometimes, this equation is referred to as the rocket equation. Notes: ( ) m v lim = 0, since m 0, and v/ t becomes finite when t 0. t 0 t Since v = v me and u = v m,e, we have v u = v me + v E, m = v m, m = v m,m = v rel momentum conserved u mass = M A train on a frictionless rail with a v machine gun firing at a rate of n bullets per second. Mass of bullet is m. momentum conserved u M v A rocket in space ejecting mass at rate of dm dt. where u: velocity of ejecting mass or bullet relative to the Earth. For both case, F ext = 0. M d v dt = dm ( u v) dt ( mn)( u v) = mn( v u) = ( ) dm dt ( u v) = dm dt ( v u) for the train, for the rocket This implies both the train and the rocket will accelerate as if there were a force (called thrust). Indeed, it is not a real external force acting on the train or the rocket but only to maintain the conservation of momentum.

88 Chapter 7 Systems of Particles 82 +ve Rocket ascending on earth by ejecting mass at a rate M g of dm dt with velcoity v rel relative to the rocket. dm dt F ext = M d v dt dm dt v rel M g = M dv ( ) dt dm dt ( v rel ) M dv dt = dm dt v rel M g v rel M M Raindrop falling and water vapor keeps on condensing on it with a rate of dm dt. v + M M v+ v g +ve F ext = M d v dt dm dt v rel F ext = M d v dt dm dt ( u moisture v rain ) M g = M dv dt dm dt dv dt + 1 M (0 v) u moisture = 0 dm dt v = g A soft and uniform string of length l and mass M is hanged in the way that its lower end is touching the table. This string is initially at rest and then suddenly released to fall freely from the top position. Find the instantaneous reaction force that the table acts on the string after it has fallen a distance y. Find also the reaction force when the whole string is just on the table.

89 Chapter 7 Systems of Particles 83 Solution Method 1: Consider the small mass element at the junction. After the string is released, it moves down under the gravity. The speed of the string becomes v when it moves down a distance y, where v 2 = 2gy. Denote the line density of the string be λ. During the short time interval t, the length of string element being stopped on the table due to collision is y. This small element is at the junction between the upper portion (i.e. the moving part) and the lower portion (i.e. the stationary part). An upward force F acts on the lower end of this moving element in order to stop it, where F has the relation F p t = (λ y) (0 v) t Hence, we have F = ( ) dy λ v dt F = λv 2, where v = dy/dt and F is negative. Notice that F is not the magnitude of force. According to Newton s three law of motion, there is a downward force, f = F, which acts on the lower portion. Thus, we obtain f = λv 2 = 2λgy. The stationary portion of the string on the table (i.e. enclosed by the green box) is at equilibrium, hence, we have R = λgy + f = λgy + 2λgy R = 3λgy When y = l, we obtain R = 3λgl = 3Mg, where M = λl is the mass of the string.

90 Chapter 7 Systems of Particles 84 Method 2: Consider F net = d p dt F net = Mg R = d [λ(l y)ẏ] dt Mg R = λlÿ λyÿ λẏ 2 Since ÿ = g, M = λl and ẏ 2 = 2gy. We have λlg R = λlg λyg 2λyg R = 3λyg When the whole sting is just on the table (i.e. y = l), R = 3Mg. Method 3: The rocket equation. Consider the stationary portion of the string (enclosed by the green box) as the interested object in the rocket equation. becomes F ext = m d v dt v dm rel dt R + λyg = m(0) (ẏ 0) λẏ R + λyg = ẏλẏ R = λyg + λẏ 2

91 Chapter 7 Systems of Particles 85 Since ẏ 2 = 2gy, we obtain R = λyg + 2λyg = 3λyg. When the string has just fallen by y = l, R = 3λlg = 3Mg. Method 4: Consider the center of mass of the string at time t. When the string falls by a distance y. The center of mass of the string measured from the reference point is y CM = = = lλy + y CM = y + l2 y 2 ( y + l y ) 2 λl ly + 1 (l + y) (l y) 2 l ly + 1 ( l 2 y 2) 2 Differentiate both sides with respect to time, we obtain Since ÿ = g and ẏ 2 = 2gy. we have 2l ẏ CM = ẏ 1 l (yẏ) ÿ CM = ÿ 1 l l (ẏ2 + yÿ ) λ (l y) ÿ CM = g 1 l (2gy + yg) ÿ CM = g 1 l (3gy) Hence, we have

92 Chapter 7 Systems of Particles 86 ÿ CM = g l (l 3y) According to Newton s second law of motion, F net = ma, we can write λlg R = λl ÿ CM R = λlg λl ÿ CM [ g ] R = λlg λl l (l 3y) R = 3λgy At the instant that the whole string is on the table, i.e. R = 3λgl = 3Mg. y = l, the reaction force A pile of string having a uniform mass density λ is staying on the top of table. One end of it is pulled by a lifting force F such that it rises vertically upward with a constant speed v 0. Find the required force F when the string is at a height y above the table. Solution Method 1: Consider the net force acting on the whole string, e.g. Fnet = d p dt. When the string rises, the two portions of the string perform differently. The upper portion moves with constant speed and it has increasing momentum as the mass of it increases, but the lower portion has zero momentum as it is always at rest. We first define our system having the entire string enclosed in the green box. Notice that the interacting forces between the two portions (i.e. at the junction of them) are internal forces instead of external forces exerted on the system. The net force acting on the whole string is F λyg, because the weight of the lower portion is balanced by the upward

93 Chapter 7 Systems of Particles 87 reaction force by the table. Therefore, F net = F λyg = d p dt F λyg = d dt (λyẏ) F λyg = λyÿ + λẏ 2 Since ẏ = v 0 = constant, we have ÿ = 0. Therefore, Method 2: By using the rocket equation F = λyg + λv 2 0 F = λ(yg + v 2 0) Consider the upper portion of the string (i.e. the portion enclosed by the green box) as the interested object in the rocket equation. F ext = m d v dt v dm rel dt We note that v = v 0 = constant, m = λy and F ext = F mg = F λyg v rel = 0 v 0 = v 0 dv dt = dv 0 dt = 0 dm dt = d(λy) dt = λẏ = λv 0 Therefore, we have F λyg = m(0) ( v 0 )λv 0, which implies F = λ (yg + v 2 0)

94 Chapter 8 Rotational Kinematics z r P y x To describe the rotation of a rigid body about a fixed axis. We can observe the motion of a fixed point P in the rigid body. The motion of P is a circular motion about the axis of rotation. y r φ P s arc length s = rφ x angular displacement φ = s r [unit: radian] 88

95 Chapter 8 Rotational Kinematics 89 y Like what is done in linear motion, average angular velocity ω av and instantaneous angular velocity ω can φ2 φ1 P at t 2 P at t 1 x be defined as: ω av = φ 2 φ 1 t 2 t 1 ω = lim t 0 = φ t φ t = dφ dt [unit: rad s 1 ] Similarly, average and angular instantaneous acceleration α av and α are defined by: α av = ω(t 2) ω(t 1 ) t 2 t 1 α = lim t 0 ω t = dω dt = ω t [unit: rad s 2 ] Angular velocity as a vector z z w y P y x P x w Right Hand Screw Rule 8.1 Rotation with constant angular acceleration Suppose α = dω dt = constant, we obtain dω = αdt. Integrating on both sides, we have ω = αt + A, where A = constant. At t = 0, ω = ω 0 = A where ω 0 is the initial angular velocity.

96 Chapter 8 Rotational Kinematics 90 ω = ω 0 + αt Therefore, ω = dφ dt = ω 0 + αt φ = ω 0 t αt2 + B At t = 0, φ = φ 0 = B where φ 0 is the initial angular displacement. φ = φ 0 + ω 0 t αt2 8.2 Relation between linear and angular variables y In a time interval t, the rotating vector r φ r s r moves through an angle φ. If t 0, s = s = r φ. x Thus the tangential velocity is s v T = lim t 0 t = r dφ dt v T = ωr Moreover, tangential acceleration is given by: a T = dv T dt = r dω dt a T = αr From previous chapters, we also know for particle undergoing circular motion with constant speed, the particle indeed accelerates toward the center (centripetal acceleration). Thus the radial acceleration is equal to: Hence, the resultant acceleration: a R = v2 T r = ω2 r a = a T + a R

97 Chapter 8 Rotational Kinematics 91 y a a T a R r x N.B. For uniform circular motion, α = 0 a = a R, pure radial acceleration!! One can check the relations stated below. A particle moves on a plane about a fixed point O with constant angular velocity and constant radius. The tangential velocity is given by v = ω r, where ω and r are the angular velocity and the position vector of the particle respectively. The centripetal acceleration of the particle is ω ( ω r).

98 Chapter 9 Rotational Dynamics 9.1 Torque Definition y F A force F was acting on a body at a point P. The torque about the point O is defined as r P x where r = OP. τ = r F y F F T r θ P pointing out of paper τ = r F sin θ x 92

99 Chapter 9 Rotational Dynamics 93 y r P θ F pointing into the paper τ = rf sin θ x F T Point O θ L τ τ = r F τ = Lmg sin θ r with direction pointing inward mg Point O τ θ τ = Lmg sin θ r with direction pointing outward mg

100 Chapter 9 Rotational Dynamics Rotational inertia and Newton s 2nd law in rotation Single particle A particle of mass m is connected to the z-axis by a massless rod of length r. From previous results (in the last chapter), a T = αr F T = ma T = mα z r y Fsinθ m θ F But F T = F sin θ r x F sin θ = mα z r z As τ z = F r sin θ, τ z = mα z r 2 N.B. Subscript z is added to specify the axis of rotation. Define I = mr 2 for single particle so that τ z = Iα z This is referred to as Newton s 2nd law for rotation. More than one particle y P Two masses m 1 and m 2 are linked by massless rods to the m 1 T 1r z-axis. m 1 and m 2 are linked to each other by a similar rod. T 1 l T 2r Rotation axis: z-axis. P is an external force. z r 1 r 2 T 2 m 2 x

101 Chapter 9 Rotational Dynamics 95 Total force on m 1 : F1 = P + T 1 + T 1r Total force on m 2 : F2 = T 2 + T 2r y Total torque on the system about z-axis: r 1 θ 1 T 1r h 2 h 1 T 2r θ 2 τ z = ( r 1 F1 ) + ( r 2 F2 ) = ( r 1 P ) + ( r 1 T 1 ) +( r }{{} 1 T 1r ) 0 + ( r 2 T 2 ) +( r }{{} 2 T 2r ) 0 ( r 1 // T 1 and r 2 // T 2 ) r 2 x = ( r 1 P ) + ( r 1 T 1r ) + ( r 2 T 2r ) z Notice that r 1 T 1r = r 1 T 1r sin θ 1 r 2 T 2r = r 2 T 2r sin θ 2 (9.1) ( r 1 T 1r ) and ( r 2 T 2r ) are in opposite directions. Let the distance between m 1 and m 2 be l. But area of triangle Om 1 m 2 is equal to h 1 = l sin θ 1 & h 2 = l sin θ h 1r 1 = 1 2 h 2r 2 l sin θ 1 r 1 = l sin θ 2 r 2 Since action-reaction forces are equal in magnitude Substitute eq. (9.2) and (9.3) into eq. (9.1), we obtain r 1 sin θ 1 = r 2 sin θ 2 (9.2) T 1r = T 2r (9.3) r 1 T 1r = r 2 T 2r

