Angular velocity and angular acceleration CHAPTER 9 ROTATION! r i ds i dθ θ i Angular velocity and angular acceleration! equations of rotational motion Torque and Moment of Inertia! Newton s nd Law for rotation Determination of the Moment of Inertia! Parallel ais theorem! Perpendicular ais theorem Rotational kinetic energy! power Rolling objects (with no slip) The arc length moved by the i th element in a rotating rigid, non-deformable disk is: ds i = r i dθ where dθ is in radians. The angular velocity of the rotating disk is defined as: ω = dθ dt, and so the linear velocity of the i th element (in the direction of the tangent) is: v i = ds i dt = r iω, where r i is the distance from the rotation ais.
If the angular velocity changes there is angular acceleration... DISCUSSION PROBLEM [9.1]: You have a friend who lives in Minnesota, and you live in Florida. As the Earth rotates, your linear velocity is hers, and your angular velocity is hers. A: less than; equal to B: equal to; greater than C: greater than; less than D: less than; greater than E: greater than; equal to The angular acceleration of the disk is: α = dω dt = d dt. dθ dt = d θ dt, and the tangential acceleration of the i th element is: a it = dv i dt = r i dω dt = r iα. But, because the ith element is traveling in a circle, it eperiences a radial (centripetal) acceleration:! r i ds i a ir (= a ic ) = v i r i = r i ω. The resultant linear acceleration is dθ θ i! a i! a it! a ir! a ir a! it! a i = a ir + a it.
Angular velocity ( ω) (vector) Dimension: ω 1 [T] (Check: v i = r i ω [L] 1 [T] = [L] [T] ). Units: rad/s Angular acceleration ( α) (vector) Dimension: α 1 [T] Linear motion a constant CONNECTION BETWEEN LINEAR AND ROTATIONAL MOTION Rotational motion α constant v = v! + at ω = ω! + αt (! ) = v! t + 1 at (θ θ! ) = ω! t + 1 αt v = v! + a(! ) ω = ω! + α(θ θ! ) You see, they re very similar Units: rad/s
r r = 0.1 m: ω! = 0: θ! = 0: t = 5 s α = 3.00 rad/s. ω =?: θ =?: a t =?: a c =? (a) ω = ω! + αt = (3.00 rad/s )(5 s) = 15.0 rad/s. Question 9.1: A disk of radius 1 cm, initially at rest, begins rotating about its ais with a constant angular acceleration of 3.00 rad/s. After 5 s, what are (a) the angular velocity of the disk, and (b) the tangential and centripetal accelerations of a point on the perimeter of the disk? (c) How many revolutions were made by the disk in those 5 s? (b) v i = r i ω = (0.1 m)(15.0 rad/s) = 1.80 m/s (linear). tangential acceleration: a t = r i α = (0.1 m)(3.00 rad/s ) = 0.36 m/s. centripetal acceleration: a c = r i ω (Check... a c = v r = (0.1 m)(15.0 rad/s) = 7.0 m/s. = (1.80 m/s) 0.1 m = 7.0 m/s.) (c) (θ θ! ) = ω! t + 1 αt = 1 (3.00 rad/s )(5 s) = 37.5 rad, n = 37.5 rad π = 5.97 rev.
