6 raymond a serway, john w jewett physics for scientists and engineers with modern physics 13

264 Chapter Linear Momentum and Collisions higher than if his back were straight As a model, consider the jumper as a thin, uniform rod of length L When the rod is straight, its center of mass is at its center Now bend the rod in a circular arc so that it subtends an angle of 90.0° at the center of the arc as shown in Figure P9.40b In this configuration, how far outside the rod is the center of mass? Section 9.6 Motion of a System of Particles 41 A 2.00-kg particle has a velocity 12.00ˆi Ϫ 3.00ˆj m/s, and a 3.00-kg particle has a velocity 11.00ˆi ϩ 6.00ˆj m/s Find (a) the velocity of the center of mass and (b) the total momentum of the system 42 The vector position of a 3.50-g particle moving in the xy S plane varies in time according to r ϭ 13ˆi ϩ 3ˆj 2t ϩ 2ˆj t At the same time, the vector position of a 5.50-g particle S varies as r ϭ 3ˆi Ϫ 2ˆi t Ϫ 6ˆj t, where t is in s and r is in cm At t ϭ 2.50 s, determine (a) the vector position of the center of mass, (b) the linear momentum of the system, (c) the velocity of the center of mass, (d) the acceleration of the center of mass, and (e) the net force exerted on the two-particle system 43 Romeo (77.0 kg) entertains Juliet (55.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet, who is in the front of the boat After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo’s cheek How far does the 80.0-kg boat move toward the shore it is facing? 44 A ball of mass 0.200 kg has a velocity of 1.50ˆi m/s; a ball of mass 0.300 kg has a velocity of Ϫ0.400ˆi m/s They meet in a head-on elastic collision (a) Find their velocities after the collision (b) Find the velocity of their center of mass before and after the collision Section 9.7 Deformable Systems 45 ⅷ For a technology project, a student has built a vehicle, of total mass 6.00 kg, that moves itself As shown in Figure P9.45, it runs on two light caterpillar tracks that pass around four light wheels A reel is attached to one of the axles, and a cord originally wound on the reel passes over a pulley attached to the vehicle to support an elevated load After the vehicle is released from rest, the load descends slowly, unwinding the cord to turn the axle and make the vehicle move forward Friction is negligible in the pulley and axle bearings The caterpillar tread does not slip on the wheels or the floor The reel has a conical shape so that the load descends at a constant low speed = challenging; vi L u (b) (a) Figure P9.47 48 On a horizontal air track, a glider of mass m carries a ⌫shaped post The post supports a small dense sphere, also of mass m, hanging just above the top of the glider on a cord of length L The glider and sphere are initially at rest with the cord vertical (Fig P9.47a shows a cart and a sphere similarly connected.) A constant horizontal force of magnitude F is applied to the glider, moving it through displacement x1; then the force is removed During the time interval when the force is applied, the sphere moves through a displacement with horizontal component x2 (a) Find the horizontal component of the velocity of the center of mass of the glider-sphere system when the force is removed (b) After the force is removed, the glider con- Figure P9.45 = intermediate; while the vehicle moves horizontally across the floor with constant acceleration, reaching final velocity 3.00ˆi m/s (a) Does the floor impart impulse to the vehicle? If so, how much? (b) Does the floor work on the vehicle? If so, how much? (c) Does it make sense to say that the final momentum of the vehicle came from the floor? If not, from where? (d) Does it make sense to say that the final kinetic energy of the vehicle came from the floor? If not, from where? (e) Can we say that one particular force causes the forward acceleration of the vehicle? What does cause it? 46 ⅷ A 60.0-kg person bends his knees and then jumps straight up After his feet leave the floor his motion is unaffected by air resistance and his center of mass rises by a maximum of 15.0 cm Model the floor as completely solid and motionless (a) Does the floor impart impulse to the person? (b) Does the floor work on the person? (c) With what momentum does the person leave the floor? (d) Does it make sense to say that this momentum came from the floor? Explain (e) With what kinetic energy does the person leave the floor? (f) Does it make sense to say that this energy came from the floor? Explain 47 ⅷ A particle is suspended from a post on top of a cart by a light string of length L as shown in Figure P9.47a The cart and particle are initially moving to the right at constant speed vi, with the string vertical The cart suddenly comes to rest when it runs into and sticks to a bumper as shown in Figure P9.47b The suspended particle swings through an angle u (a) Show that the original speed of the cart can be computed from vi ϭ 12gL 11 Ϫ cos u2 (b) Find the initial speed implied by L ϭ 1.20 m and u ϭ 35.0° (c) Is the bumper still exerting a horizontal force on the cart when the hanging particle is at its maximum angle from the vertical? At what moment in the observable motion does the bumper stop exerting a horizontal force on the cart? Ⅺ = SSM/SG; ᮡ = ThomsonNOW; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning Problems tinues to move on the track and the sphere swings back and forth, both without friction Find an expression for the largest angle the cord makes with the vertical 49 ⅷ Sand from a stationary hopper falls onto a moving conveyor belt at the rate of 5.00 kg/s as shown in Figure P9.49 The conveyor belt is supported by frictionless rollers It moves at a constant speed of 0.750 m/sS under the action of a constant horizontal external force Fext supplied by the motor that drives the belt Find (a) the sand’s rate of change of momentum in the horizontal direction, (b) the force of frictionS exerted by the belt on theSsand, (c) the external force Fext, (d) the work done by Fext in s, and (e) the kinetic energy acquired by the falling sand each second due to the change in its horizontal motion (f) Why are the answers to (d) and (e) different? 0.750 m/s a 1t ϭ 265 ve Tp Ϫ t (d) Graph the acceleration as a function of time (e) Show that the position of the rocket is x 1t ϭ v e 1Tp Ϫ t 2ln a Ϫ t b ϩ ve t Tp (f) Graph the position during the burn as a function of time 53 ⅷ A rocket for use in deep space is to be capable of boosting a total load (payload plus rocket frame and engine) of 3.00 metric tons to a speed of 10 000 m/s (a) It has an engine and fuel designed to produce an exhaust speed of 000 m/s How much fuel plus oxidizer is required? (b) If a different fuel and engine design could give an exhaust speed of 000 m/s, what amount of fuel and oxidizer would be required for the same task? This exhaust speed is 2.50 times higher than that in part (a) Explain why the required fuel mass is 2.50 times smaller, or larger than that, or still smaller Fext Figure P9.49 Section 9.8 Rocket Propulsion 50 Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics A size C5 model rocket engine has an average thrust of 5.26 N, a fuel mass of 12.7 g, and an initial mass of 25.5 g The duration of its burn is 1.90 s (a) What is the average exhaust speed of the engine? (b) This engine is placed in a rocket body of mass 53.5 g What is the final velocity of the rocket if it is fired in outer space? Assume the fuel burns at a constant rate 51 ᮡ The first stage of a Saturn V space vehicle consumed fuel and oxidizer at the rate of 1.50 ϫ 104 kg/s, with an exhaust speed of 2.60 ϫ 103 m/s (a) Calculate the thrust produced by this engine (b) Find the acceleration the vehicle had just as it lifted off the launch pad on the Earth, taking the vehicle’s initial mass as 3.00 ϫ 106 kg Note: You must include the gravitational force to solve part (b) 52 Rocket science A rocket has total mass Mi ϭ 360 kg, including 330 kg of fuel and oxidizer In interstellar space, it starts from rest at the position x ϭ 0, turns on its engine at time t ϭ 0, and puts out exhaust with relative speed ve ϭ 500 m/s at the constant rate k ϭ 2.50 kg/s The fuel will last for an actual burn time of 330 kg/(2.5 kg/s) ϭ 132 s, but define a “projected depletion time” as Tp ϭ Mi/k ϭ 360 kg/(2.5 kg/s) ϭ 144 s (which would be the burn time if the rocket could use its payload and fuel tanks, and even the walls of the combustion chamber as fuel) (a) Show that during the burn the velocity of the rocket as a function of time is given by v 1t ϭ Ϫv e ln a Ϫ t b Tp (b) Make a graph of the velocity of the rocket as a function of time for times running from to 132 s (c) Show that the acceleration of the rocket is = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ Additional Problems 54 Two gliders are set in motion on an air track A spring of force constant k is attached to the back end of the second S glider The first glider, of mass m1, has velocity v1, and the second glider, of mass m2, moves more slowly, with velocity S v2, as shown in Figure P9.54 When m1 collides with the spring attached to m2 and compresses the spring to its maximum compression xmax, the velocity of the gliders is S S S S v In terms of v1, v2, m1, m2, and k, find (a) the velocity v at maximum compression, (b) the maximum compression xmax, and (c) the velocity of each glider after m1 has lost contact with the spring v2 k m2 v1 m1 Figure P9.54 55 An 80.0-kg astronaut is taking a space walk to work on the engines of his ship, which is drifting through space with a constant velocity The astronaut, wishing to get a better view of the Universe, pushes against the ship and much later finds himself 30.0 m behind the ship Without a thruster or tether, the only way to return to the ship is to throw his 0.500-kg wrench directly away from the ship If he throws the wrench with a speed of 20.0 m/s relative to the ship, after what time interval does the astronaut reach the ship? 56 ⅷ An aging Hollywood actor (mass 80.0 kg) has been cloned, but the genetic replica is far from perfect The clone has a different mass m, his stage presence is poor, = ThomsonNOW; Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning 266 Chapter Linear Momentum and Collisions and he uses foul language The clone, serving as the actor’s stunt double, stands on the brink of a cliff 36.0 m high, next to a sturdy tree The actor stands on top of a Humvee, 1.80 m above the level ground, holding a taut rope tied to a tree branch directly above the clone When the director calls “action,” the actor starts from rest and swings down on the rope without friction The actor is momentarily hidden from the camera at the bottom of the arc, where