Rolling, Torque, and Angular Momentum

AP Physics C Rolling, Torque, and Angular Momentum Introduction: Rolling: In the last unit we studied the rotation of a rigid body about a fixed axis. We will now extend our study to include cases where the axis of rotation moves, that is, where one body experiences both rotational and translational motion at the same time. Common examples would be a yo-yo unwinding down a string or a ball rolling across the floor. Newton s Second Law of Rotation is still valid if the axis of rotation passes through the center of mass of the body and does not change its direction. Another useful concept to employ when looking at a body rotating about a moving axis is the total kinetic energy of the body. The combined kinetic energy for a body of mass m moving with a center-of-mass velocity v and rotating with angular velocity ω about an axis through the center of mass, the total kinetic energy of the body is:...where I com is the moment of inertia about the axis through the center of mass. Torque: In the previous unit we defined torque qualitatively as the tendency of a force to cause the rotation of a rigid body on which the force acts. All the particles moved in circular paths around a fixed axis. Mathematically we defined torque as the cross product of the moment or lever arm and the applied force, where the moment arm is the perpendicular distance between where the force is applied and the axis of rotation. In this unit we will expand the definition of torque to apply to a particle that moves along any path relative to a fixed point. We will explore the vector nature of torque and consider the cross product of the position vector and the force vector. Angular Momentum: Every rotational quantity that we have seen in the previous unit is the analog of some quantity of motion of a particle. The analog of momentum of a particle is angular momentum, a vector quantity denoted by L. Its relation to p, the linear momentum, is exactly the same as the relation of torque to force. The rate of change of angular momentum of a particle equals the torque of the net force acting on it. The direction of the angular momentum is given by the right-hand rule for angular velocity. The angular momentum for a system of particles is the sum of all the angular momenta of the particles. The angular momentum for a solid body is given by L = Iω. Angular momentum also forms the basis for a very important conservation principle, the principle of conservation of angular momentum. Like the conservation laws of energy and of linear momentum, this principle appears to be a universal conservation law, valid at all scales from atomic and nuclear systems to the motions of galaxies. The total angular momentum of an isolated system is constant. Performance Objectives: Upon completion of the readings and activities of the unit and when asked to respond either orally or on a written test, you will be able to: describe the rolling motion of an object as the rotational motion about an axis through the center of mass of the object while the center of mass moves in linear motion. know that the linear speed of a point on the rim of a rolling object varies from zero at the point of contact to twice the speed of the center of mass. explain that static friction provides the torque to rotate the moving object. Calculate the torque using the coefficient of static friction, the normal force and the torque arm. solve problems involving forces acting on a rolling object. show that the kinetic energy for a rolling object is the sum of the linear and angular kinetic energies of the object. Solve problems using conservation of energy. know and explain that for a body rolling down an incline the speed at the bottom depends upon a dimension less quantity β, the coefficient in the rotational inertia equation. not I, m, or R alone. define angular momentum and show that it is analogous to the definition for linear momentum. understand and explain that a particle has angular momentum relative to a point at the origin if a particle is moving with constant angular velocity in a circular path around the origin or if the particle is moving in a straight line with constant linear speed relative to the origin. determine the angular momentum for a rigid body about a fixed axis and for a system of particles. state the law of conservation of angular momentum. Use the law to solve problems for a rigid body and for a system of particles. Textbook Reference: Tipler: Chapter 9, Section 6 and Chapter 10 Glencoe Physics: Chapter 8 Problem Solving Strategies for Rotation About a Moving Axis: Recall that when Newton s Second Law was used to solve problems, an equation of motion (F net = F applied F opposing or τ net = τ applied τ opposing) was needed for each moving body. Now one body is moving with two motions so two equations will be needed, one for translational motion and one for rotational motion. Energy methods can also be used to solve problems involving the rolling motion of a rigid body. If the

