Module 27: Rigid Body Dynamics: Rotation and Translation about a Fixed Axis

Module 27: Rigid Body Dynamics: Rotation and Translation about a Fixed Axis 27.1 Introduction We shall analyze the motion o systems o particles and rigid bodies that are undergoing translational and rotational motion. There is no longer a clearly ixed axis o rotation but we shall see that is possible to describe the motion by a translation o the center o mass and a rotation about the center o mass. By choosing a reerence rame moving with the center o mass, we can analyze the rotational motion and discover that the torque about the center o mass is equal to the change in the angular momentum about the center o mass. In particular or a rigid body undergoing ixed axis rotation about the center o mass, our rotational equation o motion is similar to one we have already encountered or ixed axis rotation. We shall begin by recalling the main properties o the center o mass reerence rame 27.2 Review Center o Mass Reerence Frame Denote the position vector o the i th particle with respect to origin o reerence rame O by r i. The vector rom the origin o rame O to the center o mass o the system o particles, is deined as R cm = 1 m total The velocity o the center o mass in reerence rame O is given by N m i r i. (27.2.1) i=1 V cm = 1 m total m i v i = i p sys m total. (27.2.2) Deine the center o mass reerence rame O cm as a reerence rame moving with velocity V cm with respect to O. Denote the position vector o the i th particle with respect to origin o reerence rame O cm by r cm,i (Figure 27.1). 12/28/2010 1

Figure 27.1 Position vector o i th particle in the center o mass reerence rame. The position vector o the i th particle in the two center o mass rame is given by r cm,i = r i R cm. (27.2.3) or equivalently the position o the i th particle in the reerence rame O can be expressed as r i = r cm,i + R cm. (27.2.4) The displacement o the i th particle in the center o mass reerence rame is then given by d r cm,i = d r i d R cm. (27.2.5) The velocity o the i th particle in the center o mass reerence rame is then given by v cm,i = v i V cm (27.2.6) or equivalently the velocity o the i th particle in the reerence rame O can be expressed as 27.3 Translational Equation o Motion v i = v cm,i + V cm. (27.2.7) We shall think about the system o particles as ollows. We treat the whole system as a single point-like particle o mass m total located at the center o mass moving with the velocity o the center o mass V cm. The total external orce acting on the system acts at the center o mass and rom our earlier result we have that total F ext = d p sys dt = d dt (mtotal V cm ). (27.3.1) 27.4 Translational and Rotational Equations o Motion Recall that that we have previous shown in Module 25 that it is always true that S total = d L total S. (27.4.1) dt 12/28/2010 2

where total L S We dierentiate the LHS o Eq. (27.4.2) and ind that N = r S,cm p sys + ( r " cm,i m vcm,i i ). (27.4.2) i=1 S total = d dt ( r S,cm " p sys ) + d dt We apply the vector identity to Eq. (27.4.3) to Eq. (27.4.3) yielding $ % & N # i=1 ( r ' cm,i " m vcm,i i ) ( ). (27.4.3) d dt ( A B) = d A dt B + A d B dt, (27.4.4) total S = d r S,cm " p sys + r dt S,cm " d p sys dt d N # r + cm,i & )% " m vcm,i dt i ( + r cm,i " d. (27.4.5) N # i=1 $ ' dt (m & ) v i cm,i ) $ % ' ( i=1 The irst and third terms in Equation (27.4.5) are eliminated by noting that d r S,cm p sys = V dt cm m total V cm = 0 d r N " cm,i % N ( $ m vcm,i dt i ' = ( ( v cm,i m vcm,i i ) = 0. i=1 # & i=1 (27.4.6) Thereore the time derivative o the angular momentum about a point S, Equation (27.4.5), becomes S total = r S,cm " d p sys dt + N # r cm,i " d dt (m & ) v i cm,i ) $ % ' (. (27.4.7) i=1 In the irst term in Equation (27.4.7), the time derivative o the total momentum o the system is the total external orce, 12/28/2010 3

total F ext = d p sys dt (27.4.8) and in the second term, the orce acting on the element is F i = d dt (m v i cm,i ) (27.4.9) The expression in Equation (27.4.7) then becomes total S = r S,cm " F N total ext + #( r cm,i " F i ). (27.4.10) The irst term is the contribution to the torque about the point S i all the external orces were to act at the center o mass. The second term, "( r cm,i F i ), is the sum o the torques on the individual particles in the center o mass reerence rame. Then Equation (27.4.10) may be expressed as i=1 total S = r S,cm " F total ext N i=1 + total cm. (27.4.11) Retracing our calculating we have two conditions. For a system o particles, when we calculate the torque about a point S, we treat the system as a point-like particle located at the center o mass o the system. All the external orces F total ext act at the center o mass. We calculate the orbital angular momentum o the center o mass and determine it s time derivative and then apply orbital r S,cm F ext = d L S dt. (27.4.12) In addition, we calculate the torque about the center o mass due to all the orces acting on the particles in the center o mass reerence rame. We calculate the is equal to the time derivative o the total angular momentum o the system with respect to the center o mass in the center o mass reerence rame and then apply total cm = d L spin cm dt. (27.4.13) 27.5 Translation and Rotation o a Rigid Body Undergoing Fixed Axis Rotation 12/28/2010 4