he undergoes an elastic head-on collision with the clone, sending him over the cliff Cursing vilely, the clone falls freely into the ocean below The actor is prosecuted for making an obscene clone fall, and you are called as an expert witness at the sensational trial (a) Find the horizontal component R of the clone’s displacement as it depends on m Evaluate R (b) for m ϭ 79.0 kg and (c) for m ϭ 81.0 kg (d) What value of m gives a range of 30.0 m? (e) What is the maximum possible value for R, and (f) to what value of m does it correspond? What are (g) the minimum values of R and (h) the corresponding value of m? (i) For the actor–clone– Earth system, is mechanical energy conserved throughout the action sequence? Is this principle sufficient to solve the problem? Explain (j) For the same system, is momentum conserved? Explain how this principle is used (k) What If? Show that R does not depend on the value of the gravitational acceleration Is this result remarkable? State how one might make sense of it 57 A bullet of mass m is fired into a block of mass M initially at rest at the edge of a frictionless table of height h (Fig P9.57) The bullet remains in the block, and after impact the block lands a distance d from the bottom of the table Determine the initial speed of the bullet 59 60 61 m M h d Figure P9.57 58 A small block of mass m1 ϭ 0.500 kg is released from rest at the top of a curve-shaped, frictionless wedge of mass m2 ϭ 3.00 kg, which sits on a frictionless horizontal surface as shown in Figure P9.58a When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right as shown in the figure (a) What is the velocity of the m1 h v2 m2 m2 (a) 4.00 m/s 62 wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge? ⅷ A 0.500-kg sphere moving with a velocity given by ˆ m>s strikes another sphere of 12.00ˆi Ϫ 3.00ˆj ϩ 1.00k mass 1.50 kg that is moving with an initial velocity ˆ m>s (a) The velocity of of 1Ϫ1.00ˆi ϩ 2.00ˆj Ϫ 3.00k the 0.500-kg sphere after the collision is given by ˆ m>s Find the final velocity of 1Ϫ1.00ˆi ϩ 3.00ˆj Ϫ 8.00k the 1.50-kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic) (b) Now assume the velocity of the 0.500-kg sphere after the collision is ˆ m>s Find the final velocity 1Ϫ0.250ˆi ϩ 0.750ˆj Ϫ 2.00k of the 1.50-kg sphere and identify the kind of collision (c) What If? Take the velocity of the 0.500-kg sphere after ˆ m>s Find the the collision as 1Ϫ1.00ˆi ϩ 3.00ˆj ϩ ak value of a and the velocity of the 1.50-kg sphere after an elastic collision A 75.0-kg firefighter slides down a pole while a constant friction force of 300 N retards her motion A horizontal 20.0-kg platform is supported by a spring at the bottom of the pole to cushion the fall The firefighter starts from rest 4.00 m above the platform, and the spring constant is 000 N/m Find (a) the firefighter’s speed immediately before she collides with the platform and (b) the maximum distance the spring is compressed Assume the friction force acts during the entire motion ⅷ George of the Jungle, with mass m, swings on a light vine hanging from a stationary tree branch A second vine of equal length hangs from the same point, and a gorilla of larger mass M swings in the opposite direction on it Both vines are horizontal when the primates start from rest at the same moment George and the gorilla meet at the lowest point of their swings Each is afraid that one vine will break, so they grab each other and hang on They swing upward together, reaching a point where the vines make an angle of 35.0° with the vertical (a) Find the value of the ratio m/M (b) What If? Try the following experiment at home Tie a small magnet and a steel screw to opposite ends of a string Hold the center of the string fixed to represent the tree branch, and reproduce a model of the motions of George and the gorilla What changes in your analysis will make it apply to this situation? What If? Next assume the magnet is strong so that it noticeably attracts the screw over a distance of a few centimeters Then the screw will be moving faster immediately before it sticks to the magnet Does this extra magnet strength make a difference? ⅷ A student performs a ballistic pendulum experiment using an apparatus similar to that shown in Figure 9.9b She obtains the following average data: h ϭ 8.68 cm, m1 ϭ 68.8 g, and m2 ϭ 263 g The symbols refer to the quantities in Figure 9.9a (a) Determine the initial speed v1A of the projectile (b) The second part of her experiment is to obtain v1A by firing the same projectile horizontally (with the pendulum removed from the path) and measuring its final horizontal position x and distance of fall y (Fig P9.62) Show that the initial speed of the projectile is related to x and y by the equation (b) v1A ϭ Figure P9.58 = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ = ThomsonNOW; x 22y>g Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning Problems 267 67 A 5.00-g bullet moving with an initial speed of 400 m/s is fired into and passes through a 1.00-kg block as shown in Figure P9.67 The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 900 N/m The block moves 5.00 cm to the right after impact Find (a) the speed at which the bullet emerges from the block and (b) the mechanical energy converted into internal energy in the collision v1A y x Figure P9.62 400 m/s What numerical value does she obtain for v1A based on her measured values of x ϭ 257 cm and y ϭ 85.3 cm? What factors might account for the difference in this value compared with that obtained in part (a)? 63 ⅷ Lazarus Carnot, an artillery general, managed the military draft for Napoleon Carnot used a ballistic pendulum to measure the firing speeds of cannonballs In the symbols defined in Example 9.6, he proved that the ratio of the kinetic energy immediately after the collision to the kinetic energy immediately before is m1/(m1ϩm2) (a) Carry out the proof yourself (b) If the cannonball has mass 9.60 kg and the block (a tree trunk) has mass 214 kg, what fraction of the original energy remains mechanical after the collision? (c) What is the ratio of the momentum immediately after the collision to the momentum immediately before? (d) A student believes that such a large loss of mechanical energy must be accompanied by at least a small loss of momentum How would you convince this student of the truth? General Carnot’s son Sadi was the second most important engineer in the history of ideas; we will study his work in Chapter 22 64 ⅷ Pursued by ferocious wolves, you are in a sleigh with no horses, gliding without friction across an ice-covered lake You take an action described by these equations: 1270 kg2 17.50 m>s ˆi ϭ 115.0 kg2 1Ϫv 1fˆi ϩ 1255 kg2 1v 2fˆi v 1f ϩ v 2f ϭ 8.00 m>s (a) Complete the statement of the problem, giving the data and identifying the unknowns (b) Find the values of v1f and v2f (c) Find the work you 65 Review problem A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250-kg block on the left and a 0.500-kg block on the right Both blocks are at rest on a horizontal surface The blocks are released simultaneously so that the spring tends to push them apart Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is (a) 0, (b) 0.100, and (c) 0.462 Assume the coefficient of static friction is greater than the coefficient of kinetic friction in every case 66 Consider as a system the Sun with the Earth in a circular orbit around it Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system over a 6-month period Ignore the influence of other celestial objects You may obtain the necessary astronomical data from the endpapers of the book = intermediate; = challenging; Ⅺ = SSM/SG; ᮡ 5.00 cm v Figure P9.67 68 ⅷ Review problem There are (one can say) three coequal theories of motion: Newton’s second law, stating that the total force on a particle causes its acceleration; the work–kinetic energy theorem, stating that the total work on a particle causes its change in kinetic energy; and the impulse–momentum theorem, stating that the total impulse on a particle causes its change in momentum In this problem, you compare predictions of the three theories in one particular case A 3.00-kg object has velocity 7.00ˆj m>s Then, a total force 12.0ˆi N acts on the object for 5.00 s (a) Calculate the object’s final velocity, using the impulse–momentum theorem (b) Calculate its accelS S S eration from a ϭ S1vf Ϫ vi 2>¢t (c) Calculate its acceleraS tion from a ϭ © F>m (d) Find the object’s vector disS S S placement from ¢ r ϭ vi t Sϩ 12 at (e) Find the work done S # on the object from W ϭ F ¢r (f) Find the final kinetic 1 S #S energy from 2mv f ϭ mvf vf (g) Find the final kinetic energy from 12mv i ϩ W (h) State the result of comparing the answers to parts b and c, and the answers to parts f and g 69 A chain of length L and total mass M is released from rest with its lower end just touching the top of a table as shown in Figure P9.69a Find the force exerted by the table on the chain after the chain has fallen through a distance x as shown in Figure P9.69b (Assume each link comes to rest the instant it reaches the table.) = ThomsonNOW; x L L Ϫx (a) (b) Figure P9.69 Ⅵ = symbolic reasoning; ⅷ = qualitative reasoning 268 Chapter Linear Momentum and Collisions Answers to Quick Quizzes 9.1 (d) Two identical objects (m1 ϭ m2) traveling at the same speed (v1 ϭ v2) have the same kinetic energies and the same magnitudes of momentum It also is possible, however, for particular combinations of masses and velocities to satisfy K1 ϭ K2 but not p1 ϭ p2 For example, a 1-kg object moving at m/s has the same kinetic energy as a 4-kg object moving at m/s, but the two clearly not have the same momenta Because we have no information about masses and speeds, we cannot choose among (a), (b), or (c) 9.2 (b), (c), (a) The slower the ball, the easier it is to catch If the momentum of the medicine ball is the same as the momentum of the baseball, the speed of the medicine ball must be 1/10 the speed of the baseball because the medicine ball has 10 times the mass If the kinetic energies are the same, the speed of the medicine ball must be 1> 110 the speed of the baseball because of the squared speed term in the equation for K The medicine ball is hardest to catch when it has the same speed as the baseball 9.3 (i), (c), (e) Object has a greater acceleration because of its smaller mass Therefore, it travels the distance d in a shorter time interval Even though the force applied to objects and is the same, the change in momentum is less for object because ⌬t is smaller The work W ϭ Fd done on both objects is the same because both F and d are the same in the two cases Therefore, K1 ϭ K2 (ii), (b), (d) The same impulse is applied to both