body rolls without slipping, mechanical energy is conserved and U i + K i = U f + K f. Although there is no rotational potential energy, the potential energy of the center of mass of the rolling body is considered. Remember that a rolling body has both translational and rotational kinetic energy. 1. A string is wrapped several times around the rim of a small hoop. The hoop has radius 0.08 m and mass 1.2 kg. If the free end of the string is held in place and the hoop is released from rest, calculate a) the tension in the string while the hoop is descending; 5.88 N b) the time it takes the hoop to descend 0.5 m; 0.452 s c) the angular velocity of the rotating hoop after it has descended 0.5 m. 27.7 rad/s 2. A string is wrapped around a 3.50 kg solid disk of 0.300 meter radius. If the disk is released from rest, what is its linear acceleration downward? 6.53 m/s 2 3. A string is wrapped several times around a solid cylinder of mass 6.00 kg and radius 0.150 m. The end of the string is held stationary while the cylinder is released from rest. Find the downward linear acceleration of the cylinder and the tension in the string. 6.53 m/s 2 19.6 N 4. A string is wound around a uniform solid disk of radius R and mass M. The disk is released from rest with the string vertical and its top end tied to a fixed support as shown in the figure. As the disk descends, find: (a) the tension in the string; (b) the acceleration of the center of mass, and (c) the velocity of the center of mass. T = mg/3 a = 2g/3 v = (4gh/3) ½. (Did you notice that the linear acceleration downward for the last three problems was 2g/3? This is true for all solid disks or solid cylinders that ever were and ever will be.) The Yo-Yo: 5. A yo-yo consists of two disks of total mass M and radius R connected by a massless shaft of radius r. What is the liner acceleration of the yo-yo when released? 6. A yo-yo has a rotational inertia of 950 g-cm 2 and a mass of 120 g. Its axle radius is 3.2 mm, and its string is120 cm long. The yo-yo rolls from rest down to the end of the string. a) What is the magnitude of its linear acceleration? 0.13 m/s 2 b) How long does it take to reach the end of the string? 4.4 s As it reaches the end of the string, what are its c) linear speed, 0.55 m/s d) translational kinetic energy, 0.18 J e) rotational kinetic energy, 1.4 J f) angular speed? 170 rad/s 7. Suppose that the yo-yo in the previous problem, instead of rolling from rest, is thrown so that its initial speed down the string is 1.3 m/s. a) How long does the yo-yo take to reach the end of the string? 0.88 s As it reaches the end of the string, what are its b) total kinetic energy, 9.35 J c) linear speed, 1.41 m/s d) translational kinetic energy, 0.12 J e) angular speed, 441 rad/s f) rotational kinetic energy? 9.22 J Rolling Motion of a Rigid Body: When a body rolls, it experiences rotational motion about a moving axis. If a cylinder rolls in a straight path, the center of mass moves in a straight line while the rest of the body experiences rotational motion about the axis through the center of mass. It is convenient to look at this motion as a combination of rotation about the center of mass and the translation of the center of mass. Rolling motion is only possible if a frictional force is present between the object and the surface. The frictional force (static) is necessary to produce a net torque about the center of mass. Translational Motion + Rotational Motion = Rolling Motion Despite the presence of friction, there is no loss of mechanical energy since the contact point is at rest relative to the surface at any instant. On the other hand, if the rigid body were to slide, mechanical energy would be lost as motion progressed. Mechanical energy is conserved if a body rolls without slipping. Rolling without slipping means that: v com = Rω and a com = Rα

If an object is set into motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. Another way to look at rolling without slipping is from the point of view of a stationary observer watching a bug on the rim of a rolling bicycle wheel. The stationary observer sees the bug rotate about a stationary axis at point P. The angular speed ω of the bug about point P is the same as the angular speed of the wheel about the axis at the center of mass. The instantaneous linear velocity vectors are in a direction perpendicular to the radius drawn from the bug to the contact point P. At any instant, the point P is at rest relative to the surface since sliding does not occur. The bug on the rim moves in a complex path, called a cycloid. The linear speed of the bug varies from zero at the point of contact at the bottom of the wheel to v com = Rω at the top of the wheel. The total kinetic energy of a rolling body is the sum of the rotational kinetic energy about the center of mass, and the translational kinetic energy of the center of mass. Rolling Without Slipping Problems 8. A solid cylinder of mass 10 kg rolls without slipping on a rough surface. At the instant its center of mass has a speed of 10 m/s, determine a) the translational kinetic energy of its center of mass, b) the rotational kinetic energy about its center of mass, c) its total kinetic energy. d) What fraction of the kinetic energy is associated with the motion of translation and what fraction with the motion of rotation about the axis through the center of mass? 500 J 250 J 750 J 67 % 33 % 9. A solid sphere has a radius of 0.2 m and a mass of 150 kg. How much work is required to get the sphere rolling with an angular speed of 50 rad/s on a horizontal surface (Assume the sphere starts from rest and rolls without slipping.) 