In many applications a belt or chain is pulled from or wound onto a pulley or gear wheel... As the string (chain or belt) is removed (or added), its instantaneous velocity is the same as the tangential velocity at the rim of the wheel, providing there is no slip: v t a t i.e., v t = Rω. Also, under the same conditions, the instantaneous acceleration of the string is the same as the tangential acceleration at the rim of the wheel: i.e., a t = dv t dt v t = R dω dt = Rα. R As we saw in chapter 4 that force produces change in motion. However, force does not always produce a change in rotational motion. It is torque that produces a change in rotational motion. Consider a mass m attached to a massless rigid rod that rotates around an ais O. The force F shown will cause the mass to rotate. O The magnitude of the torque due to a force F on m is: τ = lf = (r sin θ)f, where l is called the lever arm. The lever arm is the perpendicular distance from the ais of rotation (O) to the line of action of the force. NOTE: if " F passes through O, i.e., l = 0, then τ = 0 and there will be no change in rotational motion. l r θ m " F l = r sin θ
Dimension: τ [L] [M][L] [T] = [M][L] [T] Unit: N m (vector) Form the radial and tangential components of the force: O l Then τ = lf = (r sin θ)f = r(fsin θ) i.e., τ = rf t. Note: the radial component F r, which passes through the ais of rotation, does not produce a torque and therefore it does not produce rotation; only the tangential component F t produces a torque that results in rotation. r m " F F t = F sin θ O l r θ m " F F r = F cosθ F t = F sin θ O l r θ m " F F r = F cosθ Newton s nd Law tells us that the tangential component of the force F t produces a tangential acceleration a t, Therefore, the torque is i.e., F t = ma t. τ = rf t = mra t, but, from earlier, the tangential (linear) acceleration is related to the angular acceleration α, viz: a t = rα. τ = mra t = mr α. So torque (τ) produces angular acceleration (α). A rigid object that rotates about a fied ais can be thought of as a collection of small, individual elements of mass m i that each move in a circular path of radius r i, where r i is measured from the ais of rotation.
From above, for each separate mass element, we have O r i m i τ i = m i r i α, where τ i is the net torque on the ith element. Summing over all elements, the total net torque on the object is: τ net = i τ i = i m i r i α = m i r ( i i )α = Iα, where I = m i r i i is called the MOMENT OF INERTIA. This is Newton s nd Law for rotation, i.e., τ net = Iα. A net eternal torque acting on a body produces an angular acceleration, α, of that body given by Iα, where I is the moment of inertia. (viz: F net = i F i = ma.) Dimension: I [M][L] Units: kg m (scalar) Every object has a moment of inertia about an ais of rotation. Its value depends not simply on mass but on how the mass is distributed around that ais. For a discrete collection of i objects, the moment of inertia about the rotation ais is: For a continuous object: I = I = i I i = i m i r i. Limit m i 0 i where m is a function of r. Moment of inertia... so what s that all about? m i r i = r dm,
y d l l A rod is a continuous object, so the moment of inertia is Question 9.: Find the moment of inertia of a uniform thin rod of length l and mass M rotating about an ais perpendicular to the rod and through its center. y l I = r dm = = l dm. = l The mass per unit length of the rod is M l, so the mass of the small element of length d is dm = ( M l)d. Substituting for dm, the integral becomes I = = l ( ) M l d = M l = l 3 ( ) 3 l l ( ) l3 = M l 4 ( l3 ) 4 = 1 1 Ml.
[1] Show for yourselves that the moment of inertia of a rod of mass M and length l about one end is y d 0 l I = 1 3 Ml. Question 9.3: Find the moment of inertia of the circular disk shown below, rotating about an ais perpendicular to the plane and through its center. The mass of the disk is 1.50 kg. [] Show for yourselves that the moment of inertia of a rod of mass M and length l about an ais one-third the distance from one end is y d l 3 0 l 3 10 cm 0 cm I = 1 9 Ml.