objects, so they experience the same change in momentum Object has a larger acceleration due to its smaller mass Therefore, the distance that object covers in the time interval is larger than that for object As a result, more work is done on object and K2 Ͼ K1 9.4 (a) All three are the same Because the passenger is brought from the car’s initial speed to a full stop, the change in momentum (equal to the impulse) is the same regardless of what stops the passenger (b) Dashboard, seat belt, air bag The dashboard stops the passenger very quickly in a front-end collision, resulting in a very large force The seat belt takes somewhat more time, so the force is smaller Used along with the seat belt, the air bag can extend the passenger’s stopping time further, notably for his head, which would otherwise snap forward 9.5 (a) If all the initial kinetic energy is transformed or transferred away from the system, nothing is moving after the collision Consequently, the final momentum of the system is necessarily zero and the initial momentum of the system must therefore be zero Although (b) and (d) together would satisfy the conditions, neither one alone does 9.6 (b) Because momentum of the two-ball system is conS S S served, pTi ϩ ϭ pTf ϩ pB Because the table-tennis ball bounces back from the much more massive bowling ball S S with approximately the same speed, pTf ϭ ϪpTi As a conS S sequence, pB ϭ 2pTi Kinetic energy can be expressed as K ϭ p 2/2m Because of the much larger mass of the bowling ball, its kinetic energy is much smaller than that of the table-tennis ball 9.7 (b) The piece with the handle will have less mass than the piece made up of the end of the bat To see why, take the origin of coordinates as the center of mass before the bat was cut Replace each cut piece by a small sphere located at each piece’s center of mass The sphere representing the handle piece is farther from the origin, but the product of less mass and greater distance balances the product of greater mass and less distance for the end piece as shown 9.8 (i), (a) This effect is the same one as the swimmer diving off the raft that we just discussed The vessel–passengers system is isolated If the passengers all start running one way, the speed of the vessel increases (a small amount!) the other way (ii), (b) Once they stop running, the momentum of the system is the same as it was before they started running; you cannot change the momentum of an isolated system by means of internal forces In case you are thinking that the passengers could run to the stern repeatedly to take advantage of the speed increase while they are running, remember that they will slow the ship down every time they return to the bow! 10.1 Angular Position, Velocity, and Acceleration 10.5 Calculation of Moments of Inertia 10.2 Rotational Kinematics: The Rigid Object Under Constant Angular Acceleration 10.7 The Rigid Object Under a Net Torque 10.6 Torque 10.8 Energy Considerations in Rotational Motion 10.3 Angular and Translational Quantities 10.9 Rolling Motion of a Rigid Object 10.4 Rotational Kinetic Energy The Malaysian pastime of gasing involves the spinning of tops that can have masses up to kg Professional spinners can spin their tops so that they might rotate for more than an hour before stopping We will study the rotational motion of objects such as these tops in this chapter (Courtesy Tourism Malaysia) 10 Rotation of a Rigid Object About a Fixed Axis When an extended object such as a wheel rotates about its axis, the motion cannot be analyzed by modeling the object as a particle because at any given time different parts of the object have different linear velocities and linear accelerations We can, however, analyze the motion of an extended object by modeling it as a collection of particles, each of which has its own linear velocity and linear acceleration In dealing with a rotating object, analysis is greatly simplified by assuming the object is rigid A rigid object is one that is nondeformable; that is, the relative locations of all particles of which the object is composed remain constant All real objects are deformable to some extent; our rigid-object model, however, is useful in many situations in which deformation is negligible r O P (a) P r O 10.1 Angular Position, Velocity, and Acceleration Figure 10.1 illustrates an overhead view of a rotating compact disc, or CD The disc rotates about a fixed axis perpendicular to the plane of the figure and passing through the center of the disc at O A small element of the disc modeled as a particle at P is at a fixed distance r from the origin and rotates about it in a circle of radius r (In fact, every particle on the disc undergoes circular motion about O.) It is convenient to represent the position of P with its polar coordinates (r, u), where Reference line s u Reference line (b) Figure 10.1 A compact disc rotating about a fixed axis through O perpendicular to the plane of the figure (a) To define angular position for the disc, a fixed reference line is chosen A particle at P is located at a distance r from the rotation axis at O (b) As the disc rotates, a particle at P moves through an arc length s on a circular path of radius r 269 270 Chapter 10 Rotation of a Rigid Object About a Fixed Axis r is the distance from the origin to P and u is measured counterclockwise from some reference line fixed in space as shown in Figure 10.1a In this representation, the angle u changes in time while r remains constant As the particle moves along the circle from the reference line, which is at angle u ϭ 0, it moves through an arc of length s as in Figure 10.1b The arc length s is related to the angle u through the relationship s ϭ ru uϭ In rotational equations, you must use angles expressed in radians Don’t fall into the trap of using angles measured in degrees in rotational equations u 1rad ϭ y Ꭾ,t f Ꭽ,ti uf ui x O Figure 10.2 A particle on a rotating rigid object moves from Ꭽ to Ꭾ along the arc of a circle In the time interval ⌬t ϭ tf Ϫ ti, the radial line of length r moves through an angular displacement ⌬u ϭ uf Ϫ ui Average angular speed s r (10.1b) Because u is the ratio of an arc length and the radius of the circle, it is a pure number Usually, however, we give u the artificial unit radian (rad), where one radian is the angle subtended by an arc length equal to the radius of the arc Because the circumference of a circle is 2pr, it follows from Equation 10.1b that 360° corresponds to an angle of (2pr/r) rad ϭ 2p rad Hence, rad ϭ 360°/2p Ϸ 57.3° To convert an angle in degrees to an angle in radians, we use that p rad ϭ 180°, so PITFALL PREVENTION 10.1 Remember the Radian r (10.1a) ᮣ p u 1deg 180° For example, 60° equals p/3 rad and 45° equals p/4 rad Because the disc in Figure 10.1 is a rigid object, as the particle moves through an angle u from the reference line, every other particle on the object rotates through the same angle u Therefore, we can associate the angle U with the entire rigid object as well as with an individual particle, which allows us to define the angular position of a rigid object in its rotational motion We choose a reference line on the object, such as a line connecting O and a chosen particle on the object The angular position of the rigid object is the angle u between this reference line on the object and the fixed reference line in space, which is often chosen as the x axis Such identification is similar to the way we define the position of an object in translational motion as the distance x between the object and the reference position, which is the origin, x ϭ As the particle in question on our rigid object travels from position Ꭽ to position Ꭾ in a time interval ⌬t as in Figure 10.2, the reference line fixed to the object sweeps out an angle ⌬u ϭ uf Ϫui This quantity ⌬u is defined as the angular displacement of the rigid object: ¢u ϵ u f Ϫ u i The rate at which this angular displacement occurs can vary If the rigid object spins rapidly, this displacement can occur in a short time interval If it rotates slowly, this displacement occurs in a longer time interval These different rotation rates can be quantified by defining the average angular speed vavg (Greek letter omega) as the ratio of the angular displacement of a rigid object to the time interval ⌬t during which the displacement occurs: vavg ϵ uf Ϫ ui tf Ϫ ti ϭ ¢u ¢t (10.2) In analogy to linear speed, the instantaneous angular speed v is defined as the limit of the average angular speed as ⌬t approaches zero: Instantaneous angular speed ᮣ v ϵ lim ¢tS0 ¢u du ϭ ¢t dt (10.3) Angular speed has units of radians per second (rad/s), which can be written as sϪ1 because radians are not dimensional We take v to be positive when u is increasing (counterclockwise motion in Figure 10.2) and negative when u is decreasing (clockwise motion in Figure 10.2) Section 10.1 Angular Position, Velocity, and Acceleration 271 Quick Quiz 10.1 A rigid object rotates in a counterclockwise sense around a fixed axis Each of the following pairs of quantities represents an initial angular position and a final angular position of the rigid object (i) Which of the sets can only occur if the rigid object rotates through more than 180°? (a) rad, rad (b) Ϫ1 rad, rad (c) rad, rad (ii) Suppose the change in angular position for each of these pairs of values occurs in s Which choice represents the lowest average angular speed? If the instantaneous angular speed of an object changes from vi to vf in the time interval ⌬t, the object has an angular acceleration The average angular acceleration aavg (Greek letter alpha) of a rotating rigid object is defined as the ratio of the change in the angular speed to the time interval ⌬t during which the change in the angular speed occurs: a avg ϵ vf Ϫ vi tf Ϫ ti ϭ ¢v ¢t (10.4) ᮤ Average angular acceleration ᮤ Instantaneous angular acceleration In analogy to linear acceleration, the instantaneous angular acceleration is defined as the limit of the average angular acceleration as ⌬t approaches zero: ¢v dv ϭ dt ¢tS0 ¢t a ϵ lim (10.5) Angular acceleration has units of radians per second squared (rad/s2), or simply sϪ2 Notice that a is positive when a rigid object rotating counterclockwise is speeding up or when a rigid object rotating clockwise is slowing down during some time interval When a rigid object is rotating about a fixed axis, every particle on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration That is, the quantities u, v, and a characterize the rotational motion of the entire rigid object as well as individual particles in the object Angular position (u), angular speed (v), and angular acceleration (a) are analogous to translational position (x), translational