For the special case o rigid body o mass m, we showed that with respect to a reerence rame in which the center o mass o the rigid body is moving with velocity V cm, all elements o the rigid body are rotating about the center o mass with the same angular velocity cm. For the rigid body o mass m and momentum p = m V cm, the translational equation o motion is still given by Eq. (27.3.1) which we repeat in the orm total F ext = m A cm. (27.5.1) Let s choose the z-axis as the axis o rotation that passes through the center o mass o the rigid body. We have already seen in our discussion o angular momentum o a rigid body that the angular momentum does not necessary point in the same direction as the angular velocity. However we can take the z-component o Eq. (27.4.13) ( total cm ) z = d( L spin cm ) z dt. (27.5.2) For a rigid body rotating about the center o mass with cm = cm,z ˆk, the z-component o angular momentum bout the center o mass is ( L spin cm ) z = I cm cm,z ˆk. (27.5.3) The our rotational equation o motion is 27.9 Work-Energy Theorem ( d" total cm ) z = I cm,z cm dt = I cm # cm,z. (27.5.4) For a rigid body, we can also consider the work-energy theorem separately or the translational motion and the rotational motion. Once again treat the rigid body as a pointlike particle moving with velocity V cm in reerence rame O. We can use the same technique that when treating point particles to show that the work done by the external orces is equal to the change in kinetic energy total W ext = F total d r d(m V " o ext = cm ) " d R o dt cm = m d( V cm ) " V dt cm dt o = 1 2 m " d( V cm V cm ) = 1 2 mv 2 # 1. (27.6.1) cm, 2 mv 2 = $K cm, trans o 12/28/2010 5

Thus or the translational motion " F total d r ext = 1 2 mv 2 cm, o # 1 2 mv 2 cm,. (27.6.2) For the rotational motion we go the center o mass reerence rame and the rotational work done i.e. the integral o the z-component o the torque about the center o mass with respect to d as we did or ixed axis rotational work. Then ( total ) d" d$ # o cm = I cm d" # z cm d" = I o dt # cm d$ o cm = I dt # cm d$ cm $ o cm = 1 2 I $ 2 % 1. (27.6.3) cm cm, 2 I $ 2 = &K cm cm,i rot Thereore we can combine these two separate results Eqs. (27.6.2) and (27.6.3) and get the work-energy theorem or a rotating and translating rigid body that undergoes ixed axis rotation about the center o mass. total W o, = F total d r " o ext + " (# total cm ) z d$ o & = 1 2 mv 2 + 1 cm, 2 I % 2 ) cm cm, ' ( * +, & 1 2 mv 2 cm, ' ( + 1 2 I % 2 cm cm,i ) * +. (27.6.4) = -K trans + -K rot = -K total So Eqs. (27.5.1), (27.5.4), and (27.6.4) are principles that we shall employ to analyze the motion o a rigid bodies undergoing translation and ixed axis rotation about the center o mass. (We shall call these our Rules to Live By.) 27.7 Worked Examples 27.7.1 Example Angular Impulse Two point-like objects are located at the points A, and B, o respective masses M A = 2M, and M B = M, as shown in the igure below. The two objects are initially oriented along the y-axis and connected by a rod o negligible mass o length D, orming a rigid body. A orce o magnitude F = F along the x direction is applied to the object at B at t = 0 or a short time interval t. Neglect gravity. Give all your answers in terms o M and D as needed. 12/28/2010 6

a. Describe qualitatively in words how the system moves ater the orce is applied: direction, translation and rotation. b. How ar is the center o mass o the system rom the object at point B? c. What is the direction and magnitude o the linear velocity o the center-o-mass ater the collision? d. What is the magnitude o the angular velocity o the system ater the collision? e. Is it possible to apply another orce o magnitude F along the positive x direction to prevent the system rom rotating? Does it matter where the orce is applied?. Is it possible to apply another orce o magnitude F in some direction to prevent the center o mass rom translating? Does it matter where the orce is applied? Solutions: a) An impulse o magnitude F t is applied in the + x direction, and the center o mass o the system will move in this direction. The two masses will rotate about the center o mass, counterclockwise in the igure. b) Beore the orce is applied we can calculate the position o the center o mass R cm M 2 ( / 2) ˆ ( / 2)( ˆ ArA + M BrB M D j+ M D j) = = = ( D / 6) ˆj. M + M 3M A B 12/28/2010 7