102 Chapter 9 Rotational Dynamics 96 Therefore, the total torque is τ z = r 1 P which is only dependent on external force. Torque created by internal force are cancelled out. Or τ z = i τ ext,i. Consider the total force Fi on mass m i with i = 1 or 2. Fi can be decomposed into two com- y ΣF 1T radial direction ponents, namely the tangential and the radial. r 1 m 1 ΣF 2T radial direction τ z = ( r 1 F1 ) + ( r 2 F2 ) r 2 m 2 = r 1 ( F 1T ) +r 2 ( F2T ) }{{}}{{} m 1 a 1T m 2 a 2T τ z x Radial components has no contribution to torque, i.e. τ z = r 1 (m 1 a }{{} 1T ) + r 2 (m 2 a }{{} 2T α zr 1 = (m 1 r m 2 r 2 2)α z α zr 2 ) where a it are tangential acceleration of m i and α z is the angular acceleration of the system about the z axis. Or where I = m 1 r m 2 r 2 2. For a N-particle system τ z = Iα z y m 1 Masses m 1, m 2,..., m N located at r 1, r 2,..., r N. Moment of inertia is defined by r 1 r 2 m 2 x I = i m i r 2 i r N m N

103 Chapter 9 Rotational Dynamics 97 The total torque acting on this system: τ z = i r i F i = Iα z N.B. Rigid body means the particles within the body have fixed relative position with respect to each others, i.e. as if linked by massless rods. y m 2 3 m 30 o θ 4 5 θ 1 m 3 x m 1 = 2.3 kg m 2 = 3.2 kg m 3 = 1.5 kg (a) Find moment of inertia of the system about axis perpendicular to xy plane and passing m 1, m 2 and m 3 respectively. (b) If a 4.5 N force is applied to m 2 as shown and the system is free to rotate about the axis perpendicular to the xy plane and passing through m 3. What is the angular acceleration? Solution (a) By definition, I 1 = i m i r 2 i = (2.3 kg)(0 m) 2 + (3.2 kg)(3 m) 2 + (1.5 kg)(4 m) 2 = 53 kg m 2 Similarly, I 2 = (2.3 kg)(3 m) 2 + (3.2 kg)(0 m) 2 + (1.5 kg)(5 m) 2 = 58 kg m 2 I 3 = (2.3 kg)(4 m) 2 + (3.2 kg)(5 m) 2 + (1.5 kg)(0 m) 2 = 117 kg m 2

104 Chapter 9 Rotational Dynamics 98 (b) θ = sin 1 (3/5) θ = 37 τ z = 4.5 N 5 m sin( ) = 20.7 N m But τ z = Iα z α z = τ z /I 3 = 0.18 rad s 1 in clockwise direction 9.3 Parallel axis theorem C.M. axis h z I z = I CM + Mh 2 I z = Moment of inertia rotating about z-axis, I CM = Moment of inertia rotating about the axis C.M. passing through C.M., z-axis is parallel to the C.M. axis and h is the M distance between the two parallel axes. Proof: z h z C.M. slab // to z & z axis mass m i & coordinate (, ) x i y i For the I z about the z-axis: I z = i m i r 2 i = i m i (x 2 i + y 2 i ) Let (x CM, y CM ) be the x, y coordinates of the C.M. measured from the x, y coordinate sys- y y x i y i (, ) tem. { xi = x i + x CM ( x CM, y CM ) r i z h x x y i = y i + y CM

105 Chapter 9 Rotational Dynamics 99 I z = i m i [(x i + x CM ) 2 + (y i + y CM ) 2 ] = i = i m i (x 2 i + 2x ix CM + x 2 CM + y i 2 + 2y iy CM + ycm) 2 m i (x 2 i + y 2 i ) } {{ } =I CM I z = I CM + Mh 2 Remark: +2x CM m i x i +2y CM m i y i + (x 2 CM + y 2 }{{ CM) } i i =h 2 i }{{} =Mx CM =0 }{{} =My CM =0 m i }{{} =M The following proof shows the vanishing of second and third terms in the R.H.S. of the long equation. r i = r i + r CM mi r i r CM = mi The second equation becomes r CM mi = m i r i which implies r CM mi = m i ( r i + r CM ) mi r i = 0 A further result is m i v i = 0. N.B. The axis passing through the CM is the axis that has the smallest moment of inertia as compared to other parallel axis.

106 Chapter 9 Rotational Dynamics Rotational inertia of solid bodies Discrete form: I = m i ri 2 i Integral form: I = r 2 dm A uniform rod has mass m and length L. Find the moment of inertia about an axis normal to it and through its center. Find also the moment of inertia again if the axis is shifted to one end of the rod. Solution Partition the whole rod into many infinitesimal segments with length dx and consider one of the segment at x dm = λdx, λ = density (i.e. kg m 1 ) But λ = M/L. I = +L/2 L/2 I = x 2 dm = +L/2 L/2 ( ) λ +L/2 x 3 = M 3 L/2 3L x 2 λ dx = λ +L/2 L/2 ( ) L 3 = ML2 x 2 dx By parallel axis theorem, L/2 L/2 C. M. I z = I CM + Mh 2 = 1 12 ML2 + M ( ) 2 L 2 z = 1 3 ML2

107 Chapter 9 Rotational Dynamics 101 A uniform rectangular plate of mass M and dimensions a b is rotating about an axis through the center of plate and is normal to the plate. Find the moment of inertia of this plate about the axis. z z b x dx a Solution Partition the plate into strips each having the width dx. The mass of each strip is dm = σ (adx), where σ is the density of plate The moment of inertia of the strip about the z -axis is di CM = 1 12 (dm) a2 = 1 12 σa3 dx The moment of inertia of the strip about z-axis is di = di CM + x 2 dm = 1 12 σa3 dx + aσx 2 dx, where σ = M/(ab). ( 1 di = aσ 12 a2 + x 2 The total moment of inertia about z-axis is b/2 b/2 ( ) 1 I z = di = aσ b/2 b/2 12 a2 + x 2 ( 1 = aσ 12 a2 b + 1 ) 12 b3 = a = M 12 (a2 + b 2 ) ) dx ( ) a 2 x dx = aσ 12 + x3 b/2 3 b/2 ( ) ( M 1 ab 12 a2 b + 1 ) 12 b3

108 Chapter 9 Rotational Dynamics 102 An uniform circular ring of mass M and radius R rotates about its axis. Find the moment of inertia of this ring. Solution Consider the small arc (i.e. the red one in the figure) having an subtended angle dθ. z R dm = λ (R dθ) λ = linear density (kg m) 1 Hence, its moment of inertia about the ring s center is given by I = R 2 dm = = λr 3 2π = 0 ( ) M 2πR dθ 2π 0 R 3 (2π) R 2 (λr dθ) y θ R dθ dm x I = MR 2 An uniform circular disk of radius R and mass M is rotating about the disk center. Find the moment of inertia of the disk about its axis. Solution Consider a disk as a collection of many rings.

109 Chapter 9 Rotational Dynamics 103 Think about a ring with radius r, thickness dr and surface density σ. The mass of this small ring dm = σ [π(r + dr) 2 πr 2 ] = σ [π (dr) 2 +2πr dr] }{{} higher order term 0 = σ 2πr dr Moment of inertia of the ring about an axis through its center is given by di = r 2 dm = r 2 (σ 2πr dr) = 2σ πr 3 dr Therefore, the total moment of inertia of the disk about its center is R R ( ) ( ) M R I = di = 2σ πr 3 dr = 2σ π r 3 4 dr = 2 π = 1 πr MR2 0 0 Find the moment of inertia of a uniform sphere of mass M and radius R about its axis. Solution Consider a sphere as a collection of many disks. Think about a thin disk of thickness dz and radius r in the figure, where r = R cos θ. The mass of it is dm = ρ (πr 2 ) dz = ρ (πr 2 cos 2 θ) dz. But, z = R sin θ, we have dz = R cos θ dθ and dm = ρπr 3 cos 3 θ dθ.

110 Chapter 9 Rotational Dynamics 104 The moment of inertia of the sphere I = di = r 2 dm. More specifically, 2 Thus, we obtain I = = π/2 π/2 π/2 0 = R 5 πρ = R 5 πρ = R 5 πρ r 2 2 dm = π/2 0 r 2 dm (R cos θ) 2 ρπr 3 cos 3 θ dθ π/2 0 π/2 0 π/2 0 cos 5 θ dθ (1 sin 2 θ) 2 d sin θ [ = R 5 πρ sin θ 2 sin3 θ 3 ( ) 8 = R 5 πρ 15 (1 2 sin 2 θ + sin 4 θ) d sin θ I = R 5 π ] + sin5 θ π/2 5 0 I = 2 5 MR2 M 4 3 πr Perpendicular axis theorem For a laminar, if z- axis is normal to the plane of laminar, and x and y axes lie on the laminar, then the moment of inertia about these axes are related by I z = I x + I y,

111 Chapter 9 Rotational Dynamics 105 Proof: I z = I z = I z = r 2 dm (x 2 + y 2 ) dm x 2 dm + y 2 dm I z = I y + I x Find the moment of inertia of a disk of mass m and radius r about its diameter. Solution The moment of inertia of a disk about the z-axis is I z = mr 2 /2, where the axis is normal to the plane of disk and passes through the centre. Suppose that the x and y axes lie on the disk and they cut at the disk center, I z = I x + I y = 2I D, where I x = I y = I D and I D is the required moment of inertia. Therefore, I D = mr 2 /4. A uniform laminar of mass m has an area bounded by y = x 2 and y = 2. (a) Find the moment of inertia of the laminar about the y-axis. (b) Find the moment of inertia of the laminar about the x-axis, hence, find the moment Soulution of inertia of the laminar about the z-axis. (a) Method 1: Consider the vertical strip, as shown in the figure. The moment of inertia of this strip about the y-axis is I y = di y, where di y = [(2 x 2 ) σdx] x }{{} 2 and σ is the dm surface density.

112 Chapter 9 Rotational Dynamics I y = 2m 5 I y = (2 x 2 ) σ x 2 dx 2 [ 2 ] 2 = σ 2x 2 dx x 4 dx 2 2 [ ] 4(2 3/2 ) = σ 2(25/2 ) 3 5 = 3m [ ] 4(2 3/2 8 ) 2(25/2 ) = 3m [ ] 5 Method 2: Consider the horizontal strip, as shown in the figure. The moment of inertia of this strip about the y-axis is I y = di y, where di y = 1 12 dm (2x)2 and dm = σ (2x) dy = σ (2 y) dy.