speed (v), and translational acceleration (a) The variables u, v, and a differ dimensionally from the variables x, v, and a only by a factor having the unit of length (See Section 10.3.) We have not specified any direction for angular speed and angular acceleration Strictly speaking, v and a are the magnitudes of the angular velocity and the anguS S lar acceleration vectors1 V and A, respectively, and they should always be positive Because we are considering rotation about a fixed axis, however, we can use nonvector notation and indicate the vectors’ directions by assigning a positive or negative sign to v and a as discussed earlier with regard to Equations 10.3 and 10.5 For rotation about a fixed axis, the only direction that uniquely specifies the rotational motion is the direction along the axis of rotation Therefore, the directions S S of V and A are along this axis If a particle rotates in the xy plane as in Figure 10.2, S the direction of V for the particle is out of the plane of the diagram when the rotation is counterclockwise and into the plane of the diagram when the rotation is clockwise To illustrate this convention, it is convenient to use the right-hand rule demonstrated in Figure 10.3 When the four fingers of the right hand are wrapped in the direction of rotation, the extended right thumb points in the direction of S S S S V The direction of A follows from its definition A ϵ dV>dt It is in the same direcS S tion as V if the angular speed is increasing in time, and it is antiparallel to V if the angular speed is decreasing in time PITFALL PREVENTION 10.2 Specify Your Axis In solving rotation problems, you must specify an axis of rotation This new feature does not exist in our study of translational motion The choice is arbitrary, but once you make it, you must maintain that choice consistently throughout the problem In some problems, the physical situation suggests a natural axis, such as the center of an automobile wheel In other problems, there may not be an obvious choice, and you must exercise judgment v v Although we not verify it here, the instantaneous angular velocity and instantaneous angular acceleration are vector quantities, but the corresponding average values are not because angular displacements not add as vector quantities for finite rotations Figure 10.3 The right-hand rule for determining the direction of the angular velocity vector 272 Chapter 10 Rotation of a Rigid Object About a Fixed Axis 10.2 Rotational Kinematics: The Rigid Object Under Constant Angular Acceleration When a rigid object rotates about a fixed axis, it often undergoes a constant angular acceleration Therefore, we generate a new analysis model for rotational motion called the rigid object under constant angular acceleration This model is the rotational analog to the particle under constant acceleration model We develop kinematic relationships for this model in this section Writing Equation 10.5 in the form dv ϭ a dt and integrating from ti ϭ to tf ϭ t gives Rotational kinematic equations vf ϭ vi ϩ at¬1for constant a2 ᮣ PITFALL PREVENTION 10.3 Just Like Translation? Equations 10.6 to 10.9 and Table 10.1 suggest that rotational kinematics is just like translational kinematics That is almost true, with two key differences (1) In rotational kinematics, you must specify a rotation axis (per Pitfall Prevention 10.2) (2) In rotational motion, the object keeps returning to its original orientation; therefore, you may be asked for the number of revolutions made by a rigid object This concept has no meaning in translational motion (10.6) where vi is the angular speed of the rigid object at time t ϭ Equation 10.6 allows us to find the angular speed vf of the object at any later time t Substituting Equation 10.6 into Equation 10.3 and integrating once more, we obtain u f ϭ u i ϩ vit ϩ 12at 1for constant a2 (10.7) where ui is the angular position of the rigid object at time t ϭ Equation 10.7 allows us to find the angular position uf of the object at any later time t Eliminating t from Equations 10.6 and 10.7 gives v f ϭ v i ϩ 2a 1u f Ϫ u i 2¬1for constant a2 (10.8) This equation allows us to find the angular speed vf of the rigid object for any value of its angular position uf If we eliminate a between Equations 10.6 and 10.7, we obtain u f ϭ u i ϩ 12 1vi ϩ vf 2t 1for constant a (10.9) Notice that these kinematic expressions for the rigid object under constant angular acceleration are of the same mathematical form as those for a particle under constant acceleration (Chapter 2) They can be generated from the equations for translational motion by making the substitutions x S u, v S v, and a S a Table 10.1 compares the kinematic equations for rotational and translational motion Quick Quiz 10.2 Consider again the pairs of angular positions for the rigid object in Quick Quiz 10.1 If the object starts from rest at the initial angular position, moves counterclockwise with constant angular acceleration, and arrives at the final angular position with the same angular speed in all three cases, for which choice is the angular acceleration the highest? TABLE 10.1 Kinematic Equations for Rotational and Translational Motion Under Constant Acceleration Rotational Motion About a Fixed Axis Translational Motion vf ϭ vi ϩ at vf ϭ vi ϩ at uf ϭ u i ϩ v i t ϩ vf2 ϭ vi2 ϩ 2a(uf Ϫ ui ) uf ϭ ui ϩ 12 (vi ϩ vf )t 2 at xf ϭ xi ϩ vi t ϩ 12 at vf2 ϭ vi2 ϩ 2a(xf Ϫ xi ) xf ϭ xi ϩ 12 (vi ϩ vf )t Section 10.3 E XA M P L E Angular and Translational Quantities 273 Rotating Wheel A wheel rotates with a constant angular acceleration of 3.50 rad/s2 (A) If the angular speed of the wheel is 2.00 rad/s at ti ϭ 0, through what angular displacement does the wheel rotate in 2.00 s? SOLUTION Conceptualize Look again at Figure 10.1 Imagine that the compact disc rotates with its angular speed increasing at a constant rate You start your stopwatch when the disc is rotating at 2.00 rad/s This mental image is a model for the motion of the wheel in this example Categorize The phrase “with a constant angular acceleration” tells us to use the rigid object under constant angular acceleration model Analyze Arrange Equation 10.7 so that it expresses the angular displacement of the object: Substitute the known values to find the angular displacement at t ϭ 2.00 s: ¢u ϭ u f Ϫ u i ϭ vit ϩ 12at ¢u ϭ 12.00 rad>s2 12.00 s2 ϩ 12 13.50 rad>s2 12.00 s 2 ϭ 11.0 rad ϭ 111.0 rad 157.3°>rad ϭ 630° (B) Through how many revolutions has the wheel turned during this time interval? SOLUTION Multiply the angular displacement found in part (A) by a conversion factor to find the number of revolutions: ¢u ϭ 630° a rev b ϭ 1.75 rev 360° (C) What is the angular speed of the wheel at t ϭ 2.00 s? SOLUTION Use Equation 10.6 to find the angular speed at t ϭ 2.00 s: Finalize vf ϭ vi ϩ at ϭ 2.00 rad>s ϩ 13.50 rad>s2 12.00 s ϭ 9.00 rad>s We could also obtain this result using Equation 10.8 and the results of part (A) (Try it!) What If? Suppose a particle moves along a straight line with a constant acceleration of 3.50 m/s2 If the velocity of the particle is 2.00 m/s at ti ϭ 0, through what displacement does the particle move in 2.00 s? What is the velocity of the particle at t ϭ 2.00 s? Answer Notice that these questions are translational analogs to parts (A) and (C) of the original problem The mathematical solution follows exactly the same form For the displacement, ¢x ϭ x f Ϫ x i ϭ v it ϩ 12at ϭ 12.00 m>s2 12.00 s ϩ 12 13.50 m>s2 12.00 s 2 ϭ 11.0 m and for the velocity, vf ϭ vi ϩ at ϭ 2.00 m>s ϩ 13.50 m>s2 12.00 s ϭ 9.00 m>s There is no translational analog to part (B) because translational motion under constant acceleration is not repetitive 10.3 Angular and Translational Quantities In this section, we derive some useful relationships between the angular speed and acceleration of a rotating rigid object and the translational speed and acceleration of a point in the object To so, we must keep in mind that when a rigid object Section 10.5 279 Calculation of Moments of Inertia Equation 1.1, r ϵ m/V, where r is the density of the object and V is its volume From this equation, the mass of a small element is dm ϭ r dV Substituting this result into Equation 10.17 gives Iϭ Ύ rr dV If the object is homogeneous, r is constant and the integral can be evaluated for a known geometry If r is not constant, its variation with position must be known to complete the integration The density given by r ϭ m/V sometimes is referred to as volumetric mass density because it represents mass per unit volume Often we use other ways of expressing density For instance, when dealing with a sheet of uniform thickness t, we can define a surface mass density s ϭ rt, which represents mass per unit area Finally, when mass is distributed along a rod of uniform cross-sectional area A, we sometimes use linear mass density l ϭ M/L ϭ rA, which is the mass per unit length Table 10.2 gives the moments of inertia for a number of objects about specific axes The moments of inertia of rigid objects with simple geometry (high symmetry) are relatively easy to calculate provided the rotation axis coincides with an axis of symmetry, as in the examples below Quick Quiz 10.4 A section of hollow pipe and a solid cylinder have the same radius, mass, and length They both rotate about their long central axes with the same angular speed Which object has the higher rotational kinetic energy? (a) The hollow pipe does (b) The solid cylinder does (c) They have the same rotational kinetic energy (d) It is impossible to determine E XA M P L E Uniform Rigid Rod yЈ Calculate the moment of inertia of a uniform rigid rod of length L and mass M (Fig 10.9) about an axis perpendicular to the rod (the y axis) and passing through its center of mass y dx SOLUTION Conceptualize Imagine twirling the rod in Figure 10.9 with your fingers around its midpoint If you have a meterstick handy, use it to simulate the spinning of a thin rod x O x Categorize This example is a substitution problem, using the definition of moment of inertia in Equation 10.17 As with any calculus problem, the solution involves reducing the integrand to a single variable The shaded length element dx in Figure 10.9 has a mass dm equal to the mass per unit length l multiplied by dx Iy ϭ ϭ Check this result in Table 10.2 Figure 10.9 (Example 10.4) A uniform rigid rod of length L The moment of inertia about the y axis is less than that about the yЈ axis The latter axis is examined in Example 10.6 dm ϭ ldx ϭ Express dm in terms of dx : Substitute this expression into Equation 10.17, with r ϭ x 2: L Ύ r dm ϭ Ύ L>2 ϪL>2 M x L>2 c d ϭ L ϪL>2 x2 M dx L M M dx ϭ L L 12 ML Ύ L>2 ϪL>2 x dx 280 Chapter 10 E XA M P L E Rotation of