The center o mass is a distance (2 / 3)D rom the object at B and is a distance (1/ 3)D rom the object at A. c) Because F tiˆ = 3MVcm, the magnitude o the velocity o the center o mass is then ( F t) /( 3M) and the direction is in the positive-î direction (to the right). d) Because the orce is applied at the point B, there is no torque about the point B, hence the angular momentum is constant about the point B. The initial angular momentum about the point B is zero. The angular momentum about the point B ater the impulse is applied is the sum o two terms, 0 = L = r 3 MV + L = (2 D / 3) ˆj F" tiˆ + L 0 = (2 DF" t / 3)( # kˆ ) + L B, B, cm cm cm The angular momentum about the center o mass is given by cm L = I k = (2 M ( D / 3) + M (2 D / 3) ) k = (2 / 3) MD k cm cm ˆ 2 2 ˆ 2 ˆ 12/28/2010 8

Thus the angular about the point B ater the impulse is applied is 0 = (2 " / 3)( # ˆ) + (2 / 3) 2 DF t k MD k We can solve this equation or the angular speed ˆ. = F" t MD e) No. The orce additional orce would have to be applied at a distance 2D / 3 above the center o mass, which is not a physical point o the system. ) An additional orce o the same magnitude, in the negative x direction, would result in no net orce and hence no acceleration o the center o mass. 12/28/2010 9

27.7.2 Example Person on a railroad car moving in a circle A person o mass M is standing on a railroad car, which is rounding an unbanked turn o radius R at a speed v. His center o mass is at a height o L above the car midway between his eet, which are separated by a distance o d. The man is acing the direction o motion. What is the magnitude o the normal orce on each oot? Solution: We begin by choosing a cylindrical coordinate system and drawing a ree-body orce diagram, shown below. We decompose the contact orce between the oot closest to the center o the circular motion and the ground into a tangential component corresponding to static riction 1 and a perpendicular component, N 1. In a similar ashion we decompose the contact orce between the oot urthest rom the center o the circular motion and the ground into a tangential component corresponding to static riction 2 and a perpendicular component, N 2. We do not assume that the static riction has its maximum magnitude nor due we assume that 1 = 2 or N 1 = N 2. The gravitational orce acts at the center o mass. We shall use our two dynamical equations o motion, Eq. (27.5.1) or translational motion and Eq. (27.5.4) or the rotational motion about the center o mass noting that we are considering the special case that cm = 0 because the object is not rotating about the center o mass. 12/28/2010 10

In order to apply Eq. (27.5.1), we treat the person as a point-like particle located at the center o mass and all the eternal orces act at this point. The radial component o Newton s Second Law (Eq. (27.5.1) is given by ˆr : 1 2 = m v2 R. (27.7.1) The vertical component o Newton s Second Law is given by ˆk : N 1 + N 2 mg = 0. (27.7.2) The rotational equation o motion (Eq. (27.5.4) is total cm = 0. (27.7.3) We begin our calculation o the torques about the center o mass by noting that the gravitational orce does not contribute to the torque because it is acting at the center o mass. We draw a torque diagram in the igure below showing the location o the point o application o the orces, the point we are computing the torque about (which in this case is the center o mass), and the vector r cm,1 rom the point we are computing the torque about to the point o application o the orces. The torque on the inner oot is given by cm,1 = r cm,1 " ( 1 + N $ 1 ) = # d 2 ˆr # L ' ˆk % & ( ) " (# 1ˆr + N ˆk) = $ d 1 2 N + L ' % & 1 1 ( ) ˆ*. (27.7.4) We draw a similar torque diagram or the orces applied to the outer oot. 12/28/2010 11