113 Chapter 9 Rotational Dynamics 107 Therefore, I y = = di y = dm (2x) [σ (2 y) dy] (2 y) 2 = σ y 3/2 dy = 4 15 = ( ) 2 5/2 σ ( ) ( ) 2 5/2 3m 8 2 = 2m 5 (b) Consider the shaded region which has length 2x and small height dy, where x 0. When it rotates about the x-axis, it obtains the moment of inertia di x. Since I x = di x, where di x = (dm) y 2 = (2x σ dy) y 2 = ( 2 y σ dy ) y 2 = 2y 5/2 σ dy, therefore, (c) I z = I x + I y = 12m 7 I x = + 2m 5 = 74m 35. di x = 2σ 2 0 y 5/2 dy ( ) y 7/2 2 = 2σ = 211/2 σ 7/2 0 7 ( ) = 211/2 3m 7 8 = 12m 2 7

114 Chapter 9 Rotational Dynamics Equilibrium of rigid body For a rigid body to stay at equilibrium, the following conditions must be satisfied. 1) Fext = 0 2) τ ext = 0 (about choice of reference point) Question: If we know Fext = 0 and τ = 0 about one particular point, say O, can we conclude the total torque about any other choice of point? z F i P r P r i r P O r i x y Rigid body Refer to point O, τ o = τ 1 + τ τ N = ( r 1 F 1 ) + ( r 2 F 2 ) ( r N F N ) = 0 Now, we consider the torque about another point P, τ P = ( r 1 r P ) F 1 + ( r 2 r P ) F ( r N r P ) F N = [( r 1 F 1 ) + ( r 2 F 2 ) ( r N F N )] [ r }{{} P ( F 1 + F F N )] }{{} Given condition = 0 i F i =0 = 0 If i F i = 0 (i.e. translational equilibrium established) and τ ext = 0 about a given point, then τ ext = 0 about any point and thus, equilibrium must be established.

115 Chapter 9 Rotational Dynamics 109 L 4 M L As the bar is in equilibrium, F i = 0 i m R 1 + R 2 + M g + m g = O R 1 r R1 y z r R2 R2 x or R1 + R 2 + (M + m) g = O Take upward as positive: R 1 + R 2 (M + m)g = 0 (9.4) Mg r M mg Take moment about O, τ = r R1 R 1 + r M (M g) + r R2 R 2 τ = L 2 R 1 + L 4 Mg + L 2 R 2 = 0 Therefore, R 1 R 2 1 Mg = 0 (9.5) 2 Summing up (9.4) and (9.5) gives 2R 1 = (M + m)g Mg = mg Mg R 1 = 1 2 mg Mg R 2 = R Mg = 1 2 mg Mg

116 Chapter 9 Rotational Dynamics 110 A non-uniform rod of mass m is at equilibrium at its position as shown in the figure. It has a block of mass M fixed at its middle. As the rod is non-uniform, the center of mass of the rod does not locate at its center, but instead it is displaced a distance from the center point. Find the normal forces acting on it by the frictionless wall and the rough floor. Solution frictionless wall h M ladder mass = m a 2 a 3 C.M. O rough a R 2 R 1 f O Mg mg To obtain equilibrium, i F i = 0. { R2 = f R 1 = (m + M)g Consider the torque about the axis through O and perpendicular to the paper. ( a ( a Mg + mg = R 2 h 2) ( 3) a ) ( a ) R 2 = f = Mg + mg 2h 3h

117 Chapter 9 Rotational Dynamics 111 A uniform beam of mass m is at equilibrium, as shown in the below figure. Find the reaction force exerted on the hinge by the wall. α θ L M F v O y T α θ mg F h Mg x Solution Label the x and y components of the reaction force at the hinge by F h and F v respectively. F h T cos α = 0 (9.6) F v mg Mg + T sin α = 0 (9.7) Consider the torque about the axis passing through O and perpendicular to the paper. ( ) L T L sin(α + θ) MgL cos θ mg cos θ = 0 2 g(m + m/2) cos θ T = sin(α + θ) From (9.6), From (9.7), F h = F v = (m + M)g g(m + m/2) cos θ cos α sin(α + θ) g(m + m/2) cos θ sin α sin(α + θ)..

118 Chapter 9 Rotational Dynamics Non-equilibrium situation: pure rotation A block of mass m is connected to one end of a light string, while the other end of the string is winded to a smooth pulley of mass M and radius R. Find the linear acceleration of the block and the angular acceleration of the pulley when the block is released. Solution Frictionless pulley with mass M and radius R. T R M mg T = ma, where a = d2 y dt 2 T = mg ma Torque acting on pulley, y T m mg τ = RT = Iα, If the rope runs through the pulley without slipping, where α = d2 θ dt 2 where I = 1 2 MR2 for disk rotating about its center. R (mg ma) = 1 2 MR2 α 2mg 2ma = MRα (9.8) Rθ = y R d2 θ dt = d2 y 2 dt 2 i.e. Rα = a Thus, (9.8) becomes 2mg 2mRα = MRα Hence, α = a = 2mg R(M + 2m) 2mg M + 2m

119 Chapter 9 Rotational Dynamics 113 Two blocks of mass m and 2m are connected by a light string. They are hanging over a pulley of mass M and radius R. If the system has no friction, find the linear acceleration of the blocks and the angular acceleration of the pulley when the blocks are released. Solution T 1 T 1 m mg R +ve T 2 T 2 2m 2mg Consider the equations of motion for both blocks. T 1 mg = ma (9.9) 2mg T 2 = 2ma (9.10) Total torque on pulley, i.e. τ = Iα RT 2 RT 1 = ( 1 2 MR2 ) α (9.11) The condition of rolling without slipping: Put (9.9) and (9.10) into (9.11): a = Rα (9.12) (2mg 2ma)R (ma+mg)r = 1 2 MR2 α (9.13) Substitute (9.12) into (9.13), we obtain α = 2mg 2m(Rα) m(rα) mg = 1 2 MRα mg = 1 MRα + 3mRα 2 mg MR/2 + 3mR and a = Rα = mg M/2 + 3m 9.8 Non-equilibrium situation: rotational and translational motion If i F i 0 and i τ i 0 about any axis, the motion of the rigid body has both selfrotation and the motion of the C.M. For the present course, we only focus on cases such that: a) Axis of rotation passes through C.M. b) Rotating axis always has the same direction in space.

120 Chapter 9 Rotational Dynamics 114 A solid cylinder of radius R and mass M is released from rest on an inclined surface such that it rolls down without slipping on the surface. Find the linear and angular acceleration of the cylinder. mass = M f R a C.M. N α Mgsinθ The equations of motion of the cylinder that is normal and along the motion. N = Mg cos θ Mg sin θ f = Ma C.M. (9.14) Total torque on the cylinder: Mgcosθ Put (9.15) into (9.14): θ τ = fr = Iα = 1 2 MR2 α f = 1 MRα (9.15) 2 Mg sin θ 1 2 MRα = Ma C.M. But the condition of rolling without slipping is Rα = a C.M., thus Mg sin θ 1 2 Ma C.M. = Ma C.M. a C.M. = 2 3 g sin θ and α = a C.M R = 2g 3R sin θ ω o A uniform solid cylinder of radius R and mass M is given an initial velocity ω 0 and then lowered on a uniform horizontal surface. The coefficient of kinetic friction between the cylinder and the surface is µ k. Initially the cylinder slips as it moves along the surface, but after a time t, pure rolling without slipping begins. (a) What is the velocity v CM at time t? (b) What is the value of t?

121 Chapter 9 Rotational Dynamics 115 Solution (a) During the time interval 0 to t: α N ω o a CM f Mg Consider the convention of motions (both linear and rotational) of the cylinder. The positive directions are shown in green color in the figure. The frictional force f is a non-zero constant when the cylinder is slipping. We know also that at t = 0, the C.M. of cylinder is at rest and the frictional force starts to increase the linear speed of it towards right while the rotational speed of it decreases gradually. For the rotation of the cylinder, which gives f = Ma CM, where f = µ k N = µ k Mg a CM = µ k g (9.16) Rf = Iα ( ) MR 2 R(µ k Mg) = α 2 α = 2µ kg R (9.17) Note that when the cylinder starts to move a Rα as the cylinder is slipping and it is referred to as the first stage of motion. The occurrence of pure rolling (i.e. without slipping) will starts at time t when v f = Rω f. a = v f 0 t α = ω f ( ω 0 ) t (9.18) (9.19)

122 Chapter 9 Rotational Dynamics 116 Eqs (9.16) and (9.18) give µ k g = v f t When rolling without slipping occurs, we have v f = Rω f. Thus, µ k g = Rω f, which t gives t = Rω f µ k g (9.20) From Eq. (9.19), we have αt = ω f + ω 0 (9.21) Substitute Eq. (9.17) and Eq. (9.20) into the above equation, we obtain ( ) 2µ k g Rωf = ω k + ω 0 R µ k g ω f = ω 0 3 The linear speed of cylinder at time t is v f = Rω 0 3. (b) From Eq. (9.20), the time t is Rω 0 3µ k g. Two solid cylinders are sticked together and string is winded on the cylinder with the smaller radius. Assume that the small cylinder is very light and it is negligible as compared to the large cylinder which has mass M. Find the linear and angu- α T lar accelerations of the system if one end of the string is fixed but the cylinders are allowed to fall freely. R o R a Solution Consider the linear motion of the cylinders, the equation of motion of this is given by Mg T = Ma (9.22) mass M Mg

123 Chapter 9 Rotational Dynamics 117 Consider the rotational motion of the cylinders, the equation of motion of this is given by (9.23) and (9.24) gives τ = R 0 T = R 0 M(g a) (9.23) τ = Iα = 1 2 MR2 α (9.24) R 0 M(g a) = 1 2 MR2 α α = 2(g a) R 0 R 2 (9.25) For no slipping, a = R 0 α (9.26) Substituting (9.26) into (9.25), we obtain Hence, a R 0 = 2(g a) R 0 R 2 a = 2gR2 0 R 2 + 2R 2 0 α = a R 0 = 2gR 0 R 2 + 2R 2 0 A uniform thin rod of length 2L and mass M lies horizontally on two smooth supports, as shown in the figure. The two ends of rod are labeled as A snd B. If support at A is suddenly removed, at that instant what are (a) the torque about B, (b) the vertical reaction force at B.

124 Chapter 9 Rotational Dynamics 118 Solution (a) When support A is just removed, the rod is exerted by two forces, its weight and the reaction force R at B by the right support. The torque acting on the rod about B is τ = MgL. (b) The moment of inertia of the rod about B is given by I B = 1 12 M(2L)2 + M(L 2 ) = 4 3 ML2. The angular acceleration of rod about B, that is α, is obtained by using Newton s second law for rotation. MgL = I B α MgL = 4 3 ML2 α α = 3g 4L Thus, at the instant, the linear acceleration of the center of rod is ( ) 3g a = Lα = L = 3 4L 4 g. Consider the equation of motion of rod just after support A is removed. Mg R = Ma ( ) 3 Mg R = M 4 g R = 1 4 Mg

125 Chapter 9 Rotational Dynamics 119 A uniform thin rod of length 2L and mass M has one end hinged to the floor. Initially, it is held vertical above the floor and it is disturbed to fall down. The rod makes an angle θ with the vertical at time t. The moment of inertia of the rod about one end of rod is 4 3 ML2. (a) Find θ as a function of θ. (b) Find θ as a function of θ. Solution (a) The hinge exerts a force on the rod. The components of the force along and normal to the horizontal are S and R respectively. The torque about the lower end of rod is Mg (L sin θ). Hence, we have (b) Since θ = d θ ( dt = d θ ) dθ ( dθ dt = Mg (L sin θ) = 4 3 ML2 θ θ = 3g 4L sin θ d θ ) θ, we can write dθ ( d θ ) θ = 3g dθ 4L sin θ θ 0 θ θ d θ = 3g sin θ dθ 4L 0 θ 2 θ = 3g θ 2 4L cos θ 0 θ 2 = 3g (cos θ 1) 2 4L 3g θ = (1 cos θ) 2L 0