r The moment of inertia is given by I = r dm. Consider r 1 a ring of radius r and width dr. If the mass of the object is M, the mass of the ring is dm = M πrdr π r r 1 ( ), where r 1 and r are the inner and outer radii of the object. Then, substituting for dm, = dr r I = r M πrdr π r r 1 = r 1 M r r ( 1 ) r 4 4 r ( ) = M r r ( 1 ) r = r1 r r 1 r 3 dr M r r ( 1 ) r 4 r 4 ( 1 ) M r r ( 1 ) r r ( 1 ) r + r ( 1 ) = 1 M r + r ( 1 ). I = 1 M r + r ( 1 ) = 1 (1.50 kg) ( (0.0 m) + (0.10 m) ) = 3.75 10 kg m. [1] If r 1 = 0, then I disk = 1 MR, where R is the radius of the disk ( = r ). [] For a thin hoop, r 1 r = R, then I hoop = 1 M R ( ) = MR. Providing the thicknesses of the disks are uniform, the moments of inertia do not depend on thickness. So, these epressions also apply to cylinders and tubes. R R
Sure, but, what s the significance of I? Values of the moment of inertia for simple shapes... Remember, from chapter 4... Mass a measure of resistance to a change in linear motion, e.g., how difficult it is to start or stop linear motion. Moment of Inertia a measure of resistance to a change in rotational motion, i.e., how difficult it is to start or stop rotational motion. Thin rod I = 1 1 ML L L Thin rod I = 1 3 ML Hollow cylinder I = 1 M(R + r ) Slab I = 1 1 M(a + b ) a b a Slab I = 1 3 Ma b Hollow sphere I = 3 MR Thin-walled cylinder I = MR Same torques Different moments of inertia RMM04VD1.MOV Solid cylinder I = 1 MR Solid sphere I = 5 MR
b O O a 5 kg The combined moment of inertia is I = I ruler + I mass, where I ruler and I mass are calculated about O O. Question 9.4: A 1 m ruler has a mass of 0.5 kg. A 5 kg mass is attached to the 100 cm end of the rule. What is its moment of inertia about the 0 cm end? Since a >> b we assume the meter rule is a rod (viz: Question 9.) with I = 1 3 Ma about O O. I ruler = 1 3 Ma = 1 (0.5 kg)(1 m) 3 = 0.083 kg m. The moment of inertia of the 5 kg mass about O O is I mass = ma = (5.0 kg)(1 m) = 5.0 kg m. Thus, the total moment of inertia of the ruler and mass is I ruler + I mass = 5.083 kg m.
DISCUSSION PROBLEM [9.]: Parallel ais theorem: A pair of meter rulers are placed so that their lower ends are against a wall. One of the rulers has a large mass attached to its upper end. If the meter rulers are released at the same time and allowed to fall, which one hits the floor first? A: The meter ruler with the mass. B: The meter ruler without the mass. C: They hit the floor at the same time. I cm I d Usually, the moment of inertia is given for an ais that passes through the center of mass (cm) of the object. What if the object rotates about an ais parallel to the ais through the center of mass for which we don t know the moment of inertia? The moment of inertia about a general (parallel) ais is given by: I = I cm + Md where I cm is the moment of inertia about the center of mass, M is the mass of the object and d is the distance between the parallel aes. Note: the two rotation aes must be parallel
Perpendicular ais theorem: i z y i r i m i Consider a planar object (e.g., a thin disk or sheet) in the,y plane. By definition, the moment of inertia about the z-ais (perpendicular to the plane of the object) is I z = i m i r i = m i ( i i + y i ) = m i i i + i m i y i But m i i i y = I, i.e., the moment of inertia about, and i m i y i = I y, i.e., the moment of inertia about y. Eample of a disk: I z = I + I y. Note: the object must be planar Question 9.5: Four masses at the corners of a square with side length L = m are conncted by massless rods. The masses are m 1 = m 3 = 3 kg and m = m 4 = 4 kg. Find (a) the moment of inertia about the z-ais, (b) the moment of inertia about an ais that is perpendicular to the plane of the ensemble and passes through the center of mass of the system, (c) the moment of inertia about the -ais, which passes through m 3 and m 4. y L m 1 m L z y I z = 1 MR. But, by symmetry, I = I y. z m 4 m 3 I = I y = 1 4 MR.
z y m 3 kg 4 kg m (a) Since we are dealing with discrete masses I = i I i = i m i r i. Moment of inertia about the z-ais: I z = m i r i i = (3 kg)( m) + (4 kg)( m) +(3 kg)( m) + (4 kg)(0) = 56 kg m. (b) By symmetry, the center of mass is at the center of the square. cm D 4 kg 3 kg m I cm = m i r i i = (3 kg)( m) + (4 kg)( m) +(3 kg)( m) + (4 kg)( m) = 8 kg m. Check, using the parallel ais-theorem I z = I cm + MD I cm = I z MD = (56 kg m ) (14 kg)( m) = 8 kg m. z y m 3 kg 4 kg 4 kg 3 kg m (c) Since the ensemble is planar and confined to the,y plane, we can use the perpendicular ais theorem, i.e., I z = I + I y. But, by symmetry, I = I y. I = 1 I z = 1 ( 56 kg m) Check: I = m i r i i = 8 kg m. = ( 3 kg) ( m) + ( 4 kg) ( m) = 8 kg m.