The torque on the outer oot is given by cm,2 = r cm,2 " ( 2 + N $ 2 ) = + d ' ˆr # L ˆk % & 2 ( ) " (# 2ˆr + N $ ˆk) = # d 2 2 N + L ' 2 2 % & ( ) ˆ*.(27.7.5) Notice that the orces 1, N 1, and 2 all contribute torques about the center o mass in the positive ˆ -direction while N 2 contribute torques about the center o mass in the negative ˆ -direction while N 2. According to Eq. (27.7.3) the sum o these torques about the center o mass must be zero. Thereore total cm = cm,1 + " cm,2 = d 2 N + L % 1 1 # $ & ' ˆ( + " ) d 2 N + L % # $ 2 2 & ' ˆ( = d. (27.7.6) " 2 (N ) N ) + L( + ) % 1 2 1 2 # $ & ' ˆ( = 0 Notice that the magnitudes o the two riction orces appear together as a sum in Eqs. (27.7.6) and (27.7.1). We now can solve Eq. (27.7.1) or 1 + 2 and substitute the result into Eq. (27.7.6) yielding the condition that We can rewrite this equation as d 2 (N N v2 ) + Lm 1 2 R = 0. (27.7.7) N 2 N 1 = 2Lmv2 dr. (27.7.8) We also rewrite the vertical equation o motion (Eq. (27.7.2) in the orm N 2 + N 1 = mg. (27.7.9) We now can solve or N 2 by adding together Eqs. (27.7.8) and (27.7.9) and then divide by two, 12/28/2010 12

N 2 = 1 2Lmv 2 2 # " dr + mg $ & %. (27.7.10) We now can solve or N 1 by subtracting Eqs. (27.7.8) rom (27.7.9) and then divide by two, N 1 = 1 " 2 $ # mg 2Lmv2 dr % ' &. (27.7.11) Check your result: We see that the normal orce acting on the outer oot is greater in magnitude than the normal orce acting on the inner oot. We expect this result because as we increase the speed v, we ind that at a maximum speed v max, the normal orce on the inner oot goes to zero and we start to rotate in the positive ˆ -direction, tipping outward. We can ind this maximum speed by setting N 1 = 0 in Eq. (27.7.11) resulting in v max = gdr 2L. (27.7.12) 27.7.3 Example Torque, Rotation and Translation: Yo-Yo A Yo-Yo o mass m has an axle o radius b and a spool o radius R. Its moment o inertia about the center can be taken to be I cm = (1/ 2)mR 2 and the thickness o the string can be neglected. The Yo-Yo is released rom rest. You will need to assume that the center o mass o the Yo-Yo descends vertically, and that the string is vertical as it unwinds. a) What is the tension in the cord as the Yo-Yo descends? b) What is the magnitude o the angular acceleration as the yo-yo descends and the magnitude o the linear acceleration? c) Find the angular velocity o the Yo-Yo when it reaches the bottom o the string, when a length l o the string has unwound. 12/28/2010 13

Solutions: a) As the Yo-Yo descends it rotates clockwise in the above diagram. The torque about the center o mass o the Yo-Yo is due to the tension and increases the magnitude o the angular velocity (see the igure below). The direction o the torque is into the page in the igure above (positive z -direction). Use the right-hand rule to check this, or use the cross-product deinition o torque: = r " T. (27.7.13) cm cm,t About the center o mass, r cm, T = b ˆi and T = T ˆj, so the torque is cm = r cm,t " T = (#b î) " (#T ĵ) = bt ˆk. (27.7.14) Applying Newton s Second Law in the ĵ -direction, mg T = ma y (27.7.15) Applying the torque equation or the Yo-Yo: where is the z-component o the angular acceleration. bt = I cm z (27.7.16) The z-component o the angular acceleration and the y-component o the linear acceleration are related by the constraint condition a y = b z (27.7.17) where b is the axle radius o the Yo-Yo. Substitute Eq. (27.7.17) into (27.7.15) yielding 12/28/2010 14

mg T = mb" z (27.7.18) Now solve Eq. (27.7.16) or z and substitute the result into Eq.(27.7.18), Solve Eq. (27.7.19)or the tension T, mg T = mb2 T I cm (27.7.19) mg T = # 1+ mb2 " I cm $ & % mg = mb 2 $ # 1+ " (1/ 2)mR 2 & % = # " mg 1+ 2b2 R 2 $ & % (27.7.20) b) Substitute Eq. (27.7.20) into Eq. (27.7.16) to determine the z-component o the angular acceleration, z = bt 2bg = I cm (R 2 + 2b 2 ). (27.7.21) Using the constraint condition Eq. (27.7.17), we can ind the y-component o linear acceleration is then a y = b z = 2b 2 g (R 2 + 2b 2 ) = g 1+ R 2 / 2b ; (27.7.22) 2 Note that both quantities z > 0 and so Eqs. (27.7.21) and (27.7.22) are the magnitudes o the respective quantities. For a typical Yo-Yo, the acceleration is much less than that o an object in ree all. b) Use conservation o energy to determine the angular velocity o the Yo-Yo when it reaches the bottom o the string. As in the above diagram, choose the downward vertical direction as the positive ĵ -direction and let y = 0 designate the location o the center o mass o the Yo-Yo when the string is completely wound. Choose U ( y = 0) = 0 or the zero reerence potential energy. This choice o direction and reerence means that the gravitational potential energy will be negative but increasing while the Yo-Yo descends. For this case, the gravitational potential energy is U = mg y. (27.7.23) 12/28/2010 15

