Newton s Laws of Motion

Transcription

1 Chapter 4 Newton s Second Law: in vector form Newton s Laws of Motion σ റF = m റa in component form σ F x = ma x σ F y = ma y in equilibrium and static situations a x = 0; a y = 0 Strategy for Solving Newton s Law Problems Isolate the bodies in a system. Analyze all the forces acting on each body and draw one free-body diagram for each body. Based on the free-body diagram, set up a most convenient x-y coordinate system. Break each force into components using this coordinates. For each body, sum up all the x-components of the forces to an equation: σ F x = ma x. For each body, sum up all the y-components of the forces to an equation: σ F y = ma y. Use these equations to solve for unknown quantities.

2 Type of Forces (that we will deal with in this course) Contact forces Normal Force n Friction Force റf k normal to the tangent to the contacting surface contacting surface n T റf k Tension Force T (in a rope) along the rope Spring Force റF spr opposite to the deformation റF G Action-at-a-Distance Forces (non-contacting) Gravitational Force റF G (along the line connecting the two centers of mass)

3 Superposition of Forces: Resultant and Components of Force Vectors An example of superposition of forces: y In general, the resultant, or vector sum of forces, is: x θ = tan 1 R y R x direction

4 Example: Box of mass m on a frictionless incline Given: m and q Find: n and a y n Solutions: Weight: W = mg W x = mgsinq W y = mgcosq x-axis: mgsinq = ma a = gsinq y-axis: n mgcosq = 0 n = mgcosq x q q mg Another Question: Near the bottom of the incline, the box is given an initial velocity of a known magnitude v 0 pointing up the incline. What distance will it slide before it turns around and slides downward? Solution: The acceleration is given above a = gsinq Use the kinematic equation v 2 2 v 2 1 = 2a(x 2 x 1 ) (x 2 x 1 ) = v 2 1 /2a = v 2 0 /2gsinq The distance that it will slide up the incline is v 2 0 /2gsinq.

5 Velocity: A Review of Uniform Circular Motion----Section 3.4 tangent to the circle with constant magnitude v 1 = v 2 = v Acceleration: a rad = v2, always pointing toward the center of the circle. R known as the centripetal acceleration A new concept Period (T, in s): Time for the object to complete one circle. New relationships between v, R, and T: Velocity Acceleration: Another new concept v = 2πR T a rad = v2 = 4π2 R R T 2 Frequency (f, in s -1, or, Hz): revolutions per second. Relationship between T and f: f = 1/T

6 6.2 Motion in a Vertical Circle Example 6.5 Dynamics of a Ferris wheel at a constant speed, page 158 Given: m = 60.0 kg R = 8.00 m T = 10.0 s (v = 2pR/T) Find the normal force: (a) at the top (n T ) (b) at the bottom (n B ) at the top (a) At the top n T mg = ma T = m( v2 ) R (b) At the bottom n T = mg m v2 R = m(g v2 R ) n B mg = ma B = m( v2 R ) n T = mg + m v2 R = m(g + v2 R ) at the bottom

7 A roller-coaster car of mass m = 100 kg is initially resting on an incline at a height H = 20.0 m above the bottom of the loop. It slides down and loops inside a circular track of radius R = 5.00 m. Neglect all friction. (a) What is its speed at the bottom of the loop (Point A)? Set up the y-axis. Apply the conservation of energy: mgh m(0)2 = mg m(v A) 2 v A = 2gH (b) At Point A, draw a free-body diagram. What is the normal n that the track exerts on the car? n mg = ma A = m v A 2 = m 2gH R R n = mg + 2mgH/R (c) Speed at the top of the loop (Point B)? mgh m(0)2 = mg 2R + 1 m(v 2 B) 2 v B = 2gH 4gR n A mg H (d) Free-body diagram and normal force at Point B? n mg = ma B = m v B 2 R B = m(2gh 4gR)/R n mg n = m(2gh 5gR)/R (a) Minimum H for car not to fall off the track at Point B? n = 0, mg = m v B 2 R, H = 5 2 R B A R y o

8 Gravitation and Circular Satellite Orbit If a satellite is in a circular orbit with speed v orbit, the gravitational force provides the centripetal force needed to keep it moving in a circular path. The orbital speed of a satellite G mm E r 2 = F g = F rad = m v2 r v orbit = Gm E r The period of a satellite v = 2πr T T = 2πr v = 2πr r Gm E = 2πr3/2 Gm E

9 6.4 Weight and Gravitation Acceleration near the surface of the Earth The weight of an object near the surface of the earth is: With this we find that the acceleration due to gravity near the earth's surface is:

10 Example: Problem 7, Exam II, Fall 2016 (a) A satellite of mass 80.0 kg is in a circular orbit around a spherical planet Q of radius m. The satellite has a speed 5000 m/s in an orbit of radius m. What is the mass of the planet Q? (b) Imagine that you release a small rock from rest at a distance of 20.0 m above the surface of the planet. What is the speed of the rock just before it reaches the surface? Given: About the satellite (m s = 80.0 kg, r orbit = m, v = 5000 m/s) About the planet Q(R Q = m) Find: (a) The mass of the planet Q (m Q ) (b) Speed of a rock after falling h = 20.0 m. (a) G m s m Q v r 2 = F g = F rad = m 2 s orbit r orbit m Q = r orbitv 2 G (b) First, find the gravitational acceleration g Q near the surface of the planet Q. R Q r orbit x m s g Q = G m s m Q R Q 2 g Q = G m Q R Q 2 Then, apply the kinematic equation v 2 2 = v g Q h to find v 2 with v 1 = 0.