126 Chapter 9 Rotational Dynamics 120 A uniform thin rod of mass m and length l is resting with its lower end on a frictionless floor, where the rod makes an angle θ 0 to the vertical. Then, it is released to fall down. Find the normal force exerted on the rod by the floor at the instant of release. Solution The equations of motion for translational and rotational motions. R mg = mÿ ( ) l R 2 sin θ = 1 12 ml2 θ (Moment about C) Note that the time derviatives of y = l cos θ are 2 ẏ = l 2 θ sin θ and ÿ = l 2 [ θ sin θ + θ 2 cos θ] Set the symbols for the motion of rod at t = 0 : θ = θ 0, θ = θ0, R = R 0, ÿ = ÿ 0. The equations of motion at t = 0 become R 0 mg = mÿ 0 R 0 sin θ 0 = 1 6 ml θ 0 (9.27) Also, when the rod is just released, we have θ = 0, thus ÿ 0 = l 2 θ 0 sin θ 0. From the first equation of (9.27) and the above relation. From the second equation of (9.27) R 0 mg = ml 2 θ 0 sin θ 0 (9.28) θ 0 = 6R 0 sin θ 0 ml (9.29)

127 Chapter 9 Rotational Dynamics 121 Substitute Eq. (9.29) into Eq. (9.28) R 0 mg = ml ( ) 6R0 sin θ 0 sin θ 0 (9.30) 2 ml mg R 0 = sin 2 (9.31) θ 0 A solid sphere of mass m and radius r is at rest on a horizontal platform. The platform is given a constant acceleration a such that the sphere rolls without sliding on the platform. (a) Find the linear acceleration of the sphere. (b) Find also the angular acceleration of the sphere. Solution (a) There are some points to notice about the motion of the sphere. (i) When the platform starts to move, a frictional force exerts on the sphere and it points to the right. (ii) The sphere rolls backward relative to the platform (counterclockwise as shown in the figure) because the frictional force produces a torque about the center of sphere. (iii) The center of sphere C accelerates towards the right with a C, where a C is relative to a fixed observer. It is also true that if this observer is an inertial one. f = ma C (9.32)

128 Chapter 9 Rotational Dynamics 122 (iv) Note also that a C < a, where a is the acceleration of the platform relative to a fixed observer because the center of sphere is moving back relative to the platform. Moreover, the contact point P on the sphere is accelerating to the right, relative to C. (v) As the sphere rolls without sliding on the platform, the contact point P of the sphere is not moving relative to the platform. That is to say, point P is moving with the same speed with the platform. Since both the sphere and the platform start from rest and accelerate, point P has the same acceleration as the platform. Hence, we can write a = a C + rα. Therefore, a a C = rα ( τ ) a a C = r, I where τ = rf and I = 2 5 mr2. Thus, ( ) rf a a C = r 2 5 mr2 a a C = 5f 2m, From Eq. (9.32), we have a a C = 5(ma C) 2m, a a C = 5 2 a C a C = 2 7 a Remark: The sphere rolls backward with a C a = 2 7 a a = 5 a relative to the 7 platform, i.e. to the left of the platform. (b) Refer to Eq.(9.32), the frictional force f = ma C = 2 ma, thus 7 α = τ I = rf I = r ( 2 ma) 7 2 = 5a 5 mr2 7r

129 Chapter 10 Angular Momentum 10.1 Definition Angular momentum of a point mass m about a point O is given by: z l = m r v = r p, l = r p sin θ O l r v θ y d l dt = d d r d p ( r p) = p + r dt dt dt = v (m v) + r d p dt d l dt = r d p dt = r F = τ where F is the resultant force acting on mass m. x Note that both l and τ in the above equation must be defined with respect to the same origin. y O θ b P x A particle of mass m is released from rest at point P. Torque on m with respect to O: r m θ F=mg τ = mgr sin θ = mgb Moreover, l = rm dy dt sin θ = mv y b (inward) (inward) 123

130 Chapter 10 Angular Momentum 124 As τ = dl dt, mgb = d dt (mv yb) = mb dv y dt g = d2 y dt 2 For the case of a system of particles, m 1, m 2,..., m N having angular momenta of l 1, l 2,..., l N. Total angular momentum of the system: L = i li (by definition) d L dt = i d l i dt = i τ i ( τ i = τ i,int + τ i,ext ) State without proof: Internal torques NOT contribute to change of L. d L dt = i τ i,ext The following figure shows the analogies between linear and angular momentum. Linear momentum F // p p // Angular momentum τ // L L // p F // // = t τ // = L // t F p + p p τ L+ L L p L F = p t τ = L t

131 Chapter 10 Angular Momentum 125 An interesting observation τ = r (m g) (pointing inward) But, τ = L t L τ and it will also point inward! As L is along the spinning axis, the axis will move inward Angular momentum and angular velocity z ω θ O θ r r p y A particle m is in circular motion. Now taking the reference point as O. l = r p is not parallel to ω. Indeed, not like the case in the simplest case (i.e. the origin O is also the rotating center), angular momentum l and angular velocity ω are in general not parallel. Under what other conditions will the l and ω vectors parallel? x

132 Chapter 10 Angular Momentum 126 x p ω z r p r 2 O = + r 1 y Consider two equal masses m 1 and m 2 jointed by a light rod is rotating about the center of rod. In this case, l // ω. Generally, l is parallel to ω when the rigid body is symmetric about the rotational axis. For a more general case, where m 1 m 2, l I ω (as we have the relation p = m v for linear motion). between l and ω? What is the relation From previous figure of the single particle case, l z = l sin θ = rp sin θ = r(mv) sin θ = r(mr ω) sin θ ( v = r ω) But since r = r sin θ, l z = mr 2 ω = Iω Notes: The z-component of angular momentum is equal to Iω. Though this result is only obtained from single particle, it is also true for rigid body. If the rigid body is symmetric about the rotation axis, L = I ω. For such case, L and ω are both parallel to the rotating z-axis. In general, L z = Iω. If τ = τ x î + τ y ĵ + τ zˆk, τ z = dl z dt That is to say, if I is constant and τ z = 0, ω will not be changed!

133 Chapter 10 Angular Momentum 127 Total angular momentum of the system: x R O M L z = l M + l m L z = I M }{{} ω + m }{{} v R +ve mag. +ve mag. (take pointing outward to be +ve) r m m m p= mv mg, y = mvr = 1 2 MR2 ω + mvr Total external torque: τ z = r m m g τ z = mgr τ z = dl z dt mgr = d ( ) 1 dt 2 MR2 ω + mvr = 1 2 MR2 α + mra where α is the angular acceleration of the pulley and a is the acceleration of the mass. But α = a/r. mgr = 1 2 MR2 a R + mra a = 2mg M + 2m 10.3 Conservation of angular momentum As we know τ = d L dt if the total external torque acting on the system is zero, then d L dt momentum is conserved. = 0. This implies angular

134 Chapter 10 Angular Momentum 128 1) I i I f I i > I f I i ω i = I f ω f Angular velocity increases. w i w f 2) L w Ls L w L ω : initial wheel angular momentum, L s : final student and turn-table angular momentum. stationary turn table L i = L ω Lf = L s + ( L ω ) = L s L ω = L i L s = 2 L ω 10.4 Stability of spinning object L f L i τ // L Object is symmetric about the rotating axis τ // is applied for a time period of t. L = τ // t Spinning speed is changed but not the spinning orientation.

135 Chapter 10 Angular Momentum 129 τ L f θ L i L Object is symmetric about the rotating axis τ is applied for a time period of t. L = τ t Spinning orientation is changed. tan θ = L L i = τ t L i If L i is large, θ is small for fixed τ t. That is to say, a large spinning speed or a large I have a higher stability in spinning orientation. A running bicycle does not collapse despite of it s body makes an tilting angle with the floor. The spinning wheels keep the stability.

136 Chapter 10 Angular Momentum 130 The spinning top z z Lsin θ dφ dl τ= r Mg r O Mg L CM y O θ L L+ dl y x x Consider a symmetric spinning top, i. e. L is parallel to the rotation axis (i.e. ω). External torque on the top: τ = Mgr sin θ Over a time interval t, L = τ t = Mgr t sin θ. Moreover, L = (L sin θ) φ Mgr t sin θ φ = = Mgr t L sin θ L As a fact that d L L and τ L implies d L L. The external torque only changes the direction of L but not the magnitude of L. Hence, the vector L (or the rotation axis) moves slowly about the vertical axis (called precession). Average speed of precession: ω p = φ t = Mgr L

137 Chapter 10 Angular Momentum 131 A rod of length 2l and mass m has a fine hole at its end. Initially, the rod moves with speed v on a smooth horizontal table, where v is normal to the rod. A pin is then embedded in the hole and the rod rotates about it. (a) Find the angular speed of the rod about the pin. (b) Find the impulse acting on the rod when the pin is introduced. (c) Find the force acting on the pin when the rod is rotating. Solution (a) Moment of inertia about O just before and after the pin is inserted. I O = 1 12 m(2l)2 + ml 2 = 4 3 ml2 Angular momentum about O: mvl = I 0 ω gives ( ) 4 mvl = 3 ml2 ω (b) Let the required impulse be J, where ω = 3v 4l J = change of momentum of C.M. = mωl mv ( ) 3v Therefore, we obtain J = m l mv = mv. The negative sign means that 4l 4 the impulse is opposite to the direction of v. (c) The required force T = force on the pin, pointing outward and along the rod. T = mω 2 l = m ( ) 2 3v l = 9mv2 4l 16l

138 Chapter 10 Angular Momentum 132 Two particles of mass m 1 and m 2 are connected by an inextensible and massless string of length l. The system is placed on a smooth and horizontal plane with the string being kept tight. Mass m 1 is projected along a direction which is normal to the string and the speed is v 0. (a) Find the speed of the center of mass of the system and the angular speed of the system about its center of mass. (b) Hence, find the linear speeds of the individual masses just after the attack. Solution (a) As there is no external force acting on the system, the speed of the CM does not change. The total linear momentum of the system does not change. (m 1 + m 2 ) v C = m 1 v 0 v C = m 1 v 0 m 1 + m 2 Let C be the location of C.M. of the system. To locate it, we can write { r1 + r 2 = l r 1 m 1 = r 2 m 2, where r 1 and r 2 are the distances of m 1 and m 2 from point C respectively. Hence, we have r 1 = m 2l m 1 + m 2 r 2 = m 1l m 1 + m 2

139 Chapter 10 Angular Momentum 133 At the moment of projection, the angular momentum of the system about C is ( ) m1 m 2 L C = r 1 m 1 v 0 = lv 0 = µlv 0, m 1 + m 2 where 1 µ = 1 m m 2 or µ = m 1m 2 m 1 + m 2. Just after the projection, the angular momentum of the system about C is L C = m 1 r1ω 2 + m 2 r2ω 2 ( ) 2 ( ) 2 m2 l m1 l = m 1 ω + m 2 ω m 1 + m 2 m 1 + m ( ) 2 m1 m 2 = l 2 ω m 1 + m 2 L C = µl 2 ω By the conservation of angular momentum: µlv 0 = µl 2 ω ω = v 0 l (b) The speed of m 1 : v 1 = v C + r 1 ω, which gives v 1 = m 1v 0 + m ( 2l v0 ) = v 0 m 1 + m 2 m 1 + m 2 l The speed of m 2 : v 2 = v C r 2 ω, which gives v 2 = m 1v 0 m ( 1l v0 ) = 0 m 1 + m 2 m 1 + m 2 l

140 Chapter 11 Work, Kinetic Energy and Potential Energy 11.1 Work done by a constant force y S Consider a point mass m in a time interval of t is experiencing a constant force F. During this time interval, the displacement of m is S θ F x Work done by the force on the mass: W def = F S = F S cos θ. Work can be either positive or negative. Power is defined by: P = dw dt 134

141 Chapter 11 Work, and Kinetic Energy and Potential Energy Work done by a variable force One dimensional case Suppose there is a position dependent force F (x). F Divide the whole displacement from F(x ) i x 0 x x i x i+1 positive work done negative work done F(x) Total work done for the displacement from x 0 to x N : x N x x 0 to x N into N partitions with separation x. Consider the i-th partition, x i x i+1 and in this very small interval, F is approximately constant at F (x i ). Work done in this time interval: or W (x i ) = F (x i ) x F (x) = dw dx W x0 x N = i W i = i F (x i ) x = xn F (x)dx x 0 or it is equal to the total area of the figure with positive area for positive F (x) and negative area for negative F (x).