Mechanical energy in the initial state (Yo-Yo is completely wound): the Yo-Yo is not yet moving downward or rotating, and the center o mass is located at y = 0 so the mechanical energy is zero E i = 0. (27.7.24) Mechanical energy in the inal state (Yo-Yo is completely unwound): Denote the linear speed o the Yo-Yo as v and its angular speed as (at the point y = l ). The constraint condition between v and is given by v = b, (27.7.25) consistent with Eq. (27.7.17). The kinetic energy is the sum o translational and rotational kinetic energy, where we have used I cm = (1 / 2)mR 2, E = K + U = 1 2 mv 2 + 1 2 I cm 2 " mgl = 1 2 mb2 2 + 1 4 mr2 2 " mgl. (27.7.26) There are no external orces doing work on the system, so Thus 0 = E = E i (27.7.27) 1 2 mb2 + 1 $ " # 4 mr2 % & ' = mgl. (27.7.28) Solving or, = 4gl (2b 2 + R 2 ). (27.7.29) Note: We could also use kinematics to determine the inal angular velocity by solving or the time interval t that it takes or the Yo-Yo to travel a distance l at the constant acceleration ound in Eq. (27.7.22)), t = 2l / a y = l(r2 + 2b 2 ) b 2 g (27.7.30) 12/28/2010 16

The inal angular velocity o the Yo-Yo is then (using Eq. (27.7.21) or the z-component o the angular acceleration), = " z #t = 4gl (R 2 + 2b 2 ), (27.7.31) in agreement with Eq. (27.7.29). 27.7.4 Example Cylinder rolling down an inclined plane A uniorm cylinder o outer radius R and mass M with moment o inertia about the 2 center o mass Icm = (1/ 2) M R starts rom rest and moves down an incline tilted at an angle rom the horizontal. The center o mass o the cylinder has dropped a vertical distance h when it reaches the bottom o the incline. Let g denote the gravitational constant. The coeicient o static riction between the cylinder and the surace is µ s. The cylinder rolls without slipping down the incline. The goal o this problem is to ind the magnitude o the velocity o the center o mass o the cylinder when it reaches the bottom o the incline. Solution: We shall solve this problem three dierent ways. 1. Applying the torque equation about the center o mass and the orce equation or the center o mass motion. 2. Applying the energy equation. 3. Using torque about a ixed point that lies along the line o contact between the cylinder and the surace, Applying the torque equation about the center o mass and the orce equation or the center o mass motion 12/28/2010 17

We will irst ind the acceleration and hence the speed at the bottom o the incline using kinematics. A igure showing the orces is shown below. Choose x = 0 as the point where the cylinder just starts to roll. With the unit vectors shown in the igure above, Newton s Second Law, applied in the x - and y -directions in turn, yields Mg sin " s = Ma x, (27.7.32) N + Mg cos" = 0. (27.7.33) Choose the center o the cylinder to compute the torque about (see igure below). Then, the only orce exerting a torque about the center o mass is the riction orce, and so we have R = I. (27.7.34) s cm z Use Icm (1/ 2) = a / R in Eq. (27.7.34) to solve or the magnitude o the static riction orce yielding z x 2 = M R and the kinematic constraint or the no-slipping condition = (1/ 2) Ma. (27.7.35) s x 12/28/2010 18

Substituting Eq. (27.7.35) into Eq. (27.7.32) which we can solve or the acceleration Mg sin " (1 / 2)Ma x = Ma x (27.7.36) The displacement o the cylinder is t. The x-component o the velocity v x at the bottom is v, the time interval 2 x, / 2 x t satisies 2 ax = g sin. (27.7.37) 3 x = h / sin in the time it takes to reach the bottom, x 2 x = a t. The displacement in x x = (1/ 2) a t. Ater eliminating t, we have x = v a, so the magnitude o the velocity when the cylinder reaches the bottom o the inclined plane is v x, = 2a x x = 2((2 / 3)g sin )(h / sin ) = (4 / 3)gh. (27.7.38) Note that i we substitute Eq. (27.7.37) into Eq. (27.7.35) the magnitude o the riction orce is In order or the cylinder to roll without slipping s = (1/ 3)Mg sin. (27.7.39) s µ s Mg cos". (27.7.40) So combining Eq. (27.7.39) and Eq. (27.7.40) we have the condition that (1/ 3)Mg sin " µ s Mg cos (27.7.41) Thus in order to roll without slipping, the coeicient o static riction must satisy µ s 1 tan ". (27.7.42) 3 Applying the energy equation We shall use the act that the energy o the cylinder-earth system is constant since the static riction orce does no work. Choose a zero reerence point or potential energy at the center o mass when the cylinder reaches the bottom o the incline plane. 12/28/2010 19