10,500 J 10. The center of mass of a uniform cylinder of mass M and radius R is moving with speed v on a horizontal surface. The cylinder rolls without slipping. Find the total kinetic energy of the cylinder with respect to the reference frame fixed to the horizontal surface. 0.750 Mv 2 11. A bowling ball has a mass of 8.00 kg. If it rolls down the alley, without slipping, at 7.00 m/s, calculate a) the linear kinetic energy; 196 J b) the rotational kinetic energy; 78.4 J c) the total kinetic energy. 274.4 J 12. A metal ring, mass 2.50 kg, rolls along a horizontal surface with a constant velocity of 5.25 m/s. What is the total kinetic energy of the moving ring? 69 J 13. A hoop of radius 3.0 m has a mass of 140 kg. It rolls without slipping along a horizontal floor so that its center of mass has a speed of 0.15 m/s. How much work must be done on the hoop to stop it? -3.15 J Things begin to get tricky when we look at objects rolling down inclines. You can save yourself a lot of wasted effort if you spend some time thinking about what happens when you release a basketball down a ramp. If the ramp were frictionless, the ball would just slip along and not rotate. We know that doesn't happen, so the ramp must apply a frictional force to the ball (and vice-versa). 14. The figure below shows a hollow cylinder rolling down a ramp inclined at an angle θ. a) Label the forces acting on the ball. b) Which friction force (static or kinetic) is acting on the cylinder as it rolls down the incline? c) Write a translational and a rotational equation of motion for the rolling cylinder. d) Solve for the linear acceleration down the ramp. 15. A solid cylinder of mass 4.00 kg rolls, without slipping down a 30 slope. Find (a) the acceleration, (b) the friction force, and (c) the minimum coefficient of friction needed to prevent slipping. 3.26 m/s 2 6.5 N 0.19 16. A solid sphere released at a height h above the horizontal rolls down an incline that makes an angle θ with the horizontal. a) Calculate the velocity of its center of mass at the bottom. v com = (10gh/7)½ b) Determine the linear acceleration of the center of mass of the sphere a com = (5g sinθ)/7

17. a) Determine the acceleration of the center of mass of a uniform solid disk rolling down an incline and compare this acceleration with that of a uniform hoop. b) What is the minimum coefficient of friction required to maintain pure rolling motion for the disk? It is interesting to note that both the velocity and acceleration of the center of mass are independent of the mass and radius. If the calculations were repeated for any rolling object, the acceleration of the center of mass would be less than (g sin θ), the value it would have been if the plane were frictionless and no rolling occurred. a com = ⅔ g sinθ (disk) a com = ½ g sinθ (hoop) ⅓ tanθ 18. A uniform solid disk and a uniform hoop are placed side by side at the top of a rough incline of height h. a) If they are released from rest and roll without slipping, determine their velocities when they reach the bottom. v d = (4gh/3) ½ v h = (gh) ½ b) Which object reaches the bottom first? disk 19. A uniform solid cylinder of mass M and radius 2R rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a pulley in the shape of a disk of mass M and radius R that is mounted on a frictionless axle through its center. A block of mass M is suspended from the free end of the string. The string doesn t slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. After the system is released from rest, what is the magnitude of the downward acceleration of the block? g/3 20. Using the ideas of torque and center of gravity, explain why a ball rolls down a hill. When a ball is on an incline its center of gravity is not above the point of support. The weight acts some distance from the point of support and produces a torque about the point of support. 21. A 2.0 kg disk starts at the top of a ramp 5.0 m high and rolls down to the bottom. How fast is it moving at the bottom of the ramp on the level plane? 8.08 m/s 22. A 0.47 kg sphere of 5.2 cm radius rolls at 881 cm/sec at the bottom of the hill. It rolls without slipping up the hill. How fast is it rolling on the plateau at the top of the hill? 7.04 m/s Rolling With Slipping When an object initially slides before it rolls, the nonslip condition v com = rω does not hold. Kinetic friction due to the slipping reduces the linear speed v com and increases the angular speed ω until the nonslip condition v com = rω is reached. When the not slip condition is reached the object will continue to roll without slipping. You may have observed a bowling ball released without angular velocity. It initially skids along the alley until it reaches the rolling without slipping condition and then it rolls the rest of the way down the alley. 