a Rigid Object About a Fixed Axis Uniform Solid Cylinder A uniform solid cylinder has a radius R, mass M, and length L Calculate its moment of inertia about its central axis (the z axis in Fig 10.10) z dr SOLUTION r Conceptualize To simulate this situation, imagine twirling a can of frozen juice around its central axis R L Categorize This example is a substitution problem, using the definition of moment of inertia As with Example 10.4, we must reduce the integrand to a single variable It is convenient to divide the cylinder into many cylindrical shells, each having radius r, thickness dr, and length L as shown in Figure 10.10 The density of the cylinder is r The volume dV of each shell is its cross-sectional area multiplied by its length: dV ϭ L dA ϭ L(2pr) dr Figure 10.10 (Example 10.5) Calculating I about the z axis for a uniform solid cylinder dm ϭ rdV ϭ 2prLr dr Express dm in terms of dr: Substitute this expression into Equation 10.17: Iz ϭ Ύ r 2dm ϭ Ύ r 12prLr dr2 ϭ 2prL rϭ Use the total volume pR 2L of the cylinder to express its density: Iz ϭ 12 p a Substitute this value into the expression for Iz: R Ύ r dr ϭ prLR M M ϭ V pR 2L M b LR ϭ pR 2L 2 MR Check this result in Table 10.2 What If? What if the length of the cylinder in Figure 10.10 is increased to 2L, while the mass M and radius R are held fixed? How does that change the moment of inertia of the cylinder? Answer Notice that the result for the moment of inertia of a cylinder does not depend on L, the length of the cylinder It applies equally well to a long cylinder and a flat disk having the same mass M and radius R Therefore, the moment of inertia of the cylinder would not be affected by changing its length The calculation of moments of inertia of an object about an arbitrary axis can be cumbersome, even for a highly symmetric object Fortunately, use of an important theorem, called the parallel-axis theorem, often simplifies the calculation To generate the parallel-axis theorem, suppose an object rotates about the z axis as shown in Figure 10.11 The moment of inertia does not depend on how the mass is distributed along the z axis; as we found in Example 10.5, the moment of inertia of a cylinder is independent of its length Imagine collapsing the threedimensional object into a planar object as in Figure 10.11b In this imaginary process, all mass moves parallel to the z axis until it lies in the xy plane The coordinates of the object’s center of mass are now x CM, y CM, and z CM ϭ Let the mass element dm have coordinates (x, y, 0) Because this element is a distance r ϭ 1x ϩ y from the z axis, the moment of inertia about the z axis is Iϭ Ύr dm ϭ Ύ 1x ϩ y2 ¬dm We can relate the coordinates x, y of the mass element dm to the coordinates of this same element located in a coordinate system having the object’s center of mass as its origin If the coordinates of the center of mass are x CM, y CM, and z CM ϭ Section 10.5 y Calculation of Moments of Inertia 281 dm x, y z yЈ y CM yCM Axis through CM y Rotation axis r xCM, yCM O D CM x O x CM x xЈ x (a) (b) Figure 10.11 (a) The parallel-axis theorem If the moment of inertia about an axis perpendicular to the figure through the center of mass is ICM, the moment of inertia about the z axis is Iz ϭ ICM ϩ MD2 (b) Perspective drawing showing the z axis (the axis of rotation) and the parallel axis through the center of mass in the original coordinate system centered on O, we see from Figure 10.11a that the relationships between the unprimed and primed coordinates are x ϭ xЈ ϩ x CM, y ϭ yЈ ϩ y CM, and z ϭ zЈ ϭ Therefore, Iϭ ϭ Ύ 1x¿ ϩ x Ύ 1x¿ 2 CM 2 ϩ 1y¿ ϩ y CM 2 4dm ϩ 1y¿ 2 4dm ϩ 2x CM Ύ x¿dm ϩ 2y Ύ y¿dm ϩ 1x CM CM ϩ y CM2 Ύ dm The first integral is, by definition, the moment of inertia ICM about an axis that is parallel to the z axis and passes through the center of mass The second two integrals are zero because, by definition of the center of mass, ͐ x¿dm ϭ ͐ y¿dm ϭ The last integral is simply MD because ͐ dm ϭ M and D ϭ x CM2 ϩ y CM2 Therefore, we conclude that I ϭ ICM ϩ MD E XA M P L E (10.18) ᮤ Parallel-axis theorem Applying the Parallel-Axis Theorem Consider once again the uniform rigid rod of mass M and length L shown in Figure 10.9 Find the moment of inertia of the rod about an axis perpendicular to the rod through one end (the yЈ axis in Fig 10.9) SOLUTION Conceptualize Imagine twirling the rod around an endpoint rather than the midpoint If you have a meterstick handy, try it and notice the degree of difficulty in rotating it around the end compared with rotating it around the center Categorize This example is a substitution problem, involving the parallel-axis theorem Intuitively, we expect the moment of inertia to be greater than the result ICM ϭ 12 ML2 from Example 10.4 because there is mass up to a distance of L away from the rotation axis, whereas the farthest distance in Example 10.4 was only L/2 The distance between the center-of-mass axis and the yЈ axis is D ϭ L/2 Use the parallel-axis theorem: Check this result in Table 10.2 L I ϭ ICM ϩ MD ϭ 12 ML2 ϩ M a b ϭ 2 ML 282 Chapter 10 Rotation of a Rigid Object About a Fixed Axis F sin f F r f O r F cos f Line of action f d S Figure 10.12 The force F has a greater rotating tendency about an axis through O as F increases and as the moment arm d increases The component F sin f tends to rotate the wrench about O Torque Imagine trying to rotate a door by applying a force of magnitude F perpendicular to the door surface near the hinges and then at various distances from the hinges You will achieve a more rapid rate of rotation for the door by applying the force near the doorknob than by applying it near the hinges When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis The tendency of a force to rotate an object about some S axis is measured by a quantity called torque T(Greek letter tau) Torque is a vector, but we will consider only its magnitude here and explore its vector nature in Chapter 11 Consider the wrench in Figure 10.12 that we wish to rotate around an axis perS pendicular to the page and through the center of the bolt The applied force F acts at an angle f toS the horizontal We define the magnitude of the torque associated with the force F by the expression t ϵ r F sin f ϭ Fd PITFALL PREVENTION 10.5 Torque Depends on Your Choice of Axis (10.19) S Like moment of inertia, there is no unique value of the torque on an object Its value depends on your choice of rotation axis Moment arm 10.6 ᮣ F1 d1 O d2 F2 ACTIVE FIGURE 10.13 S The force F1 tends to rotate the object counterclockwise about an axis S through O, and F2 tends to rotate it clockwise Sign in at www.thomsonedu.com and go to ThomsonNOW to change the magnitudes, directions, and Spoints of S application of forces F1 and F2 and see how the object accelerates under the action of the two forces where r is the distance between the rotation axis and the point of application of F , S and d is the perpendicular distance from the rotation axis to the line of action of F (The line of action of a force is an imaginary line extending out both ends of the S vector representing the force The dashed line extending from the tail of in FigF S ure 10.12 is part of the line of action of F.) From the right triangle in Figure 10.12 that has the wrench as its hypotenuse, we see that d ϭ r sin f The quantity d is S called the moment arm (or lever arm) of F S In Figure 10.12, the only component of F that tends to cause rotation of the wrench around an axis through O is F sin f, the component perpendicular to a line drawn from the rotation axis to the point of application of the force The horizontal component F cos f, because its line of action passes through O, has no tendency to produce rotation about an axis passing through O From the definition of torque, the rotating tendency increases as F increases and as d increases, which explains why it is easier to rotate a door if we push at the doorknob rather than at a point close to the hinges We also want to apply our push as closely perpendicular to the door as we can so that f is close to 90° Pushing sideways on the doorknob (f ϭ 0) will not cause the door to rotate If two or more forces act on a rigid object as in Active Figure 10.13, each tends S to produce rotation about the axis at O In this example, tends to rotate the F S object clockwise and F1 tends to rotate it counterclockwise We use the convention that the sign of the torque resulting from a force is positive if the turning tendency of the force is counterclockwise and negative if the turning tendency is S clockwise For example, in Active Figure 10.13, the torque resulting fromS F1, which has a moment arm d1, is positive and equal to ϩF1d1; the torque from F2 is negative and equal to ϪF2d2 Hence, the net torque about an axis through O is a t ϭ t1 ϩ t2 ϭ F1d1 Ϫ F2d2 Torque should not be confused with force Forces can cause a change in translational motion as described by Newton’s second law Forces can also cause a change in rotational motion, but the effectiveness of the forces in causing this change depends on both the magnitudes of the forces and the moment arms of the forces, in the combination we call torque Torque has units of force times length—newton meters in SI units—and should be reported in these units Do not confuse torque and work, which have the same units but are very different concepts Quick Quiz 10.5 (i) If you are trying to loosen a stubborn screw from a piece of wood with a screwdriver and fail, should you find a screwdriver for which the handle is (a) longer or (b) fatter? (ii) If you are trying to loosen a stubborn bolt from a piece of metal with a wrench and fail, should you find a wrench for which the handle is (a) longer or (b) fatter? Section 10.7 E XA M P L E 283 The Rigid Object Under a Net Torque The Net Torque on a Cylinder A one-piece cylinder is shaped as shown in Figure 10.14, with a core section protruding from the larger drum The cylinder is free to rotate about the central axis shown in the drawing A rope wrapped around the drum, which has radius R1, S exerts a force T1 to the right on the cylinder A rope wrapped around the core, S which has radius R2, exerts a force T2 downward on the cylinder y T1 R1 (A) What is the net torque acting on the cylinder about the rotation axis (which is the z axis in Fig 10.14)? SOLUTION R2 x O z Conceptualize Imagine that the cylinder in Figure 10.14 is a shaft in a machine S T The force could be applied by a drive belt wrapped around the drum The S force T1 could be applied by a friction brake at the surface of the core T2 Figure 10.14 (Example 10.7) A solid cylinder pivoted about the zSaxis through O The moment armS of T1 is R1, and the moment arm of