11 More generally, work is the product of the component of the force in the direction of displacement and the magnitude s of the displacement. Forces applied at angles must be resolved into components. W is a scalar quantity that can be positive, zero, or negative. If W > 0 (W < 0), energy is added to (taken from) the system. Equivalent Representations: W = F s = F cos s = F s cos = Fs

12 Example 7.2 on Page 186: Sliding on a Ramp Package of mass m slides down a frictionless incline of height h and angle b. Find: The total work done by all the relevant forces. Work done by the gravity force (the weight): Component of the weight parallel to displacement = mgsinb. Work done: (mgsinb)s = mg[s(sinb)] = mgh Work done by the normal force = 0 since the component of the normal force parallel to displacement = 0 Total work done by all the relevant forces = mgh

13 y x Applications of Force and Resultant Work

14 What if the net amount of work done on an object is not zero? The work-energy theorem: 7.3 Work and Kinetic Energy The kinetic energy K of a particle with mass m moving with speed v is K = 1 2 mv2. It is the energy related to the motion of the particle. During any displacement of the particle, the net amount of work done by all the external force on the particle is equal to the change in its kinetic energy. Although the kinetic energy K is always positive, W total may be positive, negative, or zero (energy added to, taken away, or left the same, respectively). If W total = 0, then the kinetic energy does not change and the speed of the particle remains constant.

15 Examples of Force and Potential Energy Force Type Work Done by the Force (i f ) Potential Energy Gravity F g = mg W = mgy i mgy f = U i U f U grav = mgy Spring (elastic) F spring = kx W = 1 2 kx i kx f 2 = U i U f U el = 1 2 kx2

16 7.6 Conservation of Energy For a conservative force, the work done is related to the change in the potential energy: W = U i U f. Apply the work-energy theorem, we obtain: U i U f = K f K i or U i + K i = U f + K f. The expression in red is the conservation of energy, when the force is a conservative force. The total mechanical energy E is the sum of the potential energy and kinetic energy: E = U + K

17 Useful Expressions for Applications With gravity force: E = U grav + K = mgy mv2 Conservation of energy: mgy i + 1 mv 2 i 2 = mgy f + 1 mv 2 f 2 With spring force: Conservation of energy: With both g&s forces: E = U el + K = 1 2 kx mv2 1 kx 2 i mv 2 i 2 = 1 kx 2 f mv 2 f 2 E = U grav + U el + K = mgy ky mv2 Conservation of energy: mgy i ky i mv i 2 = mgy f ky f mv f 2

18 When both conservative and nonconservative forces act on an object (U grav,i + U el,i mv i 2 ) + W other = (U grav,f +U el,f mv f 2 )

19 8.2 Conservation of Momentum Consider a system consisting of a number of particles (A, B, ) External Forces: Forces acting on objects in the system by objects outside the system. Net External Force: the sum of the external forces. Internal Forces: Forces acting on particles in the system by particles in the system. Net Internal Force: the sum of the internal forces, which must be zero. Conservation of Momentum: If the net external force is zero, the momentum of the system is conserved: In the component form: Examples 8.3 and 8.4 on page 226 Example 8.5 on page 227 റp A,i + റp B,i + = റp A,f + റp B,f +. p A,i.x + p B,i,x + = p A,f,x + p B,f,x + p A,i,y + p B,i,y + = p A,f,y + p B,f,y +

20 8.3 Inelastic Collision Momentum is conserved but not the Kinetic Energy Completely inelastic collision along a straight line: m A v A,i,x + m B v B,i,x = (m A + m B )v f,x 8.4 Elastic Collision Both the Momentum and the Kinetic Energy are conserved Consider the elastic collision of two particles along a straight line. Conservation of Momentum: Conservation of Kinetic Energy: m A v A,i + m B v B,i = m A v A,f + m B v B,f 1 m 2 Av 2 A,i + 1 m 2 Bv 2 B,i = 1 m 2 Av 2 A,i + 1 m 2 2 Bv B,i

21 8.5 Impulse When a constant force റF acts on an object over a period of time t, the impulse of the force is റJ = റF t f t i = റF t If the force is not a constant, then റJ = റF avg t Since we have റF avg = m റa avg = m v = റp t റJ = റp = റp f റp i t,

22 Center of Mass Position General definition of the center of mass position: y Vector form റr cm = m A റr A +m B റr B +m C റr C + m A +m B +m C + Component form x cm = m Ax A +m B x B +m C x C + m A +m B +m C + റr C o റr B റr A x y cm = m Ay A +m B y B +m C y C + m A +m B +m C +