Then the initial potential energy is For the given moment o inertia, the inal kinetic energy is U i = Mgh. (27.7.43) 1 2 1 2 K = M vx, + Icm z, 2 2 1 1 = M v + (1/ 2) MR ( v / R ) 2 2 3 2 = M vx, 4 2 2 2 x, x,. (27.7.44) Setting the inal kinetic energy equal to the initial gravitational potential energy leads to 3 Mgh M v x 4 2 =,. (27.7.45) The magnitude o the velocity o the center o mass o the cylinder when it reaches the bottom o the incline is in agreement with Eq. (27.7.38). v x, = (4 / 3)gh, (27.7.46) Using torque about a ixed point that lies along the line o contact between the cylinder and the surace Choose a ixed point that lies along the line o contact between the cylinder and the surace. Then the torque diagram is shown below. 12/28/2010 20

The gravitational orce M g = Mg sin î + Mg cos ĵ acts at the center o mass. The vector rom the point P to the center o mass is given by r P,mg = d P î R ĵ, so the torque due to the gravitational orce about the point P is given by P, Mg = r P, Mg " M g = (d P î # R ĵ) " ( Mg sin$ î + Mg cos$ ĵ) = (d P Mg cos$ + RMg sin$) ˆk. (27.7.47) The normal orce acts at the point o contact between the cylinder and the surace and is given by N = N ĵ. The vector rom the point P to the point o contact between the cylinder and the surace is r P,N = d P î. So the torque due to the normal orce about the point P is given by P,N = r P,N " N = (d P î) " (#N ĵ) = #d P N ˆk. (27.7.48) Substituting Eq. (27.7.33) or the normal orce into Eq. (27.7.48) yields Thereore the sum o the torques about the point P is P,N = "d P Mg cos# ˆk. (27.7.49) P = P, Mg + P,N = (d P Mg cos" + RMg sin") ˆk # d P Mg cos" ˆk = Rmg sin" ˆk.(27.7.50) The angular momentum about the point P is given by L P = L cm + r P,cm M V cm = I cm " z ˆk + (dp î # R ĵ) (Mv x ) î = (I cm " z + RMv x ) ˆk The time derivative o the angular momentum about the point P is then. (27.7.51) 12/28/2010 21

d L P dt = (I cm z + RMa x ) ˆk. (27.7.52) Thereore the torque equation becomes P = d L P, (27.7.53) dt RMgsin ˆk = (I cm " z + RMa x )ˆk. (27.7.54) Using the act that I cm = (1 / 2)MR 2 and x = ax / R, the z-component o Eq. (27.7.54) becomes RMg sin = (1/ 2) MRa + Rma = (3/ 2) MRa. (27.7.55) x x x We can now solve Eq. (27.7.55) or the x-component o the acceleration in agreement with Eq. (27.7.37). a = (2 / 3) g sin, (27.7.56) x 12/28/2010 22

27.7.5 Example Bowling Ball A bowling ball o mass m and radius R is initially thrown down an alley with an initial speed v 0, and it slides without rolling but due to riction it begins to roll. The moment o inertia o the ball about its center o mass is I cm = (2 5)mR 2. Using conservation o angular momentum about a point (you need to ind that point), ind the speed v o the bowling ball when it just start to roll without slipping? Solution: We begin by coordinates or our angular and linear motion. Choose an angular coordinate increasing in the clockwise direction. Choose positive ˆk unit vector pointing into the page. Then the angular velocity vector is deined to be a d" = z ˆk = ˆk dt and the angular acceleration vector is deined to be = z ˆk = d 2 " dt 2 ˆk. Choose the positive î unit vector pointing to the right in the igure. Then the velocity o the center o mass is given by v cm = v cm,x î = dx cm î dt and the acceleration o the center o mass is given by a cm = a cm,x î = d 2 x cm dt 2 î. The ree body orce diagram is shown in the igure below. 12/28/2010 23

At t = 0, when the ball is released, v cm,0 = v 0 î and 0 = 0, so the wheel is skidding and hence the riction orce on the wheel due to the sliding o the wheel on the surace opposes the motion is kinetic riction and hence acts in the negative î -direction. Our rule to live by or rotational motion is that S = d L S dt (57) In order or angular momentum about some point to remain constant throughout the motion, the torque about that point must also be zero throughout the motion. Recall that the torque about a point S is deine as S = # i r S,i " F i where r S,i is the vector orm the point S to where the ith orce F i acts on the object. As the ball moves down the alley, the contact point will move, but the riction orce will always be parallel to the line o contact between the bowling bowl and the surace. So, i we pick any ixed point S along the line o contact between the bowling bowl and the surace then S, k = r S, k " k = 0 because r S, k and k are anti-parallel. The gravitation orce acts at the center o mass hence the torque due to gravity about the point S is S,mg = r S,mg " m g = dmg ˆk where d is the distance rom the point S tot eh contact point between the wheel and the ground. The torque due to the normal orce about the point S is 12/28/2010 24