142 Chapter 11 Work, and Kinetic Energy and Potential Energy 136 x m equilibrium position x = 0 F(x) F = kx The restoring force is F = kx and the work done by the system when the mass moves from A to B: W A B = = xb x A xb x A F (x) dx kx dx = 1 2 k(x2 B x 2 A) x F = kx A mass is attached to the lower end of a light spring which is fixed to the ceiling. The green dotted line shows the unsketched position of the light spring when there is no mass under it. The motion of mass is governed by two forces, i.e. gravity and the restoring force of spring. Therefore, the net force acting on the mass when the spring extends from its natural length is F (y) = mg ky, where y is the coordinate of the mass. The work done by the system when the mass moves from A to B: W A B = = yb y A yb y A F dy ( mg ky) dy = mg(y B y A ) 1 2 k(y2 B y 2 A)

143 Chapter 11 Work, and Kinetic Energy and Potential Energy Two dimensional case y r(t) r F(r(t)) r(t) Trajectory of a particle is given by: r(t) = f x (t)î + f y (t)ĵ r(t + t) Force at any point r is given by: x F (t) = F x ( r)î + F y ( r)ĵ Consider a particle moving from r(t) to r(t + t) during the time interval t. If t 0, force experienced by particle in this time interval is constant and F ( r(t)). Work done in this small time interval with displacement r: W = F ( r(t)) r (Only the tangential force component contributes!!) W = F d r A mass m is hanged by a string with length L initially. A force F which is always horizontal is applied to lift the mass up to an angle φ. During the process, the mass moves with constant speed so small that the centripetal force can be neglected. Find the work done by the force F. y φ L φ m T φ F F x mg

144 Chapter 11 Work, and Kinetic Energy and Potential Energy 138 Solution If centripetal force approaches zero, a x = 0 and a y = 0. y F T sin φ = 0 and T cos φ mg = 0 F = mg tan φ φ φ S Consider the displacement S from φ φ + φ: W = F S = F tang S but also: = F x x x+ x x But x = L sin φ dx = L cos φ dφ W = mg tan φ L cos φ dφ = mgl sin φ dφ Hence, W = φm 0 mgl sin φ dφ = mgl (1 cos φ m ) 11.3 Work-energy theorem O x i v i F x x f v f Work done on the mass by the force: x F x = m dv dt = mdv dx dx dt Consider a particle m displaces from x i to x f. During this displacement, the x-component of the net force acting on m is F x. = mv dv dx (11.1) W = = = xf x i xf x i vf v i F x dx mv dv dx (using eq. (11.1)) dx mv dv W = 1 2 m(v2 f v 2 i ) Define kinetic energy K = 1 2 mv2. We obtain W = K f K i = K. If W is positive, v f > v i and K > 0. If W is negative, v f < v i and K < 0.

145 Chapter 11 Work, and Kinetic Energy and Potential Energy 139 N.B. In inertia frames having relative motion, the absolute value of kinetic energy are not the same, but the theorem W = K holds in all inertia frames Work done and kinetic energy in rotational motion y F φ ds dθ P r x If the rigid body is to displace from θ i to θ f, Consider a rigid body moving through an angle dθ about the rotational z-axis with a force acting on point P. Work done by the force: dw = (F sin φ) ds = F sin φ rdθ = τ z dθ where τ z is the z-component of the torque about O. If the torque is constant, W = θf θ i W = τ z θ τ z dθ Power: P = dw dt = τ z dθ dt = τ zω. w Consider each point of the rigid body, say m 1, m 2,..., m N. Kinetic energy of particle i is given by: v 2 m 2 r 2 r 1 v1 m 1 K i = 1 2 m ivi 2 = 1 2 m i(r i ω) 2 = 1 2 m iri 2 ω 2 Total rotational kinetic energy of the rigid body: K = i K i = 1 2 m iri 2 ω 2 i ( ) = 1 m i ri 2 2 i ω 2 K = 1 2 Iω2 I is the moment of inertia of the rigid body about the rotational axis.

146 Chapter 11 Work, and Kinetic Energy and Potential Energy A combination of rotational and translational motions In previous sections, we have considered cases of pure translational (i.e. movement of C.M. of rigid body) or pure rotational (about a fixed axis) motion. Now we turn into case such that both the CM is moving and the rigid body is rotating. y mn r n r CM r n CM Consider a rigid body consisted of particles m 1, m 2,..., m N. Total K.E. of the body: K = 1 m i vi 2 (11.2) 2 Note that r i = r CM + r i v i = v CM + v i x where v i = velocity of mass i with respect to the Earth s frame, v CM = velocity of the body s center of mass with respect to the Earth s frame, v i = velocity of mass i with respect to the body s center of mass. From (11.2), we obtain K = 1 m i ( v CM + v 2 i) ( v CM + v i) = 1 m i (vcm v CM v i + v i2 ) 2 But consider the second term: v CM (m i v i) = i i As v CM = ( i m i v i )/M, i i v CM (m i v i m i v CM ) = v CM m i v i MvCM 2 i i i v CM (m i v i) = v CM M v CM Mv 2 CM = 0 And the third term: 1 2 i m i v i 2 = 1 2 m i (r iω) 2 = 1 2 Iω2 where ω is the angular velocity about an axis passing through the center of mass. K = 1 2 Mv2 CM Iω2 1st term: Translational term of the C.M. as if there is no rotation. i 2nd term: Rotational term with rotation about the axis passing through the C.M. as if the rotational axis does not move.

147 Chapter 11 Work, and Kinetic Energy and Potential Energy Kinetic energy in collision For elastic collision, K = 0, i. e. K f = K i. For inelastic collision, K < 0, i. e. K f < K i. For complete inelastic collision, the two colliding objects stick together after collision. A moving particle has speed u and mass m 2. It collides on a stationary particle which has mass m 1. If the collision is elastic, find the final speeds of the particles. m 2 u m 1 Solution By the conservation laws, we have From (11.3), m 1 v 1 + m 2 v 2 = m 2 u (11.3) 1 m 2 1v m 2 2v2 2 = 1m 2 2u 2 (11.4) v 1 = m 2u m 2 v 2 (11.5) m 1 Substitute (11.5) into (11.4), we get: m 1 ( m2 u m 2 v 2 m 1 ) 2 + m 2 v 2 2 = m 2 u 2 m 2 2u 2 + m 2 2v 2 2 2m 2 2uv 2 + m 1 m 2 v 2 2 = m 1 m 2 u 2 (m 1 + m 2 ) v 2 2 2m 2 uv 2 + (m 2 m 1 ) u 2 = 0 Solve for v 2 and then v 1, we find ( ) 2m2 v 1 = u m 1 + m 2 and v 2 = 11.7 Conservative force ( m2 m 1 m 1 + m 2 ) u. Potential energy is only defined for conservative force in which a particle moving under the force influence has constant mechanical energy. s of conservative force:

148 Chapter 11 Work, and Kinetic Energy and Potential Energy 142 1) Spring 2) Gravitational force 3) Coulomb force of non-conservative force - friction. Definition A conservative force is a force such that if a particle moves under the influence of this force, the work done by the force on moving the particle from an arbitrary point A to another arbitrary point B would be the same along any arbitrarily chosen path, e.g. path 1 and path 2, as shown in the figure. y 1 B A 2 In other words, work done of a conservative force to move an object along a closed path (i.e. starting and ending at the same point) is zero. x Remark: There are some other rigorous definitions using the mathematical language, but they are out of the scopes of this course, e.g. 1) F ( r) is conserved if and only if there exists a scalar function φ( r) such that φ( r) = F ( r), where φ is referred to as the potential function of the force, and is a gradient operator. 2) F ( r) is conserved if and only if F = 0.

149 Chapter 11 Work, and Kinetic Energy and Potential Energy Potential energy Consider a particle moves in the influence of a conservative force, which is position dependent, i. e. F (x). Now the particle displaces from x i to x f, potential difference U is defined: U = U f U i = W where W is the work done by the force during the displacement x i to x f. Or xf U = U(x f ) U(x i ) = F (x) dx x i If for a particular reference point x 0, the potential energy is defined as zero, i.e. U(x 0 ) def = 0. The inverse of the above equation gives s (a) The spring-mass system x U(x) = F (x) dx x 0 F (x) = du dx Thus F = kx x m equilibrium position x = 0, U = 0 F = kx Take the equilibrium position to be x = 0 so that U(0) = 0. U(x) U(0) = U(x) = du dx = 1 k(2x) = kx = F 2 x 0 x 0 U(x) = 1 2 kx2 F (x) dx ( kx) dx

150 Chapter 11 Work, and Kinetic Energy and Potential Energy 144 (2) An object under the force of gravity y Take U(0) = 0. y F = mg U(y) U(0) = U(y) = y 0 y 0 U(y) = mgy F (y) dy ( mg) dy y = 0, U = 0 Thus du dy = mg = F 11.9 Conservation of mechanical energy U = U f U i = W (11.6) U i v i U f v f initial position final position But W = x f x i F (x) dx is the work done by the force in the journey from x i x f. From previous chapter, W = xf Substitute (11.7) into (11.6), we have x i F (x)dx = 1 2 m(v2 f v 2 i ) = K f K i = K (11.7) U f U i = K i K f U i + K i = U f + K f U = K In an isolating system whose only conservative force exists, mechanical energy of a particle conserves.

151 Chapter 11 Work, and Kinetic Energy and Potential Energy One dimensional conservative system U(x) E 4 E 3 E 2 E 1 E 0 K(x f ) U(x ) f K(x ) g U(x ) g x a x b x c x d x e x f x g x Particle experienced a conservative force field with potential energy U(x). F (x) = du dx At x = x a, x d, x f, x g, F = 0. x = x d, x g : stable equilibrium - slightly displaced particle experiences a restoring force x = x f : unstable equilibrium - displaced particle experiences a force in the same direction as displacement x = x a : neutral equilibrium - displaced particle experiences no force U(x) mv2 = E, where E is the conserved total energy. If E = E 4 as shown in the previous figure, E 4 = K(x g ) + U(x g ) at x = x g E 4 = K(x f ) + U(x f ) at x = x f If the energy of the particle E is different, it will have different behavior as follows: 1) If E = E 0, particle stays stationary at x = x d.

152 Chapter 11 Work, and Kinetic Energy and Potential Energy 146 2) If E = E 1, particle stays in the region x c x x e. 3) If E = E 2, particle may stay in the two valleys. However if it is in one of the valley. 4) If E = E 3, particle can stay in the region x > x b. 5) If E E 4, particle can be anywhere. If U(x) is known, it is possible to work out the particle s position.