T2 is R2 Categorize This example is a substitution problem in which we evaluate the net torque using Equation 10.19 S The torque due to T1 about the rotation axis is ϪR1T1 (The sign is negative S because the torque tends to produce clockwise rotation.) The torque due to T2 is ϩR2T2 (The sign is positive because the torque tends to produce counterclockwise rotation of the cylinder.) Evaluate the net torque about the rotation axis: a t ϭ t1 ϩ t2 ϭ R 2T2 Ϫ R 1T1 As a quick check, notice that if the two forces are of equal magnitude, the net torque is negative because R1 Ͼ R2 Starting from rest with both forces of equal magnitude acting on it, the cylinder would rotate clockwise because S S T1 would be more effective at turning it than would T2 (B) Suppose T1 ϭ 5.0 N, R1 ϭ 1.0 m, T2 ϭ 15.0 N, and R2 ϭ 0.50 m What is the net torque about the rotation axis, and which way does the cylinder rotate starting from rest? SOLUTION Substitute the given values: # a t ϭ 10.50 m2 115 N2 Ϫ 11.0 m2 15.0 N2 ϭ 2.5 N m Because this net torque is positive, the cylinder begins to rotate in the counterclockwise direction 10.7 The Rigid Object Under a Net Torque In Chapter 5, we learned that a net force on an object causes an acceleration of the object and that the acceleration is proportional to the net force These facts are the basis of the particle under a net force model whose mathematical representation is Newton’s second law In this section, we show the rotational analog of Newton’s second law: the angular acceleration of a rigid object rotating about a fixed axis is proportional to the net torque acting about that axis Before discussing the more complex case of rigid-object rotation, however, it is instructive first to discuss the case of a particle moving in a circular path about some fixed point under the influence of an external force Consider a particle of mass m rotating in a circle of radius r under the influence S S of a tangential net force © Ft and a radial net force © Fr as shown in Figure 10.15 The radial net force causes the particle to move in the circular path with a cenS tripetal acceleration The tangential force provides a tangential acceleration at , and a Ft ϭ ma t ⌺ Ft m ⌺ Fr r Figure 10.15 A particle rotating in a circle under the influence of a net S S tangential force © Ft A net force © Fr in the radial direction also must be present to maintain the circular motion 284 Chapter 10 Rotation of a Rigid Object About a Fixed Axis S y The magnitude of the net torque due to © Ft on the particle about an axis through the center of the circle is dFt a t ϭ a Ft r ϭ 1mat 2r Because the tangential acceleration is related to the angular acceleration through the relationship at ϭ r a (Eq 10.11), the net torque can be expressed as dm r O a t ϭ 1mra 2r ϭ 1mr 2a x Figure 10.16 A rigid object rotating about an axis through O Each mass element dm rotates about the axis with the same angular acceleration a Recall from Equation 10.15 that mr is the moment of inertia of the particle about the z axis passing through the origin, so that a t ϭ Ia (10.20) That is, the net torque acting on the particle is proportional to its angular acceleration, and the proportionality constant is the moment of inertia Notice that ͚ t ϭ Ia has the same mathematical form as Newton’s second law of motion, ͚ F ϭ ma Now let us extend this discussion to a rigid object of arbitrary shape rotating about a fixed axis as in Figure 10.16 The object can be regarded as an infinite number of mass elements dm of infinitesimal size If we impose a Cartesian coordinate system on the object, each mass element rotates in a circle about the origin S and each has a tangential acceleration at produced by an external tangential force S d Ft For any given element, we know from Newton’s second law that dFt ϭ 1dm2at S The torque dt associated with the force d Ft acts about the origin and is given by dt ϭ r dF t ϭ a t r dm Because at ϭ r a, the expression for dt becomes dt ϭ ar dm Although each mass element of the rigid object may have a different translaS tional acceleration at , they all have the same angular acceleration a With this in mind, we can integrate the above expression to obtain the net torque ©t about an axis through O due to the external forces: atϭ Ύ ar Ύ dm ϭ a r dm where a can be taken outside the integral because it is common to all mass elements From Equation 10.17, we know that ͐r dm is the moment of inertia of the object about the rotation axis through O, and so the expression for ͚ t becomes Torque is proportional to angular acceleration ᮣ a t ϭ Ia (10.21) This equation for a rigid object is the same as that found for a particle moving in a circular path (Eq 10.20) The net torque about the rotation axis is proportional to the angular acceleration of the object, with the proportionality factor being I, a quantity that depends on the axis of rotation and on the size and shape of the object Equation 10.21 is the mathematical representation of the analysis model of a rigid object under a net torque, the rotational analog to the particle under a net force Finally, notice that the result ͚ t ϭ Ia also applies when the forces acting on the mass elements have radial components as well as tangential components That is because the line of action of all radial components must pass through the axis of rotation; hence, all radial components produce zero torque about that axis Quick Quiz 10.6 You turn off your electric drill and find that the time interval for the rotating bit to come to rest due to frictional torque in the drill is ⌬t You replace the bit with a larger one that results in a doubling of the moment of iner- Section 10.7 The Rigid Object Under a Net Torque 285 tia of the drill’s entire rotating mechanism When this larger bit is rotated at the same angular speed as the first and the drill is turned off, the frictional torque remains the same as that for the previous situation What is the time interval for this second bit to come to rest? (a) ⌬t (b) ⌬t (c) ⌬t (d) 0.5 ⌬t (e) 0.25 ⌬t (f) impossible to determine E XA M P L E Rotating Rod A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane as in Figure 10.17 The rod is released from rest in the horizontal position What are the initial angular acceleration of the rod and the initial translational acceleration of its right end? L Pivot SOLUTION Mg Conceptualize Imagine what happens to the rod in Figure 10.17 when it is released It rotates clockwise around the pivot at the left end Figure 10.17 (Example 10.8) A rod is free to rotate around a pivot at the left end The gravitational force on the rod acts at its center of mass Categorize The rod is categorized as a rigid object under a net torque The torque is due only to the gravitational force on the rod if the rotation axis is chosen to pass through the pivot in Figure 10.17 We cannot categorize the rod as a rigid object under constant angular acceleration because the torque exerted on the rod and therefore the angular acceleration of the rod vary with its angular position S Analyze The only force contributing to the torque about an axis through the pivot is the gravitational force Mg exerted on the rod (The force exerted by the pivot on the rod has zero torque about the pivot because its moment arm is zero.) To compute the torque on the rod, we assume the gravitational force acts at the center of mass of the rod as shown in Figure 10.17 L t ϭ Mg a b Write an expression for the magnitude of the torque due to the gravitational force about an axis through the pivot: (1) Use Equation 10.21 to obtain the angular acceleration of the rod: aϭ Mg 1L>22 3g t ϭ ϭ I 2L ML a t ϭ La ϭ Use Equation 10.11 with r ϭ L to find the initial translational acceleration of the right end of the rod: 2g Finalize These values are the initial values of the angular and translational accelerations Once the rod begins to rotate, the gravitational force is no longer perpendicular to the rod and the values of the two accelerations decrease, going to zero at the moment the rod passes through the vertical orientation What If? What if we were to place a penny on the end of the rod and then release the rod? Would the penny stay in contact with the rod? Answer The result for the initial acceleration of a point on the end of the rod shows that at Ͼ g An unsupported penny falls at acceleration g So, if we place a penny at the end of the rod and then release the rod, the end of the rod falls faster than the penny does! The penny does not stay in contact with the rod (Try this with a penny and a meterstick!) The question now is to find the location on the rod at which we can place a penny that will stay in contact as both begin to fall To find the translational acceleration of an arbitrary point on the rod at a distance r Ͻ L from the pivot point, we combine Equation (1) with Equation 10.11: at ϭ r a ϭ 3g 2L r 286 Chapter 10 Rotation of a Rigid Object About a Fixed Axis For the penny to stay in contact with the rod, the limiting case is that the translational acceleration must be equal to that due to gravity: at ϭ g ϭ 3g 2L r r ϭ 23L Therefore, a penny placed closer to the pivot than two-thirds of the length of the rod stays in contact with the falling rod, but a penny farther out than this point loses contact CO N C E P T UA L E XA M P L E Falling Smokestacks and Tumbling Blocks When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground as shown in Figure 10.18 Why? SOLUTION As the smokestack rotates around its base, each higher portion of the smokestack falls with a larger tangential acceleration than the portion below it according to Equation 10.11 The angular acceleration increases as the smokestack tips farther Eventually, higher portions of the smokestack experience an acceleration greater than the acceleration that could result from gravity alone; this situation is similar to that described in Example 10.8 It can happen only if these portions are being pulled downward by a force in addition to the gravitational force The force that Figure 10.18 (Conceptual Example causes that to occur is the shear force from lower portions of the smokestack 10.9) A falling smokestack breaks at some point along its length Eventually, the shear force that provides this acceleration is greater than the smokestack can withstand, and the smokestack breaks The same thing happens with a tall tower of children’s toy blocks Borrow some blocks from a child and build such a tower Push it over and watch it come apart at some point before it strikes the floor E XA M P L E Angular Acceleration of a Wheel A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle as in Figure 10.19 A light cord wrapped around the wheel supports an object