23. A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane, with initial speed v com, 0 = 8.5 m/s and initial angular speed ω 0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.21. The kinetic frictional force acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. a) When speed v com has decreased enough and angular speed ω had increased enough, the ball stops sliding and then rolls smoothly. What then is v com in terms of ω? b) During the sliding, what is the ball s linear acceleration? 2.1 m/s 2 c) During the sliding, what is the ball s angular acceleration? 47 rad/s 2 d) How long does the ball slide? 1.2 sec e) How far does the ball slide? 8.6 m f) What is the speed of the ball when smooth rolling begins? 6.1 m/s 24. A bowling ball is given an initial speed v 0 on an alley such that it initially slides without rolling. The coefficient of friction between the ball and the alley is μ. Show that at the time rolling without slipping occurs (a) the velocity of the ball s center of mass is 5v 0/7 and (b) the distance it has traveled is 12v 02/49μg. Torque Revisited: Up to this point we calculated the torque for a rigid body that rotated about a fixed axis. All the points in that body were moving in a circle about the axis with a moment or torque arm equal to the radius of the circle. The same definition of torque can be applied to an individual particle that moves along any path relative to a fixed point (rather than a fixed axis). The position of the particle is given by the position vector r. The fixed point could be the origin or some other coordinate. 25. The position vector r of a particle points along the positive direction of the z axis. In what direction is the force producing the torque, if the torque on the particle is (a) zero, (b) in the negative x direction, and (c) in the negative y direction? ± z + y x 26. Show that, if r and F lie in a given plane, the torque τ = r F has no component in that plane.

27. What is the magnitude and direction of the torque about the origin on a plum located at coordinates ( 2.0 m,0, 4.0 m) due to a force F whose only component is a) F x = 6.0 N b) F x = 6.0 N c) F z = 6.0 N d) F z = 6.0 N 24j m-n -24j m-n 12j m-n -12j m-n 28. What is the magnitude and direction of the torque about the origin on a particle located at coordinates (0. 4.0 m, 3.0 m) due to a) F 1 = (2.0 N, 0, 0) 10 m-n 53 wrt y-axis b) F 2 = (0, 2.0 N, 4.0 N) 22 m-n in negative x direction 29. The figure below shows two particles A and B at xyz coordinates (1 m, 1 m, 0) and (1 m, 0, 1 m). Acting on each particle are three numbered forces, all of the same magnitude and each directed parallel to an axis. a) force F 1 = (3.0 N)i (4.0 N)j + (6.0 N)k? (4, 6, 6) m-n b) force F 2 = (-3.0 N)i (4.0 N)j (5.0 N)k? (26, 3, 18) m-n c) the vector sum F 1 + F 2? (30, 3, 24) m-n d) Repeat part (c) about a point with coordinates (3.0 m, 2.0 m, 4.0 m) instead of about the origin. 4 m-n Angular Momentum and Angular Impulse: The angular momentum of a rigid body rotating about a fixed axis is defined as L = Iω. For a single particle the angular momentum relative to any point would be: L = r x p or L = m(r x v)...where m is the mass of the particle, r is the position vector from the point to the particle and v is the translational velocity. The product of the torque and the time interval during which it acts is called the angular impulse, J θ. The angular impulse acting on the body causes a change in the angular momentum of the body about the same axis. For a torque that varies with time, the angular impulse is defined as: a) Which of the forces produce a torque about the origin (0, 0, 0) that is directed parallel to y? 5 and 6 b) Rank the forces according to the magnitudes of the torques they produce on the particles about the origin, greatest first. 1and 4 tie, then all the rest tie. 30. A force: F = (2 N)i (3N)j acts on a particle whose position vector relative to the origin is: r 0 = (-0.3 m)i + (0.5 m)j a) Calculate the torque about the origin produced by this force. ( 0.10 m-n) k b) What is the angle between the directions of vectors F and r? 2.7 If the position of the particle is given relative to the origin and you are asked to calculate the torque about a different coordinate point, you have to calculate a new position vector relative to that point: r = r o p, where r is the new position vector, r o is the position relative to the origin, and the torque is to be calculated about p. 31. A force: F = (4 N)i + (5 N)k acts on a particle whose position vector relative to the origin is r 0 = (0.3 m)i (0.5 m) j. Calculate the torque produced by this force a) about the origin. -25i 15j + 2k m-n b) about a point p = (2.0m, 0, 3.0m). 