153 Chapter 12 Conservation of Energy If external force acting on the system is not zero, the conservation of energy becomes K + U = W ext, where W ext is the work done on the system by the external force. W spring U spring + K W spring K W grav W grav K + U grav K + U grav +U spring Earth Earth Earth Earth System = Mass System = Mass + System = Mass + System = Mass + K = Spring Earth Earth + Spring W spring + W grav K + U spring = K + U grav = K + U grav + W grav W spring U spring = 0 147

154 Chapter 12 Conservation of Energy 148 A thin rod of mass m and length l is held vertical on the top of a horizontal floor, where the rod s lower end is hinged to a fixed joint on the floor. The rod is released and it hits the floor when it s angular speed is ω. Obtain ω by the following methods. (a) The Newton s second law of motion for rotation. (b) The work-energy theorem for a torque. (c) The conservation of mechanical energy. Solution (a) The moment of inertia of the rod about its end is given by I, where I = 1 12 ml2 + m( l 2 )2 = 1 3 ml2 Newton s second law of motion for rotation states that τ = I θ. Hence, we have 3g 2l 3g 2l mg( l 2 ) sin θ = 1 3 ml2 θ π/2 0 3g sin θ 2l = θ 3g 2l sin θ = θ ( d θ dθ ) ( cos θ) sin θ dθ = π/2 0 3g 2l ω = ω 0 = ω2 2 = ω2 2 3g l θ d θ

155 Chapter 12 Conservation of Energy 149 (b) Work done by the torque equals to the change of kinetic energy. π/2 0 π/2 0 τ dθ = K.E. mg ( l 2 ) sin θ dθ = 1 2 Iω2 0 mg ( l π/2 2 ) sin θ dθ = 1 2 (1 3 ml2 ) ω 2 3g l 0 ( cos θ) π/2 ω = (c) Conservation of mechanical energy: 0 = ω 2 3g l P.E. + K.E. = 0 {0 mg ( l 2 )} + {1 2 (1 3 ml2 ) ω 2 0} = 0 g = 1 3 lω2 ω = 3g l 12.1 Internal energy in a system of particle K + U + E int = W ext where E int is the change in internal energy of the system. Internal energy is the K.E. associated with the random motion of atoms and molecules (usually related to the object temperature), or the P.E. associated with forces between atoms. E int = K int + U int Looking deeper into the cases of rigid body F ext = Ma CM for rigid body F ext Notice that F ext may not act on the C.M. Consider the force acting for a short period and in the period, C.M. displaces by dx CM. Multiply both sides by dx CM, we CM dx CM CM have F ext dx CM = Ma CM dx CM

156 Chapter 12 Conservation of Energy 150 It is just a product of a force and the displacement dx CM. There is NO physical meaning such as work done! Notice that the definition of work done F d x refers to an applied force where it applies to a point mass and the mass is displaced by d x. F ext dx CM = Ma CM dx CM = M dv CM v CM dt dt F ext dx CM = Mv CM dv CM Consider the C.M. displaces from x i to x f and its velocity change from v CM,i to v CM,f. xf x i F ext dx CM = 1 2 Mv2 CM,f 1 2 Mv2 CM,i = K CM,f K CM,i ( K CM def = 1 2 Mv2 CM ) or F ext s CM = K CM if F ext is constant. This is the Center of Mass (COM) energy equation, where s CM is the displacement of the center of mass. The COM equation is not the work-energy theorem for a particle. s CM is the center of mass displacement but not the displacement of the point that the force acts on. K + U + E int = W ext This is the Conservation of Energy (COE) equation Some examples of conservation of energy 1) A sliding block is stopped on a horizontal table with friction. Center of mass (COM) energy equation: fs CM = 1 2 Mv2 CM Conservation of energy (COE) equation: W f = 1 2 Mv2 CM + E int,block 2) Pushing a stick on a horizontal frictionless table. S F ext CM S CM CM

157 Chapter 12 Conservation of Energy 151 Center of mass (COM) energy equation: F ext s CM = 1 2 Mv2 CM Conservation of energy (COE) equation: F ext s = 1 2 Mv2 CM Iω2 If F ext is acted on center of mass, s = s CM F ext s = F ext s CM = 1 2 Mv2 CM 3) Ball rolling down an inclined plane without slipping f S CM Mg θ Center of mass (COM) energy equation: (Mg sin θ f) s CM = 1 2 Mv2 CM Conservation of energy (COE) equation: Mg s CM sin θ = 1 }{{} 2 Mv2 CM Iω2 M g acts on CM Notice that the frictional force does no work in the COE eq. as the instantaneous point of contact between the ball and the plane does not move.

158 Chapter 12 Conservation of Energy 152 Two men are pushing each other as shown in the figure. The man m 2 is pushed away from the man m 1 by straightening their arms and the force between them is F. (a) What is the speed of m 2 just after losing contact? (b) What is the change in internal energies for m 1 and m 2? m m 2 m 2 is pushed to move forward 1 frictionless floor Solution (a) Consider m 2 as the interested system, COM equation of it is F s CM = K CM = 1 2 m 2vCM,m 2 2 where s CM is the displacement of the center of mass of m 2. 2F scm v CM,m2 = (b) For m 2, the COE equation is m 2 K + E int,m2 = W ext { K = KCM = F s CM where. W ext = F s Note that s is the total extension of m 1 s hand (i.e. the displacement of m 2 s hand when a force F is acting on it, where s s CM ). E int,m2 = F s F s cm For m 1, COE equation is E int,m1 = W ext = F s ( F opposite to s)

159 Chapter 12 Conservation of Energy 153 A uniform sphere of radius r and mass m rolls without sliding on the inner surface of a fixed and large spherical hollow of radius R, where R > r. The line joining OC makes an acute angle θ with the lower vertical line through O, where O and C are the center of the hollow and the sphere respectively. The moment of inertia of the sphere is (2/5)mr 2. Show that the equation of motion of the sphere is given by θ + 5g sin θ = 0. 7(R r) Obtain your answer by the following methods. (a) Conservation of mechanical energy. (b) Newton s second law of motion. Show further that the sphere oscillates with a period 2π 7(R r) 5g when θ is small. Solution (a) Set point O be the reference zero of the gravitational potential. P.E. = mg(r r) cos θ K.E. = 1 2 I φ m(r r)2 θ2

160 Chapter 12 Conservation of Energy 154 (b) where I is the moment of inertia of the sphere. By the conservation of mechanical energy, we know that K.E. + P.E. = E, where E is a constant. Therefore, 1 2 I φ m(r r)2 θ2 mg(r r) cos θ = E Since the sphere rolls without slipping, we have arc BP equals arc QP, i.e. Rθ = r(φ + θ). Differentiate on both sides with time, we obtain R θ = r( φ + θ). Rearrange it, we notice an useful relation Therefore, 1 2 ( 2 5 mr2 Simplify it, we have ) ( R r r ( ) R r φ = θ r ) 2 θ m(r r)2 θ2 mg(r r) cos θ = E 7 10 (R r)2 θ2 g(r r) cos θ = E Differentiate on both sides with time ( ) 7 2 (R r) 2 θ θ + g(r r) θ sin θ = (R r) θ + g sin θ = 0 Hence, we obtain θ + 5g 7(R r) sin θ = 0

161 Chapter 12 Conservation of Energy 155 By the Newton s law of motion { fr = I φ (rotational motion) f mg sin θ = m(r r) θ (linear motion) A negative sign appears in the first equation, as the direction of torque is opposite to the measurement of φ. Eliminate f, we have I φ r mg sin θ = m(r r) θ Substitute the moment of inertia of the sphere, i.e. I = 2 5 mr2, and the relation (R r) θ = r φ into the above expression, we have 2 ( ) 5 mr2 R r θ m(r r) r r θ mg sin θ = (R r) θ + g sin θ = 0 θ + 5g 7(R r) sin θ = 0 When θ is small, sin θ θ. Therefore, the equation of motion of sphere becomes θ + 5g 7(R r) θ = 0 This is in the form of S.H.M., e.g. θ + ω 2 θ = 0, where ω = 7(R r) period T = 2π/ω, we have T = 2π. 5g Remark: 5g. As the 7(R r) If the sphere rolls without slipping, arc BP equals arc QP. Rθ = r (θ + φ) (R r) θ = rφ (R r) θ = r φ

162 Chapter 13 Gravitation 13.1 Newton s law of universal gravitation F 21 m 1 m 2 r21 : position of m 2 relative to m 1 r 21 F 21 : force experienced by m 2 due to m 1 F 21 = Gm 1m 2 ˆr r m 1 r 12 F 12 m 2 r 12 : position of m 1 relative to m 2 F 12 : force experienced by m 1 due to m 2 F 12 = Gm 1m 2 ˆr r Gravitation near the Earth s surface If we assume the Earth to be a stationary uniform sphere, m M E = Mass of the Earth R E = Radius of the Earth M E R E Gravitation pull on the mass m: F = GM Em, i.e. g RE 2 0 = GM E RE 2 156

163 Chapter 13 Gravitation Effect of Earth s rotation Consider a mass m hanged by a string and the string is deviated from the true vertical line through center of Earth. F c = m g }{{} 0 + T true vertical direction Consider the force diagram in the right figure. Use the Cosine Law, we can write T 2 = F 2 c + (mg 0 ) 2 2F c mg 0 cos φ Note that T = mg eff where g eff is the effective measured gravity. For the case at the equator, i.e. φ = 0. T 2 = Fc 2 + (mg 0 ) 2 2F c mg 0 T 2 = (mg 0 F c ) 2 T = mg eff = mg 0 F c T = mg eff = mg 0 mω 2 R E Therefore, when φ = 0 g eff = g 0 ω 2 R E Let α be the angle between the string and the real vertical axis.