of mass m Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord M O SOLUTION R Conceptualize Imagine that the object is a bucket in an old-fashioned wishing well It is tied to a cord that passes around a cylinder equipped with a crank for raising the bucket After the bucket has been raised, the system is released and the bucket accelerates downward while the cord unwinds off the cylinder T T Categorize The object is modeled as a particle under a net force The wheel is modeled as a rigid object under a net torque Analyze The magnitude of the torque acting on the wheel about its axis of rotation is t ϭ TR, where T is the force exerted by the cord on the rim of the wheel (The gravitational force exerted by the Earth on the wheel and the normal force exerted by the axle on the wheel both pass through the axis of rotation and therefore produce no torque.) m mg Figure 10.19 (Example 10.10) An object hangs from a cord wrapped around a wheel Section 10.8 Energy Considerations in Rotational Motion 287 a t ϭ Ia Write Equation 10.21: 112¬¬a ϭ Solve for a and substitute the net torque: TR at ϭ I I a Fy ϭ mg Ϫ T ϭ ma Apply Newton’s second law to the motion of the object, taking the downward direction to be positive: (2) Solve for the acceleration a : aϭ mg Ϫ T m Equations (1) and (2) have three unknowns: a, a, and T Because the object and wheel are connected by a cord that does not slip, the translational acceleration of the suspended object is equal to the tangential acceleration of a point on the wheel’s rim Therefore, the angular acceleration a of the wheel and the translational acceleration of the object are related by a ϭ Ra Use this fact together with Equations (1) and (2): (3) a ϭ Ra ϭ mg Ϫ T TR ϭ m I mg Solve for the tension T: (4) Tϭ Substitute Equation (4) into Equation (2) and solve for a: (5) aϭ Use a ϭ Ra and Equation (5) to solve for a: aϭ g a ϭ R R ϩ 1I>mR2 Finalize ϩ 1mR 2>I2 g ϩ 1I>mR 2 We finalize this problem by imagining the behavior of the system in some extreme limits What If? What if the wheel were to become very massive so that I becomes very large? What happens to the acceleration a of the object and the tension T? Answer If the wheel becomes infinitely massive, we can imagine that the object of mass m will simply hang from the cord without causing the wheel to rotate We can show that mathematically by taking the limit I S ϱ Equation (5) then becomes aϭ g ϩ 1I>mR 2 S which agrees with our conceptual conclusion that the object will hang at rest Also, Equation (4) becomes Tϭ mg ϩ 1mR >I2 mg S 1ϩ0 ϭ mg which is consistent because the object simply hangs at rest in equilibrium between the gravitational force and the tension in the string 10.8 Energy Considerations in Rotational Motion Up to this point in our discussion of rotational motion in this chapter, we focused primarily on an approach involving force, leading to a description of torque on a rigid object In Section 10.4, we discussed the rotational kinetic energy of a rigid 288 Chapter 10 Rotation of a Rigid Object About a Fixed Axis object Let us now extend that initial energy discussion and see how an energy approach can be useful in solving rotational problems We begin by considering the relationship between the torque acting on a rigid object and its resulting rotational motion so as to generate expressions for power and a rotational analog to the work–kinetic energy theorem Consider the rigid S object pivoted at O in Figure 10.20 Suppose a single external force is applied at F S S P, where F lies in the plane of the page The work done on the object by F as its point of application rotates through an infinitesimal distance ds ϭ r du is F f ds du P r dW ϭ F # ds ϭ 1F sin f2r du S O S S Figure 10.20 A rigid object rotates about an axis through O under the S action of an external force F applied at P where F sin f is the tangential component of F, or, in other words, the component S of the force along the displacement Notice that the radial component vector of F does no work on the object because it is perpendicular to the displacement of the S point of application of F S Because the magnitude of the torque due to F about an axis through O is defined as r F sin f by Equation 10.19, we can write the work done for the infinitesimal rotation as (10.22) dW ϭ t du S The rate at which work is being done by F as the object rotates about the fixed axis through the angle du in a time interval dt is dW du ϭt dt dt Because dW/dt is the instantaneous power ᏼ (see Section 8.5) delivered by the force and du/dt ϭ v, this expression reduces to Power delivered to a rotating rigid object ᏼϭ ᮣ dW ϭ tv dt (10.23) This equation is analogous to ᏼ ϭ Fv in the case of translational motion, and Equation 10.22 is analogous to dW ϭ Fx dx In studying translational motion, models based on an energy approach can be extremely useful in describing a system’s behavior From what we learned of translational motion, we expect that when a symmetric object rotates about a fixed axis, the work done by external forces equals the change in the rotational energy of the object To prove that fact, let us begin with ͚ t ϭ Ia Using the chain rule from calculus, we can express the net torque as dv dv du dv a t ϭ Ia ϭ I dt ϭ I du dt ϭ I du v Rearranging this expression and noting that ͚ t du ϭ dW gives a tdu ϭ dW ϭ Iv dv Integrating this expression, we obtain for the total work done by the net external force acting on a rotating system Work–kinetic energy theorem for rotational motion ᮣ aWϭ Ύ vf Iv dv ϭ 12Iv f Ϫ 12Iv i (10.24) vi where the angular speed changes from vi to vf Equation 10.24 is the work–kinetic energy theorem for rotational motion Similar to the work–kinetic energy theorem in translational motion (Section 7.5), this theorem states that the net work done by external forces in rotating a symmetric rigid object about a fixed axis equals the change in the object’s rotational energy This theorem is a form of the nonisolated system model discussed in Chapter Work is done on the system of the rigid object, which represents a transfer of energy across the boundary of the system that appears as an increase in the object’s rotational kinetic energy Section 10.8 Energy Considerations in Rotational Motion 289 TABLE 10.3 Useful Equations in Rotational and Translational Motion Rotational Motion About a Fixed Axis Translational Motion Angular speed v ϭ du/dt Angular acceleration a ϭ dv/dt Net torque ͚ t ϭ Ia If vf ϭ vi ϩ at a ϭ constant c uf ϭ ui ϩ vi t ϩ 12 at vf ϭ vi ϩ 2a(uf Ϫ ui) Translational speed v ϭ dx/dt Translational acceleration a ϭ dv/dt Net force ͚ F ϭ ma If vf ϭ vi ϩ at a ϭ constant c xf ϭ xi ϩ vi t ϩ 12 at vf ϭ vi ϩ 2a(x f Ϫ x i ) uf Work W ϭ Ύ t du Work W ϭ ui Ύ xf Fx dx xi Rotational kinetic energy KR ϭ 21 Iv2 Power ᏼ ϭ tv Angular momentum L ϭ Iv Net torque ͚ t ϭ dL/dt Kinetic energy K ϭ 12 mv Power ᏼ ϭ Fv Linear momentum p ϭ mv Net force ͚ F ϭ dp/dt In general, we can combine this theorem with the translational form of the work–kinetic energy theorem from Chapter Therefore, the net work done by external forces on an object is the change in its total kinetic energy, which is the sum of the translational and rotational kinetic energies For example, when a pitcher throws a baseball, the work done by the pitcher’s hands appears as kinetic energy associated with the ball moving through space as well as rotational kinetic energy associated with the spinning of the ball In addition to the work–kinetic energy theorem, other energy principles can also be applied to rotational situations For example, if a system involving rotating objects is isolated and no nonconservative forces act within the system, the isolated system model and the principle of conservation of mechanical energy can be used to analyze the system as in Example 10.11 below Finally, in some situations an energy approach does not provide enough information to solve the problem and it must be combined with a momentum approach Such a case is illustrated in Example 10.14 in Section 10.9 Table 10.3 lists the various equations we have discussed pertaining to rotational motion together with the analogous expressions for translational motion The last two equations in Table 10.3, involving angular momentum L, are discussed in Chapter 11 and are included here only for the sake of completeness E XA M P L E 1 Rotating Rod Revisited A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end (Fig 10.21) The rod is released from rest in the horizontal position (A) What is its angular speed when the rod reaches its lowest position? SOLUTION Conceptualize Consider Figure 10.21 and imagine the rod rotating downward through a quarter turn about the pivot at the left end Also look back at Example 10.8 This physical situation is the same Categorize As mentioned in Example 10.8, the angular acceleration of the rod is not constant Therefore, the kinematic equations for rotation (Section 10.2) cannot be used to solve this example We categorize the system of the rod and the Earth as an isolated system with no nonconservative forces acting and use the principle of conservation of mechanical energy Ei ϭ U ϭ MgL/2 O L/2 CM Ef ϭ KR ϭ ϪI v 2 Figure 10.21 (Example 10.11) A uniform rigid rod pivoted at O rotates in a vertical plane under the action of the gravitational force 290 Chapter 10 Rotation of a Rigid Object About a Fixed Axis Analyze We choose the configuration in which the rod is hanging straight down as the reference configuration for gravitational potential energy and assign a value of zero for this configuration When the rod is in the horizontal position, it has no rotational kinetic energy The potential energy of the system in this configuration relative to the reference configuration is MgL/2 because the center of mass of the rod is at a height L/2 higher than its position in the reference configuration When the rod reaches its lowest position, the energy of the system is entirely rotational energy 12Iv 2, where I is the moment of inertia of the rod about an axis passing through the pivot Kf ϩ Uf ϭ Ki ϩ Ui Write a conservation of mechanical energy equation for the system: 2 Iv Substitute for each of the energies: Solve for v and use I ϭ 13ML2 (see Table 10.2) for the rod: vϭ ϩ ϭ ϩ 12MgL MgL B I ϭ MgL B 13ML2 ϭ 3g BL (B) Determine the tangential speed of the center of mass and the tangential speed of the lowest point on the rod when it is in the vertical position SOLUTION Use Equation 10.10 and the result from part (A): Because r for the lowest point on the rod is twice what it is for the center of mass, the lowest point has a tangential speed twice that of the center of mass: v CM ϭ r v ϭ L vϭ 2 23gL v ϭ 2vCM ϭ 23gL Finalize The initial configuration