Question 9.6: Four thin rods, each of length l and mass M, are arranged to form a square, in the,y plane, as shown. If the origin of the aes is at the center of the square, (a) using the parallel ais theorem, show that I z = 4 3 Ml. (b) Hence find I and I y. z l l y (a) Using the parallel ais theorem, we have for each rod I z = I cm + Md, where I cm is the moment of inertia through the center of mass of each rod and d = l. I z (total) = 4 1 1 ml + m l = 4 4 3 ml. l z l y (b) Since the object is planar we can use the perpendicular ais theorem, i.e., I z = I + I y. But because of symmetry I = I y. I = I y = 1 I z = 3 ml.
We saw in chapter 6 that a linearly moving object has translational kinetic energy... an object rotating about an ais has rotational kinetic energy... ω O r i v i m i If a rigid object is rotating with angular velocity ω, the kinetic energy of the i th element is: K i = 1 m i v i = 1 m ir i ω, since v i = r i ω. So, the total rotational kinetic energy is: 1 K = i K i = m i r i ω i = 1 Iω. * K rot = 1 Iω is the analog of K trans = 1 mv. A torque is required to rotate (or slow down) an object... but torque involves force... When force is applied over a distance, work is done, given by: O dw =! F d! s = F.dscos γ = F cos γ.ds = F t.ds = F t.rdθ = τ.dθ (J or N.m). Power is the rate at which the torque does work, i.e., P = dw = τ dθ dt dt = τω (watts). Note: P is the instantaneous power. dw = τ.dθ is the analog of dw = F.ds. P = τω is the analog of P = Fv.! r ds dθ! F F t γ! F θ F r γ + θ = 90 " F t = F cos γ
Question 9.7: The 3.9 liter V-8 engine fitted to a 488GTB Ferrari develops 560 ft lb of torque at 3000 rev/min. What is the power developed by the engine with these parameters? ( 1 ft lb = 1.36 N m.) From earlier, power P = τω. Convert the torque to N m and angular velocity to rad/s, and i.e., τ = 1.36 560 = 76 N m, ω = π(3700 rev/min) 60 s/min = 387.5 rad/s. P = (76 N m)(387.5 rad/s) =.95 10 5 watts. But 746 watts = 1 HP P =.95 105 watts 746 watts/hp = 396 HP.
(a) To find the rotational kinetic energy we need to know the angular velocity (ω). Treating the grindstone as a solid disk I = 1 MR = 1 (16.0 kg)(0.50 m) =.0 kg m. Question 9.8: An electric motor eerts a constant torque of 10.0 N m to the shaft of a grindstone with mass 16.0 kg and radius 0.50 m. If the system starts from rest, find (a) the rotational kinetic energy of the grindstone after 8.0 s, (b) the work done by the motor during this time, and (c) the average power delivered by the motor. Since τ = Iα, α = τ I = 10 N m kg m = 5.0 rad/s. The grindstone starts from rest ( ω! = 0) so ω = αt = (5.0 rad/s )(8.0 s) = 40 rad/s. K = 1 Iω = 1 ( kg m )(40 rad/s) = 1600 J. (b) There are two ways to determine the work done by the motor. (i) By the work-kinetic energy theorem we would epect the motor to have done 1600 J of work. (ii) We can use the epression W = τθ, but we need to find θ, i.e., the angle through which the grindstone has turned in 8.0 s. θ = ω! t + 1 αt = 1 (5.0 rad/s )(8.0 s) = 160 rad
W = τθ = (10.0 N m)(160 rad) = 1600 J. (c) The average power is the total work done divided by the total time interval, i.e., P av = ΔW = 1600 J Δt 8.0 s = 00 W. Note: the epression P = τω we derived earlier is actually the instantaneous power. We cannot use that epression here as ω is not constant. The instantaneous power actually increases linearly from zero at t = 0 to (10.0 N m) (40 rad/s) = 400 W at t = 8.0 s. If a rigid object is suspended from an arbitrary point O and is free to rotate about that point, it will turn until the center of mass is vertically beneath the suspension point. y r cm cm O cm Mg If the y-direction is vertical and the suspension point is not at the center of mass, the object will eperience a net torque given by τ = Mg cm, where cm is the component of the center of mass. Therefore, the object will rotate until cm = 0, i.e., the suspension point is direction above the center of mass.