S,N = r S,N " m g = #dn ˆk with the same moment arm d. Because the wheel is not accelerating in the ĵ -direction, rom Newton s Second Law, we note that Thereore mg N = 0 S,N + S,mg = d(mg " N ) ˆk = 0 Hence, there is no torque about the point S and the angular momentum about the point S is constant, L S,0 = L S, (58) The initial angular momentum about the point S is only due to the translation o the center o mass, L S,0 = r S,cm,0 m v cm,0 = m Rv cm,0 ˆk. (59) The inal angular momentum about the point S has both a translational and rotational contribution L S, = r S,cm, m v cm, + I cm " = m Rv cm, ˆk + I cm " z, ˆk, (60) When the wheel is rolling without slipping, v cm, = R z, (61) 12/28/2010 25

and also I cm = (2 / 5)m R 2. Thereore the inal angular momentum about the point S is L S, = (m R + (2 / 5)m R)v cm, ˆk = (7 / 5)m Rvcm, ˆk. (62) Equating the z-components in Equations (59) and (62) yields which we can solve or m Rv cm,0 = (7 / 5)m Rv cm,, (63) v cm, = (5 / 7)v cm,0. (64) We could also solve this problem by calculating the analyzing the translational motion and the rotational motion about the center o mass. Gravity exerts no torque about the center o mass, and the normal component o the contact orce has a zero moment arm; the only orce that exerts a torque is the rictional orce, with a moment arm o R (the orce vector and the radius vector are perpendicular). The rictional orce should be in the negative direction, to the let in the igure above. From the right-hand rule, the direction o the torque is into the page, and hence in the positive z-direction. Equating the z-component o the torque to the rate o change o angular momentum, cm = R k = I cm " z, (65) where k is the magnitude o the kinetic riction orce and is the z-component o the angular acceleration o the bowling ball. Note that Equation (65) results in a positive angular acceleration, which is consistent with the ball tending to rotate as indicated in the igure. The riction orce is also the only orce in the horizontal direction, and will cause an acceleration o the center o mass, Note that the acceleration will be negative, as expected. a cm,x = k / m; (66) Now we need to consider the kinematics. The bowling ball will increase its angular speed as given in Equation (65) and decrease its linear speed as given in Equation (66); 12/28/2010 26

z (t) = " z t = R k I cm t v cm,x (t) = v cm,0 # k m t. (67) As soon as the ball stops slipping, the kinetic riction no longer acts, static riction is zero, and the ball moves with constant angular and linear velocity. Denote the time when this happens as t ; at this time, Eq. (61) holds and the relations in Equation (67) become R 2 k I cm t v cm,0 k m t = v cm, = v cm,. (68) We can now solve the irst equation in Eq. (68) or t and ind that t = I cm k R 2 v cm,. (69) We now substitute Eq. (69) into the second equation in Eq. (68) and ind that v cm, v cm, = v cm,0 k m I cm k R 2 v cm, = v cm,0 I cm m R 2 v cm, (70) The second equation in (70) is easily solved or v cm, = v 0 1+ I cm / mr 2 = 5 7 v cm,0, (71) agreeing with Eq. (64) where we have used I cm = (2 / 5)m R 2 or a uniorm sphere. 27.7.6 Example Rotation and Translation: Object and Stick Collision A long narrow uniorm stick o length l and mass m lies motionless on ice (assume the ice provides a rictionless surace). The center o mass o the stick is the same as the geometric center (at the midpoint o the stick). The moment o inertia o the stick about 12/28/2010 27

its center o mass is I cm. A puck (with putty on one side) has the same mass m as the stick. The puck slides without spinning on the ice with a speed o v 0 toward the stick, hits one end o the stick, and attaches to it. You may assume that the radius o the puck is much less than the length o the stick so that the moment o inertia o the puck about its center o mass is negligible compared to I cm. a) How ar rom the midpoint o the stick is the center o mass o the stick-puck combination ater the collision? b) What is the linear velocity o the stick plus puck ater the collision? c) Is mechanical energy conserved during the collision? Explain your reasoning. d) What is the angular velocity o the stick plus puck ater the collision? e) How ar does the stick's center o mass move during one rotation o the stick? Solution: In this problem we will calculate the center o mass o the puck-stick system ater the collision. There are no external orces or torques acting on this system so the momentum o the center o mass is constant beore and ater the collision and the angular momentum about the center o mass o the puck-stick system is constant beore and ater the collision. We shall use these relations to compute the inal angular velocity o the puckstick about the center o mass. We note that the mechanical energy is not constant because the puck collides completely inelastically with the stick. 12/28/2010 28