in this example is the same as that in Example 10.8 In Example 10.8, however, we could only find the initial angular acceleration of the rod Applying an energy approach in the current example allows us to find additional information, the angular speed of the rod at another instant of time E XA M P L E Energy and the Atwood Machine Two cylinders having different masses m1 and m2 are connected by a string passing over a pulley, as shown in Active Figure 10.22 The pulley has a radius R and moment of inertia I about its axis of rotation The string does not slip on the pulley, and the system is released from rest Find the translational speeds of the cylinders after cylinder descends through a distance h, and find the angular speed of the pulley at this time R SOLUTION Conceptualize We have already seen examples involving the Atwood machine, so the motion of the objects in Active Figure 10.22 should be easy to visualize Categorize Because the string does not slip, the pulley rotates about the axle We can neglect friction in the axle because the axle’s radius is small relative to that of the pulley Hence, the frictional torque is much smaller than the net torque applied by the two cylinders provided that their masses are significantly different Consequently, the system consisting of the two cylinders, the pulley, and the Earth is an isolated system with no nonconservative forces acting; therefore, the mechanical energy of the system is conserved Analyze We define the zero configuration for gravitational potential energy as that which exists when the system is released From Active Figure 10.22, we see m2 h h m1 ACTIVE FIGURE 10.22 (Example 10.12) An Atwood machine with a massive pulley Sign in at www.thomsonedu.com and go to ThomsonNOW to change the masses of the hanging cylinders and the mass and radius of the pulley and see how the cylinders move Section 10.9 Rolling Motion of a Rigid Object 291 that the descent of cylinder is associated with a decrease in system potential energy and that the rise of cylinder represents an increase in potential energy Kf ϩ Uf ϭ Ki ϩ Ui Write a conservation of energy equation for the system: Substitute for each of the energies: Use vf ϭ Rvf to substitute for vf : 12m 1v f ϩ 12m 2v f ϩ 12Iv f 2 ϩ 1m 1gh Ϫ m 2gh2 ϭ ϩ 2 m 1v f Solve for vf : Use vf ϭ Rvf to solve for vf : ϩ 12m 2v f ϩ 12 a m1 ϩ m2 ϩ (1) vf ϭ c vf ϭ vf R ϭ I v f ϭ m 2gh Ϫ m 1gh R2 I b v f ϭ m 2gh Ϫ m 1gh R2 1m Ϫ m 2gh m ϩ m ϩ I>R d 1>2 1m Ϫ m 2gh 1>2 d c R m ϩ m ϩ I>R Finalize Each cylinder can be modeled as a particle under constant acceleration because it experiences a constant net force Think about what you would need to to use Equation (1) to find the acceleration of one of the cylinders and reduce the result so that it matches the result of Example 5.9 Then it and see if it works! 10.9 Rolling Motion of a Rigid Object Henry Leap and Jim Lehman In this section, we treat the motion of a rigid object rolling along a flat surface In general, such motion is complex For example, suppose a cylinder is rolling on a straight path such that the axis of rotation remains parallel to its initial orientation in space As Figure 10.23 shows, a point on the rim of the cylinder moves in a complex path called a cycloid We can simplify matters, however, by focusing on the center of mass rather than on a point on the rim of the rolling object As shown in Figure 10.23, the center of mass moves in a straight line If an object such as a cylinder rolls without slipping on the surface (called pure rolling motion), a simple relationship exists between its rotational and translational motions Consider a uniform cylinder of radius R rolling without slipping on a horizontal surface (Fig 10.24) As the cylinder rotates through an angle u, its center of mass R u s s ϭR u Figure 10.23 One light source at the center of a rolling cylinder and another at one point on the rim illustrate the different paths these two points take The center moves in a straight line (green line), whereas the point on the rim moves in the path called a cycloid (red curve) Figure 10.24 For pure rolling motion, as the cylinder rotates through an angle u its center moves a linear distance s ϭ R u 292 Chapter 10 Rotation of a Rigid Object About a Fixed Axis moves a linear distance s ϭ R u(see Eq 10.1a) Therefore, the translational speed of the center of mass for pure rolling motion is given by PITFALL PREVENTION 10.6 Equation 10.25 Looks Familiar Equation 10.25 looks very similar to Equation 10.10, so be sure to be clear on the difference Equation 10.10 gives the tangential speed of a point on a rotating object located a distance r from a fixed rotation axis if the object is rotating with angular speed v Equation 10.25 gives the translational speed of the center of mass of a rolling object of radius R rotating with angular speed v v CM ϭ ds du ϭR ϭ Rv dt dt (10.25) where v is the angular speed of the cylinder Equation 10.25 holds whenever a cylinder or sphere rolls without slipping and is the condition for pure rolling motion The magnitude of the linear acceleration of the center of mass for pure rolling motion is a CM ϭ dv CM dv ϭR ϭ Ra dt dt (10.26) where a is the angular acceleration of the cylinder Imagine that you are moving along with a rolling object at speed vCM, staying in a frame of reference at rest with respect to the center of mass of the object As you observe the object, you will see the object in pure rotation around its center of mass Figure 10.25a shows the velocities of points at the top, center, and bottom of the object as observed by you In addition to these velocities, every point on the object moves in the same direction with speed vCM relative to the surface on which it rolls Figure 10.25b shows these velocities for a nonrotating object In the reference frame at rest with respect to the surface, the velocity of a given point on the object is the sum of the velocities shown in Figures 10.25a and 10.25b Figure 10.25c shows the results of adding these velocities Notice that the contact point between the surface and cylinder in Figure 10.25c has a translational speed of zero At this instant, the rolling object is moving in exactly the same way as if the surface were removed and the object were pivoted at point P and spun about an axis passing through P We can express the total kinetic energy of this imagined spinning object as K ϭ 12IP v (10.27) where IP is the moment of inertia about a rotation axis through P Because the motion of the imagined spinning object is the same at this instant as our actual rolling object, Equation 10.27 also gives the kinetic energy of the rolling object Applying the parallel-axis theorem, we can substitute IP ϭ ICM ϩ MR into Equation 10.27 to obtain K ϭ 12ICMv ϩ 12MR 2v Using vCM ϭ R v, this equation can be expressed as Total kinetic energy of a rolling object K ϭ 12ICMv ϩ 12Mv CM2 ᮣ (10.28) 2 ICMv The term represents the rotational kinetic energy of the cylinder about its center of mass, and the term 12Mv CM2 represents the kinetic energy the cylinder PЈ CM v ϭRv PЈ v ϭR v v ϭ0 P (a) Pure rotation CM PЈ v CM v CM v CM P (b) Pure translation CM v ϭ v CM ϩ R v ϭ 2v CM v ϭ v CM v ϭ0 P (c) Combination of translation and rotation Figure 10.25 The motion of a rolling object can be modeled as a combination of pure translation and pure rotation Section 10.9 Rolling Motion of a Rigid Object would have if it were just translating through space without rotating Therefore, the total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass This statement is consistent with the situation illustrated in Figure 10.25, which shows that the velocity of a point on the object is the sum of the velocity of the center of mass and the tangential velocity around the center of mass Energy methods can be used to treat a class of problems concerning the rolling motion of an object on a rough incline For example, consider Active Figure 10.26, which shows a sphere rolling without slipping after being released from rest at the top of the incline Accelerated rolling motion is possible only if a friction force is present between the sphere and the incline to produce a net torque about the center of mass Despite the presence of friction, no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant (On the other hand, if the sphere were to slip, mechanical energy of the sphere–incline– Earth system would be lost due to the nonconservative force of kinetic friction.) In reality, rolling friction causes mechanical energy to transform to internal energy Rolling friction is due to deformations of the surface and the rolling object For example, automobile tires flex as they roll on a roadway, representing a transformation of mechanical energy to internal energy The roadway also deforms a small amount, representing additional rolling friction In our problem-solving models, we ignore rolling friction unless stated otherwise Using vCM ϭ R v for pure rolling motion, we can express Equation 10.28 as K ϭ 12ICM a K ϭ 12 a 293 M R v x h u vCM ACTIVE FIGURE 10.26 A sphere rolling down an incline Mechanical energy of the sphere– Earth system is conserved if no slipping occurs Sign in at www.thomsonedu.com and go to ThomsonNOW to roll several objects down the hill and see how the final speed depends on the type of object v CM b ϩ 2Mv CM2 R ICM ϩ M b v CM2 R2 (10.29) For the sphere–Earth system, we define the zero configuration of gravitational potential energy to be when the sphere is at the bottom of the incline Therefore, conservation of mechanical energy gives K f ϩ Uf ϭ K i ϩ Ui ICM 2a R ϩ M b v CM2 ϩ ϭ ϩ Mgh v CM ϭ c 2gh ϩ 1ICM>MR 2 d 1>2 (10.30) Quick Quiz 10.7 A ball rolls without slipping down incline A, starting from rest At the same time, a box starts from rest and slides down incline B, which is identical to incline A except that it is frictionless Which arrives at the bottom first? (a) The ball arrives first (b) The box arrives first (c) Both arrive at the same time (d) It is impossible to determine E XA M P L E Sphere Rolling Down an Incline For the solid sphere shown in Active Figure 10.26, calculate the translational speed of the center of mass at the bottom of the incline and the magnitude of the translational acceleration of the center of mass SOLUTION Conceptualize Imagine rolling the sphere down the incline Compare it in your mind to a book sliding down a frictionless incline You probably have experience with objects rolling down inclines and may be tempted to think that the sphere would move down the incline faster than the book You not, however, have experience with objects sliding down frictionless inclines! So, which object will reach the bottom first? (See Quick Quiz 10.7.)