164 Chapter 13 Gravitation 158 Using Sine Law and let R be the distance between the particle and the z-axis. sin α sin(π α φ) = F c mg 0 g 0 sin α = sin(α + φ) ω 2 R g 0 sin α = sin α cos φ + cos α sin φ ω 2 R g 0 = cos φ + cot α sin φ ω 2 R g 0 = cot φ + cot α ω 2 R sin φ cot α = g 0 ω 2 R cos φ ω 2 R sin φ Or we can write, cot α = 2 (g 0 ω 2 R E cos 2 φ) ω 2 R E sin 2φ Using the Sine law again, sin α = sin φ F c T sin α mω 2 R E cos φ = sin φ mg eff sin α = R Eω 2 g eff sin α = R Eω 2 2g eff sin φ cos φ sin 2φ As α is small, α sin α = R Eω 2 thus 2g eff sin 2φ. The maximum of α occurs when φ = π 4 and α max = R Eω 2 2g eff

165 Chapter 13 Gravitation Gravitational force due to an uniform spherical shell Theorem 1: An uniform spherical shell attracts an external particle as if all the mass of the shell was concentrated at the center. Theorem 2: An uniform spherical shell exerts no force on a particle located inside the sphere Gravitational potential energy r a M F m a dr b r b Recall: U = U f U i = W if. Consider a mass m displaces from a to b. W ab = b a = GMm F d r = [ 1 ] rb r r a rb r a GMm dr r 2 If r b > r a, W ab < 0; if r b < r a, W ab > 0. [ 1 = +GMm 1 ] r b r a Notice that W ab is the work done by the gravitational force in bringing m from a to b. In bringing the mass m from a to b, [ 1 U = U b U a = W ab = GMm 1 ] r b r a

166 Chapter 13 Gravitation 160 Now, we take r a = r, U a = U(r) and r b, where U( ) = 0. ( U( ) U(r) = GMm 1 ) r U(r) = GMm r = W r Escape speed A particle having initial speed of v 0 is launched on the surface of the Earth. For the particle to escape from the Earth s gravitational force field, it is energetic possible for it to travel to infinity. Therefore, 1 2 mv2 0,min + ( GMm ) R E = 0 2GM v 0,min = 13.6 Potential energy of many-particle system R E m 2 r 12 r 23 m 1 r13 m 3 ( Gm1 m 2 U = + Gm 1m 3 + Gm ) 2m 3 r 12 r 13 r 23 It is the formation energy of the mass system started from nothing. Particles are then introduce one by one from infinity to specific positions by external agent. Thus, the energy required by an external agent to take these three particles to separated infinity: E = U

167 Chapter 13 Gravitation Energy consideration of satellite motion r M m ω Consider a satellite orbiting a planet. U = GMm r K = 1 2 mv2 = 1 2 m(ωr)2 = 1 2 mω2 r 2 If the gravitational force provides the centripetal force, GMm r 2 = mω 2 r GM r = ω 2 r 2 K = 1 GMm 2 r E = K + U = 1 GMm GMm 2 r r = GMm 2r

168 Chapter 14 Simple Harmonic Motions 14.1 Hooke s law A light and unsketched spring has one end fixed while the next end connected to a block of mass m. The block is acted by a force such that the spring extends by a length A. The system is set free to move thereafter. The governing force on the block is the restoring force F which obeys the Hooke s law, F = k x, where k is a positive constant, or the force constant, x is the displacement of block. Notice that the restoring force is always opposite to the displacement of block. When the spring is extended, the force is a tensive one; when the spring is shortened, the force is a compressive one. Both forces tend to recover the natural length of spring. The equation of motion of block becomes kx = mẍ ẍ = k m x The acceleration is linearly proportional to the displacement of the object, but the directions of them are opposite to each other. It is referred to as the simple harmonic motion (SHM), ẍ = ω 2 x 162

169 Chapter 14 Simple Harmonic Motions 163 where ω 2 = k m and ω = k/m is the angular speed of the oscillation. The period of oscillation is T = 2π/ω = 2π m/k. Since ẍ = ẋ dẋ, we have dx v 0 ẋ dẋ dx = ω2 x x ẋ dẋ = ω 2 x dx v 2 = ω 2 (A 2 x 2 ) The speed is a maximum, ω 2 A 2, when the spring returns to the natural length, i.e. unsketched and uncompressed. The restoring force at this instant is zero. When the block is at the extreme positions, x = ±A, the block is stationary and is said to be located at the amplitude of oscillation. The acceleration so obtained has the largest magnitude, ω 2 A, and the restoring force is the greatest, ka. A One can verify that the simplest solutions of ẍ = ω 2 x could be x = A cos ωt or x = A sin ωt. The selection of them depends on the initial conditions. If the initial time t = 0 is measured when the block is at the positive extreme position and it begins to return, x = A cos ωt. If the initial time t = 0 is measured when the block is at the equilibrium position and it is moving along the positive direction, x = A sin ωt. If the initial time t = 0 is measured at an arbitrary instant, an extra phase angle is inserted, e.g x = A sin(ωt + φ). A simple pendulum has a bob of mass m connected to one end of a light string of length L, while the next end is attached to a fixed point P. The system is displaced slightly, it oscillates to-and-fro along a small arc of a circle of center P. Find the equation of motion of the system if the oscillation is small.

170 Chapter 14 Simple Harmonic Motions 164 Solution For small oscillation, the angle θ is small, the tangent force along the arc of circle is roughly horizontal, it is given by mg sin θ mgθ. It is nearly horizontal, as θ is small. On the other hands, the horizontal displacement of bob from O is x = L tan θ Lθ. Therefore, the equation of motion becomes mgθ = mẍ mg ( x L ) = mẍ ẍ = g L x The angular speed ω = g/l, and the period of oscillation is T = 2π/ω = 2π L/g. The liquid in a uniform U-tube is disturbed such that the liquid oscillates inside it. At time t, the liquid in one limb is higher than the equilibrium line by x. Find the equation of motion of the liquid if friction is ignored. The total length of liquid in the tube is 2h.

171 Chapter 14 Simple Harmonic Motions 165 Solution The excess pressure on the whole liquid equals the pressure difference on the two sides of S at the bottom of U-tube, i.e. Excess pressure = excess height liquid density g = 2xρg. Since pressure = force per unit area, we have the force on liquid = pressure area of cross-section of the tube = (2xρg) A. The mass of liquid inside the tube is 2hAρ. Therefore, the equation of motion of the liquid is 2xρgA = 2hAρẍ. The negative sign appears as the direction of the restoring force is opposite to the displacement of liquid. After simplification, we obtain ẍ = g h x = ω2 x, where ω 2 = g/h. The period of oscillation is T = 2π/ω = 2π h/g Potential energy and kinetic energy in a springmass system In a spring-mass system, the spring is extended from its natural length by using an applied force F. If the extension is x, the total work done by the force is x F dx = x 0 0 kx dx = 1 2 kx2. The displacement x could be negative when the spring is being compressed. However, the expression is the same for extension and compression. This work done also gives the

172 Chapter 14 Simple Harmonic Motions 166 potential enegy stored in the spring. Thus, the total mechanical energy of an oscillating mass connected with it is 1 2 mv kx2 = 1 2 mω2 A 2 = 1 2 ka2 = E = constant. We have applied the relation v 2 = ω 2 (A 2 x 2 ). In the figure, U and K are the potential energy and kinetic energy respectively. Three identical spring having force constant k = 2 N/m and natural length 1 m lie on the medians of an equilateral triangle on a smooth and horizontal plane. Each of them has one end fixed on the vertex of the triangle and the next end attaches to a particle of mass m at the centroid G of the triangle. The spring are unsketched as shown in the figure. The mass is displaced to point P and then released. What will be the kinetic energy of m when it returns to the centroid?

173 Chapter 14 Simple Harmonic Motions 167 Solution Initially, the springs occupy no elastic potential energy because they are unsketched. When the mass is displaced to P, spring a is extended, and spring b and c are compressed. The length of extension, GP = 0.5 m and the length of compression is given by 1 CP = 1 1 cos 30 o = m. By the conservation of mechanical energy, we have The kinetic energy of m at G = The total elastic potential energy stored at P Therefore, K.E. = 1 2 kx2 a kx2 b kx2 c K.E. = 1 2 (2N/m) (0.5m) (2N/m) (0.134m)2 + 1 (2N/m) (0.134m)2 2 K.E. = Joule Two massless springs, each with force constant k and unstretched length l 0 are connected in a straight line as shown in the left figure. Find an expression for the work done required to move the point of attachment between the two springs a perpendicular distance x from the equilibrium point. Show that the work done for such movement is given by kx4 when x << l 0. [Hint: (1 + y) p = 1 + py + Solution p (p 1) y l 2 0 p (p 1) (p 2) y 3 +, where p is real.] Denote the extension of each spring by e when the joint of springs is displaced by x. e = l0 2 + x 2 l 0 As the horizontal component of the restoring force of each spring cancels each other, the ( applied force F = 2ke sin θ = 2k ( l0 2 x + x 2 l 0 ) ). The work done by the l x 2

174 Chapter 14 Simple Harmonic Motions 168 force is W = x 0 W = 2k F dx ( x 0 x ) l 0 x dx l x 2 W = kx 2 2kl 0 l x 2 + 2kl 2 0 Using the hint, we obtain the binomial expression of l x 2 l x 2 = l 0 {1 + ( ) } 2 1/2 x l 0 ( = l x 2 1 ) x l0 2 8 l0 4 When x l 0, { ( W = kx 2 2kl 0 l x l0 2 8 )} x 4 + 2kl l Therefore, W = kx4. 4l Damped oscillation When there is a damping force exerted on the oscillator, e.g. Fd = b v, where b is a positive damping constant. The system is said to be linearly damped. The equation of motion becomes kx bẋ = mẍ, i.e. mẍ + bẋ + kx = 0 After solving this equation (not the scope of this course), we obtain ( b x = A 0 e (b/2m) t cos(ω t + δ), and ω = ω 0 1 2mω 0 where A 0 is the initial amplitude and ω 0 = k/m is the natural frequency without damping. For weak damping, b/(2mω 0 ) << 1 and ω ω 0. The dashed curve in the figure shows the variation of amplitudes, e.g. A = A 0 e (b/2m) t. ) 2

175 Chapter 14 Simple Harmonic Motions 169 We notice that if the damping constant b increases gradually, the angular frequency ω vanishes at the critical value b c = 2mω Resonance If a back and forth force is given to the oscillator, the system is said to be driven oscillations. In general, driving any system at a frequency near its natural frequency results in large oscillations. The large response is known as resonance. The figure below shows some resonance curves. When the damping is small and the frequencies are close to the natural frequency, f 0, the amplitude of oscillation can become very large. When the damping is large, the amplitude becomes small and it appears as a broad peak near f 0.

176 Chapter 15 Fluids 15.1 Static equilibrium in fluids: Pressure and depth Consider a cylindrical container of cross-section area A, filled with fluid a height h. The top surface is at an atmospheric pressure P atm. At the bottom of the container (point b), the downward force is F atm plus the weight of the fluid. F b = F atm + mg P b = F b A = F atm A + mg A = P atm + ρv g A where ρ is the density of the fluid. Hence, we obtain P b = P atm + ρgh. Remark: You may note that one day while swimming below the surface of a river, you let out a small bubble of air from your mouth. As the bubble rises toward the water surface, its diameter increases. The reason is simple, since the bubble rises the pressure in the surrounding water decreases. The volume of air bubble thus expands. A U-shaped tube is filled mostly with water, but a small amount of vegetable oil has been added to one side. The density of the water is 1000 kg/m 3, and the density of the vegetable oil is 920 kg/m 3. If the depth of the oil is 5.00 cm, what is the difference in level h between the top of the oil on one side of the U and the top of the water on the other side? 170

177 Chapter 15 Fluids 171 Solution Note that the pressure at point A is the same as that at point B. Hence we can write P atm + ρ 1 gh 1 = P atm + ρ 2 gh 2 that is ρ 1 h 1 = ρ 2 h 2, or h 1 = ρ 2h 2 ρ 1. Plugging in the densities of water and oil as well as the height of oil in the U tube. We obtain h 1 = 4.60 cm. Since h = h 2 h 1 = 5.00 cm 4.60 cm = 0.40 cm Pascal s principle Recall that if the atmospheric pressure is P atm, the pressure at a depth h below the fluid surface is P = P atm + ρgh. Suppose now, that the atmospheric pressure is replaced by a pressure P atm + P, the pressure at the depth h is P = P atm + P + ρgh. Thus, by increasing the pressure at the top of the fluid by an amount, say, P, we have increased it by the same amount everywhere in the fluid.