2.5i + 20.5j + 2k 32. What is the torque about the origin on a pebble located at coordinates (3.0 m, 2.0 m, 4.0 m) due to 33. Calculate the angular momentum of a uniform sphere of radius 0.20 m and mass 4.0 kg if it is rotating about an axis along a diameter at 6.0 rad/s. 0.384 kg-m 2 /sec 34. What is the angular momentum of the hour hand on a clock, about an axis through the center of the clock face, if the clock hand has a length of 25.0 cm and a mass of 20.0g? Take the hour hand to be a slender rod rotating about one end. 6.06 x 10-8 kg-m 2 /sec 35. A solid wooden door 1.0 m wide and 2.0 m high is hinged along one side and has a total mass of 50.0 kg. Initially open and at rest, the door is struck at its center with a hammer. During the blow, an average force of 2000.0 N acts for 0.01 seconds. Find the angular velocity of the door after the impact. 0.60 rad/s 36. A man of mass 70.0 kg is standing on the rim of a large disk that is rotating at 0.500 rev/s about an axis through its center. The disk has mass 120.0 kg and radius 4.00 m. Calculate the total angular momentum of the man-plus-disk system. 6530 kg-m 2 /s 37. The figure shows a rigid hoop of radius R and mass m, a square made off our thin bars each of length R and mass m. The rigid structure rotates at a constant speed about a vertical axis with a period of rotation of 2.5 sec. Assuming R = 0.50 m and m= 2.0 kg, calculate (a) the structure's rotational inertia about the axis of rotation and (b) its angular momentum about that axis.

38. A rock of mass 2.0 kg has a constant velocity of 12m/s. When it is at point P as shown in the figure, what is its angular momentum relative to point 0? (Remember constant velocity means no change in speed or direction.) 116 kg-m 2 /s a) Rank the particles according to the magnitudes of their angular momentum about point O, greatest first. 1 and 3 tie, next 2 and 4 tie, then 5 b) Which particles have negative angular momentum about point O? 39. Two objects are moving as shown in the figure to the right. What is their total angular momentum about point 0? 9.8 kg-m 2 /s 40. In the figure to the right, a particle P with mass 2.0 kg has position vector r of magnitude 3.0 m and velocity v of magnitude 4.0 m/s. A force F of magnitude 2.0 N acts on the particle. All three vectors lie in the xy plane oriented as shown. About the origin, what are(a) the angular momentum of the particle and (b) the torque acting on the particle? 1.2 kg-m/s 3.0 m-n 41. At a certain time, a 0.25 kg object has a position vector r = 2.0i - 2.0k, in meters. At that instant, its velocity in meters per second is v = 5.0i + 5.0k, and the force in newtons acting on it is F = 4.0j. a) What is the angular momentum of the object about the origin? 0 b) What torque acts on it? 42. A 4.0 kg particle moves in an xy-plane. At the instant when the particle's position and velocity are r = (2.0i + 4.0j) m and v = 4.0j m/s, the force on the particle is F = 3.0i N. At this instant, determine a) the particle's angular momentum about the origin, b) the particle's angular momentum about the point (0, 4) c) the torque acting on the particle about the origin, and d) the torque acting on the particle about the point (0, 4) 32 kg-m 2 /s k + 32 kg-m 2 /s k 12 m-n k 0 43. In part a of the figure below, particles 1 and 2 move around point O in opposite directions, in circles with radii 2 m and 4 m. In part b, particles 3 and 4 travel in the same direction, along straight lines at perpendicular distances of 4 m and 2 m from point O. Particle 5 moves directly away from O. All five particles have the same mass and the same constant speed. Conservation of Angular Momentum: Conservation of angular momentum states that when the net external torque on a system is zero, the angular momentum of the system remains constant. This principle of conservation of angular momentum ranks with the principles of conservation of linear momentum and conservation of energy as one of the most fundamental of physical laws. 44. On an old-fashioned rotating piano stool, a man sits holding a pair of dumbbells at a distance of 0.60 m from the axis of rotation of the stool. He is given an angular velocity of 5.0 rad/s, after which he pulls the dumbbells in until they are only 0.2 m distant from the axis. The man's moment of inertia about the axis of rotation is 5.0 kg-m 2 and may be considered constant. Each dumbbell has a mass of 5.0 kg and may be considered a point mass. Friction is negligible. a) What is the initial angular momentum of the system? 43.0 kg-m 2 /s b) What is the angular velocity of the system after the dumbbells are pulled in toward the axis? 7.96 rad/s c) Compute the kinetic energy of the system before and after the dumbbells are pulled in. Account for the difference, if any. 108 J 171 J 45. A puck on a frictionless air-hockey table has a mass of 0.05kg and is attached to a cord passing through a hole in the table surface, as in the figure. The puck is originally revolving at a distance of 0.20 m from the hole, with an angular velocity of 3.0 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the puck revolves to 0.10 m. The puck may be considered a point mass. a) What is the new angular velocity? 12 rad/s b) Find the change in kinetic energy of the puck. 0.027 J c) How much work was done by the person who pulled the cord?