pivot U 1 = mg l : K 1 = 0 cm mg l U = 0 : K = 1 Iω (a) Use Newton s nd Law for rotation, i.e., τ = Iα, where τ = mg l and I = ( 1 3)ml. Then α = τ I = 3g l, which is the initial angular acceleration of the center of mass. The linear acceleration of the 100 cm end is then Question 9.9: The zero end of a 1 m ruler of mass 0.5 kg is attached to a frictionless pivot, the other end is free to rotate in the vertical plane. If the ruler is released from rest in the horizontal position, what is (a) the initial acceleration of the 100 cm end of the ruler, and (b) the linear speed of the 100 cm end as the ruler passes through the vertical? a = lα = 3g = 14.7 m/s, which is greater than g! Also note, the torque τ varies as the ruler swings down. (b) To find the speed of the 100 cm end as it passes the vertical we use the conservation of mechanical energy, i.e., U 1 + K 1 = U + K. Taking the zero of the gravitational potential energy at the point where the center of mass is at its lowest point, then mg l + 0 = 0 + 1 Iω.
ω v cm v = 0 v cm Consider a ball, cylinder, wheel or disc) rolling on a surface without slipping. ω = mgl I = 3g l. The linear velocity of the 100 cm end as it passes the vertical is then v = lω = 3gl = 3(9.81 m/s )(1.0 m) = 5.4 m/s. The point in contact with the surface has zero instantaneous velocity relative to the surface. The velocity of the cm is v cm ( = rω = v), The velocity of a point at the top is v cm (= v). Since the object has the same linear velocity as the cm, i.e., v, the total kinetic energy is: K tot = 1 mv + 1 Iω translational + rotational where ω = v r. So, as a ball rolls down a hill... v, ω Gravitational potential energy translational energy + rotational energy
Consider rolling a sphere, cylinder and a hoop down an incline. Do they have the same velocity at the bottom? h v For each object, conservation of energy gives: mgh = 1 mv + 1 Iω, and with no slip v = Rω, where R is the radius of the object. After manipulation we find: v gh =. 1 + I mr For maimum velocity: I smallest value. So, in a race between objects rolling down a slope, the order would be (1) sphere, () cylinder, (3) hoop, and is completely independent of m and R! v gh = 1 + I mr For a solid sphere I = 5 mr. v = gh 1 +, i.e., v = 10gh 5 7. RMA07VD.MOV Since this is independent of m and R, a bowling ball and a pool ball would have the same speeds at the bottom of an incline, despite their different masses and radii!
Question 9.10: A uniform solid ball of mass M and radius R rolls without slipping down an incline at an angle θ to the horizontal. Find (a) the frictional force acting at the point of contact with the surface, and (b) the acceleration of the center of mass of the ball, in terms of M, R, g and θ. (a) Draw the free body diagram of all the forces acting on the ball. Use Newton s nd Law down the incline: N O f R θ Mg Mg sin θ f = Ma cm... (i) where a cm is the acceleration of the center of mass. To get an epression for a cm, we take torques about O, the center of the ball. (Note: the normal force N and the weight force Mg do not contribute to the torque as their lines of action pass through O.) τ = fr = Iα, i.e., α = fr I, NOTE: there must be friction at the point of contact otherwise the ball would slide down the incline! θ where I is the moment of inertia of the ball and α its angular acceleration. With no slip a cm = Rα = fr I...... (ii) Substituting for a cm in equation (i), we get Mg sin θ f = M fr I = MfR 5 MR = 5 f.