a) With respect to the center o the stick, the center o mass o the stick-puck combination is (neglecting the radius o the puck) d cm = m d + m d stick stick puck puck m(l / 2) = m stick + m puck m + m = l 4. (27.7.72) b) During the collision, the only net orces on the system (the stick-puck combination) are the internal orces between the stick and the puck (transmitted through the putty). Hence, linear momentum is conserved. Initially only the puck had linear momentum p 0 = mv 0. Ater the collision, the center o mass o the system is moving with speed v. Equating initial and inal linear momenta, mv 0 = (2m)v v = v 0 2. (27.7.73) The direction o the velocity is the same as the initial direction o the puck s velocity. Note that the result o part a) was not needed or part b); i the masses are the same, Equation (27.7.73) would hold or any mass distribution o the stick. c) The orces that deorm the putty do negative work (the putty is compressed somewhat), and so mechanical energy is not conserved; the collision is totally inelastic. d) Choose the center o mass o the stick-puck combination, as ound in part a), as the point about which to ind angular momentum. This choice means that ater the collision there is no angular momentum due to the translation o the center o mass. Beore the collision, the angular momentum was entirely due to the motion o the puck, L = r p = ( l / 4)( mv ) kˆ, (27.7.74) 0 puck 0 0 where ˆk is directed out o the page in the igure above. Ater the collision, the angular momentum is 12/28/2010 29

L = I cm " kˆ, (27.7.75) where I c m is the moment o inertia about the center o mass o the stick-puck combination. This moment o inertia o the stick about the new center o mass is ound rom the parallel axis theorem, and the moment o inertia o the puck is m(l / 4) 2, and so I c m = I c m, stick + I c m, puck = (I cm + m(l / 4) 2 ) + m(l / 4) 2 = I cm + ml 2 8. (27.7.76) Inserting this expression into Equation (27.7.75), equating the expressions or L 0 L and solving or yields = and m(l / 4) I cm + ml 2 / 8 v 0. (27.7.77) I the stick is uniorm, I cm = ml 2 / 12 and Equation (27.7.77) reduces to = 6 5 v 0 l. (27.7.78) It may be tempting to try to calculate angular momentum about the contact point, where the putty hits the stick. I this is done, there is no initial angular momentum, and ater the collision the angular momentum will be the sum o two parts, the angular momentum o the center o mass o the stick and the angular moment about the center o the stick, L = r p + I ". (27.7.79) cm cm cm There are two crucial things to note: First, the speed o the center o mass is not the speed ound in part b); the rotation must be included, so that v cm = v 0 / 2 " (l / 4). Second, the direction o r cm p cm with respect to the contact point is, rom the right-hand rule, into the page, or the kˆ -direction, opposite the direction o. This is to be expected, as the sum in Equation (27.7.79) must be zero. Adding the ˆk -components (the only components) in Equation (27.7.79), (l / 2)m(v 0 / 2 " (l / 4)) + I cm " = 0. (27.7.80) Solving Equation (27.7.80) or yields Equation (27.7.77). 12/28/2010 30

This alternative derivation should serve two purposes. One is that it doesn t matter which point we use to ind angular momentum. The second is that use o oresight, in this case choosing the center o mass o the system so that the inal velocity does not contribute to the angular momentum, can prevent extra calculation. It s oten a matter o trial and error ( learning by misadventure ) to ind the best way to solve a problem. e) The time o one rotation will be the same or all observers, independent o choice o origin. This act is crucial in solving problems, in that the angular velocity will be the same (this was used in the alternate derivation or part d) above). The time or one rotation is the period T = 2 / " and the distance the center o mass moves is x cm = v cm T = 2 v cm " v = 2 0 / 2 # m(l / 4) & % $ I cm + ml 2 ( / 8' v 0 = 2 I + ml 2 / 8 cm. m(l / 2) Using I cm = ml 2 / 12 or a uniorm stick gives (27.7.81) x cm = 5 6 l. (27.7.82) 12/28/2010 31

MIT OpenCourseWare http://ocw.mit.edu 8.01SC Physics I: Classical Mechanics For inormation about citing these materials or our Terms o Use, visit: http://ocw.mit.edu/terms. 12/28/2010 32