178 Chapter 15 Fluids 172 Pascal s Principle states that an external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid. An example of Pascal s Principle is the hydraulic lift, which is sketched in the figure. The cross-sectional area A 2 is larger than A 1. Suppose the piston 1 in the left cylinder is pushed by a downward force F 1, the extra pressure other than the atmospheric pressure at the fluid surface is P = F 1 /A 1. This extra pressure is transmitted everywhere in the fluid. At the right piston, that is piston 2, an extra pressure of the same amount is exerted on it. And the force acting upward on piston 2 is F 2 = ( P ) A 2 Note that the greater the area of piston 2 the greater will be the upward force. As a result, it is a good machine for lifting objects. Substituting the increase in pressure in the above relation F 2 = ( F 1 A 1 ) A 2 = ( A 2 A 1 ) F 1 > F 1. For an incompressible fluid, the volume of fluid moved in the left cylinder should be the same as that in the right cylinder. One has A 1 d 1 = A 2 d 2. Hence, we obtain F 2 = ( A 2 ) F 1 = ( d 1 ) F 1 or we we can conclude that the force is inversely proportional to the A 1 d 2 displacement of the piston. F 1 F 2 = d 2 d Archimedes principle and buoyancy A fluid exerts a net upward force on any object it surrounds. This is referred to as a buoyant force. Archimedes s Principle states that an object wholely or partially immersed in a fluid is buoyed up by a force equal in magnitude to the weight of the fluid displaced by the object. What fraction of the total volume of an iceberg is exposed? Solution The weight of the iceberg is W i = ρ i V i g, where V i and ρ i are the volume and the density of the iceberg respectively.

179 Chapter 15 Fluids 173 The volume of the submerged portion of the iceberg relates to the buoyant force F = ρ w V w g, where V w and ρ w are the volume and the density of water which is displaced by the iceberg. At equilibrium, the upward and downward forces balance each other F = W i. Hence, ρ w V w g = ρ i V i g and we obtain the volume ratio V w V i = ρ i ρ w. Plugging in the densities of ice (917 kg/m 3 ) and water (1000 kg/m 3 ) into the above equation, we conclude that 8.3 % volume of ice is above water. A piece of plastic with a density of 706 kg/m 3 is tied with a string to the bottom of a water-filled flask. The plastic is completely immersed, and has a volume of m 3. What is the tension in the string? Solution

180 Chapter 15 Fluids 174 The upward force is the buoyant force, ρ w V g. The downward forces are the tension of string plus the weight of plastic, T + ρ p V g. As the system is in equilibrium, the upward and downward forces balance each other e.g. T + ρ p V g ρ w V g = 0, which gives the tension of string T = ρ w V g ρ p V g Plugging in the densities of plastic and water, the volume of plastic and the gravitational acceleration 9.8 ms 2. The tension of string is N Continuity of fluid flow Imagine a fluid flows in a cylindrical pipe, the mass passing through the large pipe in a given time, t, must also flow past the small pipe in the same time. The masses of fluid flow in the large pipe and the small pipe are m 1 = ρ 1 V 1 = ρ 1 A 1 v 1 t m 2 = ρ 2 V 2 = ρ 2 A 2 v 2 t respectively. As m 1 = m 2, we obtain ρ 1 A 1 v 1 = ρ 2 A 2 v 2. If the fluid is incompressible, ρ 1 = ρ 2, we thus obtain A 1 v 1 = A 2 v Bernoulli s equation The pressure acting on a moving fluid does work on it that appears as a net change in the kinetic and / or potential energy of the system; that is, W = K.E. + P.E.

181 Chapter 15 Fluids 175 Consider the tube in the following figure. The left fluid element enters the inlet and the right fluid element leaves the outlet. The pressure-force F 1 = P 1 A 1 acts on the left fluid element, pushing it in the direction of motion. The amount of work on the left fluid element is F 1 l 1. At the same time interval, the fluid external to the left fluid element pushes this element with a leftward force F 2 = P 2 A 2, which is opposite to the displacement of this element. Thus, the surrounding fluid is doing a negative work done on the right fluid element, i.e. F 2 l 2. The net work done on the fluid element is W = F 1 l 1 F 2 l 2 = P 1 A 1 l 1 P 2 A 2 l 2 = P 1 A 1 (v 1 t) P 2 A 2 (v 2 t) = V (P 1 P 2 ) Since the amount of fluid moved in a time interval t is V, which is given by V = A 1 v 1 t = A 2 v 2 t. Hence, we obtain the work done due to the pressure difference W = m ρ (P 1 P 2 ). The change in kinetic energy of the fluid in the tube is K.E. = 1 2 m (v2 2 v1). 2 The change in gravitational potential energy arises because the shifting of all the molecules in the tube, P.E. = mg (y 2 y 1 ).

182 Chapter 15 Fluids 176 By using the conservation of energy, W = K.E. + P.E., we can write m ρ (P 1 P 2 ) = 1 2 m (v2 2 v 2 1) + mg (y 2 y 1 ). After simplification, we obtain the Bernoulli s equation P ρv2 1 + ρgy 1 = P ρv2 2 + ρgy 2. Each term has the dimension of energy per unit volume, or energy density. That is the the sum of pressure-energy density (P ) arising from the internal forces on the moving fluid, the kinetic-energy density ( 1 2 ρv2 ) and the potential-energy density (ρgy) is a constant. The sum is sometimes called the net energy density of the fluid. Remark: If the pipe is horizontal, we have P ρv2 1 = P ρv2 2. In other words, P ρv2 is a constant. Greater the flow rate of fluid at a certain point, lesser will be the pressure acting on it. As a practical example, the lifting force on an aerofoil is the result of the pressure difference exerted on it. Another example is the net upward force on the roof when wind blows across the roof of a house. The lower pressure over the roof is accounted by the Bernoulli s equation. Water flows through a garden hose that goes up a step 20.0-cm height. If the water pressure is 143 kpa at the bottom of the step, what is its pressure at the top of the step? Given that the cross-sectional area of the hose on top of the step is half that at the bottom of the step and the speed of the water at the bottom of the step is 1.20 m/s.

183 Chapter 15 Fluids 177 Solution Assume that water is incompressible, the continuity equation states that A 1 v 1 = A 2 v 2 Hence the velocity of water at the top of the step is m/s = 2.40 m/s. Using the Bernoulli s equation P ρv2 1 + ρgy 1 = P ρv2 2 + ρgy 2. which gives P 2 = P ρ (v2 1 v 2 2) + ρg (y 1 y 2 ). Plugging in P 1 = 143 kpa, v 1 = 1.20 m/s, y 1 = 0 m, v 2 = 2.40 m/s, y 2 = 0.20 m, and ρ = 1000 kg/m 3, we obtain the pressure at the top of the step, P 2 = 139 kpa. Find the velocity of water as it emerges from the tip of a tank as shown in figure. Solution This is a typical problem solved by the Bernoulli s equation. Plugging in the data, e.g. P 1 = P atm, y 1 = h and v 1 = 0; P 2 = P atm, y 2 = 0.

184 Chapter 15 Fluids 178 The equation P ρv2 1 + ρgy 1 = P ρv2 2 + ρgy 2 becomes P atm ρ (0)2 + ρgh = P atm ρv2 2 + ρg (0). Hence, ρgh = 1 2 ρ v2 and thus v 2 = 2gh Surface tension (a) Liquid drop A molecule in the interior of a fluid experiences attractive forces of equal magnitude in all directions, giving a net force of zero. A molecule near the surface of the fluid experiences a net attractive force toward the interior of the fluid. This causes the surface to be pulled inward, resulting in a surface of minimum area. Consider a drop of liquid, surface tension plays an important role for the formation of it. (b) water surface Molecules in the surface of a liquid are farther apart than those in the body of the liquid, i.e. the surface layer has a lower density than the liquid in bulk. (c) Molecular explanation The intermolecular forces in a liquid are similar to that in a solid. They are both attractive and repelling and the forces balance when the spacing between molecules has its equilibrium value. However, when the separation is greater than the equilibrium value (r 0 ), the attractive force between molecules exceeds the repelling force. This is the situation with the more widely spaced surface layer molecules of a liquid. The attractive forces on either side due to their neighbors which put them in a state of tension and thus

185 Chapter 15 Fluids 179 the surface behaves like an elastic skin or membrane. When a small force is applied to the liquid surface it tends to stretch, resisting penetration. (d) Definition of surface tension Consider a straight line of length l in the surface of a liquid. If the force acting at right angles to this line and in the surface is F, then the surface tension γ of the liquid is defined by γ = F l In words, γ is the force per unit length acting in the surface perpendicular to one side of a line in the surface. It is temperature dependent and has a unit of Nm 1. At 20 o C, for water γ = Nm 1 and for mercury γ = Nm 1. It should be noted that a thin film of soap or a bubble of soap has two surfaces, but the water drop has one surface. For example, the wire AB in the following figure is kept at rest by an external force F such that the surface tension force is balanced. As a film of soap has two surfaces and so the width of film contributing surface tension force is 2l and F = 2γl Pressure difference across a curved surface Consider a soap bubble in the following figure. The inward forces on the bubble are contributed by the atmospheric pressure P and the surface tension force. But these

186 Chapter 15 Fluids 180 forces are balanced by the force which is given by the pressure inside the bubble. As a result, this pressure exceeds the atmospheric pressure by p, and the magnitude of it is given by P + p. It is noted that the pressures inside (P + p) or outside (P ) the bubble provide horizontal forces P πr (2γπr) = (P + p) πr 2 4γπr = pπr 2 p = 4γ r Taking γ for a soap solution as Nm 1, the excess pressure inside a bubble of radius 1.0 cm is p = Nm m = 10 Pa. If the soap bubbles of different radii are blown separately using the apparatus as shown in figure. Now, taps T 1 and T 2 are opened, what is your observation? Solution The smaller bubble A will decrease its size and the larger bubble expands. Equilibrium will be attained when the two bubbles have the same radii. That bubble A becomes a curved film and bubble B has a larger radius Capillary rise formula Consider a fluid in a capillary tube, the fluid rises up to a height h due to the pressure difference across the curved fluid surface in the tube. Assume that the meniscus (fluid surface) is in the form of spherical shape and the pressure difference at E and D is

187 Chapter 15 Fluids 181 p = 2γ r. Of course, p E is greater than p D, since E is curved downward. The point C is at a depth h under D, the pressure at C, i.e. p C, exceeds p D by ρgh. p C = p D + ρgh = p E 2γ r + ρgh As p E is at an atmospheric pressure, p C = p atm 2γ r + ρgh. At equilibrium, p C = p A = p B = p atm. Hence we can write p C = p atm 2γ r + ρgh = p atm or ρgh = 2γ r, The height of fluid h is given by h = 2γ rρg. Remark: A U-tube with different limb diameters is shown in figure. At equilibrium, water with level difference is observed in the two limbs. Given that the surface tension of water is Nm 1 and its density is 1000 kgm 3. If the contact angle is zero between water surface and the limbs, what is the difference in water level.

188 Chapter 15 Fluids 182 Solution Since the surfaces of water is curved downward, we conclude that p B > p A and p D > p C. Mathematically, we have p B p A = 2γ r = 2 ( ) /2 p D p C = 2γ r = 2 ( ) /2 = 56 Pa = 140 Pa But, p B = p D = p atm, the pressures p A and p C can be rewritten as p A = (p atm 56) Pa, and p C = (p atm 140) Pa. Plugging in the expressions of p A and p C, we obtain (p atm 56) (p atm 140) = ρgh, which gives ρgh = 84 Pa. The difference in water levels is given by h = 84 Pa ρg = 84 Pa (1000 kgm 3 ) (9.8 ms 2 ) = 8.6 mm.