Mg sin θ = 5 f + f = 7 f, i.e., f = Mg sin θ. 7 Note that this is a static frictional force (because there is no slip). (b) Substituting for f in equation (ii), we get a cm = fr I = Mg sin θ 7 5 MR R = 5 gsin θ. 7 You can show yourselves that if we put I = βmr, where β = 1 for a hoop, β = 1 for a disk, etc., then a general epression for the static frictional force acting on an object rolling down an incline, with no slip, and the acceleration of the center of mass are: Mg sin θ f = 1 + β 1 and a cm = gsin θ 1 + β. Question 9.11: Suppose you can choose wheels of any design for a soapbo derby race car. If the total weight of the vehicle is fied, which type of wheel design should you choose if you want to have the best chance to win the race?
The total mechanical energy of the car is: translational kinetic energy + rotational kinetic energy. Conservation of energy gives: Mgh = K trans + K rot = 1 Mv + 1 Iω, where M is the total mass of the car and I is the moment of inertia of the wheels. With four wheels of radius r, for eample, we have, using the no-slip condition, ( ) v r K rot = 4 1 Iω = βmr Mgh = 1 Mv + βmv, i.e., v = Mgh ( M + 4βm). ( ) = βmv, So, for maimum speed (v) at the bottom of the hill (and greater average speed) with M fied, we want m and β to be as small as possible. Note: that the result does not depend on the radius of the wheels (r). So, solid wheels are a good choice with the mass concentrated as close to the ale as possible. F h R Question 9.1: A cue ball is struck by a horizontal cue a distance h above the center of the ball. If the cue ball is to roll without slipping, what is h? Epress your answer in terms of the radius R of the ball. You can assume that the frictional force of the table on the ball is negligible compared with the applied force F.
F h R f The net torque about the center is τ = Fh fr. From earlier, if there is no-slip then v cm = Rω, i.e., a cm = dv cm dt = R dω dt = Rα. Using Newton s nd Law F + f = ma cm. If F >> f, then F ma cm and τ Fh = Iα. F m = a cm = Rα = R Fh I, i.e., h = For a solid sphere I = 5 mr. h > 5 R h = 5 R. top spin I mr. h < 5 R back spin SLIP m 30 kg 0 kg m 1 m Question 9.13: The 30 kg mass, shown above, is released from rest, from a distance of m above the ground. Modeling the pulley as a uniform disk with a radius of 10 cm and mass 5 kg, find (a) the speed of the 30 kg mass just before it strikes the ground, (b) the angular velocity of the pulley at that instant, (c) the tensions in the two strings, and (d) the time it takes for the 30 kg block to reach the ground. Assume the bearings in the pulley are frictionless and there is no slip between the string and the pulley.
ω ω 0 kg m 1 m 30 kg m v m 1 m v 0 kg m 1 m 30 kg m v m 1 m v We have both translational and rotational motion. The moment of inertia of the pulley is: I = 1 mr = 1 (5 kg)(0.10 m) = 0.05 kg m. (a) Using conservation of energy, as m hits the ground m gh = 1 m 1v + m 1 gh + 1 m v + 1 Iω. The angular velocity of the pulley ω = v R, so (30 kg)(9.81 m/s )( m) = 1 (0 kg)v +(0 kg)(9.81 m/s )( m) + 1 (30 kg)v + 1 (0.05 kg m v ) (0.10 m), i.e., 589 = 10v + 39 +15v +1.3v v = 197 =.74 m/s. 6.3 (b) ω = v R =.74 = 7.4 rad/s. 0.10 (c) T 1 T m 1 g a m g What is the upward acceleration of m 1? a v = v! + ah. a = v (.74 m/s) = h ( m) Draw the free-body diagrams for the masses. = 1.88 m/s.
T 1 T m 1 g a m g a Then T 1 m 1 g = m 1 a, i.e., T 1 = m 1 (g + a) = 34 N. Also T m g = m ( a), i.e., T = m (g a) = 38 N. (d) ( y y! ) = h = 1 at. t = h a = ( m) = 1.46 s. 1.88 m/s