46. A turntable rotates about a fixed vertical axis, making one revolution in 10.0 s. The moment of inertia of the turntable about this axis is 1200 kg-m 2. A man of mass 80.0 kg, initially standing at the center of the turntable, runs out along a radius. What is the angular velocity of the turntable when the man is 2.0 m from the center? 0.496 rad/s 47. In models of stellar evolution, one scenario is for stars to undergo sudden gravitational collapse in which the radius of the star decreases drastically while most of the star's mass is retained. Consider the hypothetical collapse of our sun which has a mass of 1.99 x 10 30 kg and a mean radius of 6.96 x 10 8 m. The sun, a ball of gas does not rotate as a rigid body, however for this problem use the present period for equatorial rotation on its axis of about 26 days. If the sun undergoes gravitational collapse to a new radius of 32 km (about 20 miles) while keeping its present mass, calculate its new revolution rate in rev/s. (This illustrates one possible model for the formation of pulsars.) 211 rev/s 48. A large wooden turntable of radius 2.0 m and total mass 120 kg is rotating about a vertical axis through its center, with an angular velocity of 3.0 rad/s. From a very small height a sandbag of mass 100 kg is dropped vertically onto the turntable, at a point near the outer edge. a) Find the angular velocity of the turntable after the sandbag is dropped. 1.12 rad/s b) Compute the kinetic energy of the system before and after the sandbag is dropped. before: 1080 J after: 405 J c) Why are these kinetic energies not equal? 49. A solid wooden door 1.0 m wide and 2.0 m high is hinged along one side and has a total mass of 50.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud of mass 0.5 kg, traveling 10 m/s just before impact. Find the final angular velocity of the door. Is the moment of inertia of the mud significant? 0.149 rad/s 50. The outstretched arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When her arms are brought in and wrapped around her body to execute the spin, they can be considered a thin-walled hollow cylinder. If her original angular velocity is 6.28 rad/s, what is her final angular velocity? Her arms have a combined mass of 8.0 kg. When outstretched they span 1.8 m; when wrapped, they form a cylinder of radius 25 cm. (A disk and a cylinder rotating about an axis through the center have the same moment of inertia.) The moment of inertia of the remainder of her body is constant and equal to 3.0 kg-m 2. 9.26 rad/s 51. A uniform thin rod of length 0.50 m and mass 4.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.0 g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle of 60º with the rod as shown in the figure. If the bullet lodges in the rod and the angular velocity of the rod is 10 rad/s immediately after the collision, what is the bullet's speed just before impact? 1300 m/s 52. Two 2.00 kg balls are attached to the ends of a thin rod of negligible mass, 50.0 cm long. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal, as shown is the figure below, a 50.0 g wad of wet putty drops onto one of the balls, hitting it with a speed of 3.00 m/s and then sticking to it. a) What is the angular speed of the system just after the putty wad hits? 0.148 rad/s b) What is the ratio of the kinetic energy of the entire system after the collision to that of the putty wad just before? 0.0123 c) Through what angle will the system rotate until it momentarily stops? 181 Object Rotational Inertia I com Percentage of Energy in Translation Rotation Hoop MR 2 50% 50% Disk 1/2 MR 2 67% 33% Sphere 2/5 MR 2 71% 29% General βmr 2 Where β is the coefficient in the rotational inertia equation