3once each second. The gentleman bug’s angular speed is A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolutiononce each second. The gentleman bug’s angular speed is1. half the ladybug’s.2. the same as the ladybug’s.3. twice the ladybug’s.4. impossible to determineAnswer: 2. Both insects have an angular speed of 1 rev/s.
4A ladybug sits at the outer edge of a merry-go-round, that is turning and slowing down. At the instant shown in the figure, the radial component of the ladybug’s (Cartesian) acceleration is1. in the +x direction.2. in the –x direction.3. in the +y direction.4. in the –y direction.5. in the +z direction.6. in the –z direction.7. zero.Answer: 2. The radial component of the Cartesian acceleration of a rotatingobject points toward the axis of rotation.
5A ladybug sits at the outer edge of a merry-go-round that is turning and slowing down. The tangential component of the ladybug’s (Cartesian) acceleration is1. in the +x direction.2. in the –x direction.3. in the +y direction.4. in the –y direction.5. in the +z direction.6. in the –z direction.7. zero.Answer: 4.The tangential velocity of the ladybug is in the +y direction. Becausethe merry-go-round is slowing down, the tangential acceleration ofthe ladybug is in the –y direction.
6A ladybug sits at the outer edge of a merry-go-round that is turning and is slowing down. The vector expressing her angular velocity is1. in the +x direction.2. in the –x direction.3. in the +y direction.4. in the –y direction.5. in the +z direction.6. in the –z direction.7. zero.Answer: 5. The direction of the angular velocity vector is given by theright-hand rule.
7A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?Answer: 1.The normal force of the wall on the rider provides the centripetalacceleration necessary to keep her going around in a circle. The downwardforce of gravity is equal and opposite to the upward frictional force on her.
8(a) the frame of the merry-go-round or (b) that of Earth? 1. (a) only Consider two people on opposite sides of a rotating merry-go-round. One of them throws a ball toward the other. In which frame of reference is the path of the ball straight whenviewed from above:(a) the frame of the merry-go-round or(b) that of Earth?1. (a) only2. (a) and (b)—although the paths appear to curve3. (b) only4. neither; because it’s thrown while in circular motion, theball travels along a curved path.Answer: 3. If the reference frame is inertial, the path of the ball must bestraight. Of the two frames, only (b) is inertial, and so only in (b) is thepath straight.
9same force, is the torque you exert increased? 1. yes 2. no You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with thesame force, is the torque you exert increased?1. yes2. noAnswer: 2. Because the force you are applying is unchanged and the perpendiculardistance between the line of action and the pivot point (thelever arm) is likewise unchanged, the torque you apply does not increase.(Pulling at an angle only decreases the lever arm.)
10You are using a wrench and trying to loosen a rusty nut You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? List in order of descending efficiency the following arrangements:Answer: To increase a torque, you can increase either the appliedforce or the moment arm. Here the force is the same in all four situations,and so this question boils down to comparing moment arms.
114. The answer depends on the rotational inertia of the dumbbell. A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?1. (a)2. (b)3. no difference4. The answer depends on the rotational inertia of the dumbbell.Answer: 3. Because the force acts for the same time interval in both cases,the change in momentum ΔP must be the same in both cases, and thus thecenter-of-mass velocity must be the same in both cases.
124. The answer depends on the rotational inertia of the dumbbell. A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?1. (a)2. (b)3. no difference4. The answer depends on the rotational inertia of the dumbbell.Answer: 2. If the center-of-mass velocities are the same, so the translationalkinetic energies must be the same. Because dumbbell (b) is also rotating,it also has rotational kinetic energy.
13Imagine hitting a dumbbell with an object coming in at speed v, first at the center, then at one end. Is the center-of-mass speed of the dumbbell the same in both cases?1. yes2. noAnswer: 2. The moving object comes in with a certain momentum. If ithits the center, as in case (a), there is no rotation, and this collision is justlike a one-dimensional collision between one object of inertial mass mand another of inertial mass 2m. Because of the larger inertial mass of thedumbbell the incoming object bounces back.In (b), the dumbbell, starts rotating. The incoming object encounters less“resistance” and therefore transfers less of its momentum to the dumbbell,which will therefore have a smaller center-of-mass speed than in case (a).
14stick if it is balanced by a support force at the 0.25-m mark? A 1-kg rock is suspended by a massless string from one end of a 1-m measuring stick. What is the mass of the measuringstick if it is balanced by a support force at the 0.25-m mark?kgkg3. 1 kg4. 2 kg5. 4 kg6. impossible to determineAnswer: 3. Because the stick is a uniform, symmetric body,we can considerall its weight as being concentrated at the center of mass at the 0.5-m mark.Therefore the point of support lies midway between the two masses, andthe system is balanced only if the total mass on the right is also 1 kg.
15A box, with its center-of-mass off-center as indicated by the dot, is placed on an inclined plane. In which of the four orientations shown, if any, does the box tip over?Answer: C only. In order to tip over, the box must pivot about its bottomleft corner. Only in case C does the torque about this pivot (due to gravity)rotate the box in such a way that it tips over.
16Consider the situation shown at left below Consider the situation shown at left below. A puck of mass m, moving at speed v hits an identical puck which is fastened to a pole using a string of length r. After the collision, the puck attached to the string revolves around the pole. Suppose we now lengthen the string by a factor 2, as shown on the right, and repeat the experiment. Compared to the angular speed in the first situation, the new angular speed is1. twice as high2. the same3. half as much4. none of the aboveAnswer: 3. If we ignore the effect of the string during the collision betweenthe two pucks, the incoming puck comes to rest and the puck on thestring gets a speed v. Because of the string, however, this puck can’t movein a straight line, but is instead constrained to moving in a circle. Becauseits speed is v, it takes a time interval 2πr/v to complete one revolutionaround the axis. The angular speed of the puck is v = ωr, and so ω = v/r.When we lengthen the string by a factor of 2, the circle made by the puckgets twice as large. Because the collision still gives the puck the samespeed, it now takes the puck twice as long to complete one revolution. Itsangular speed is therefore ω = v/2r.
172. larger because she’s rotating faster. A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational inertia andher angular speed increases so that her angular momentum is conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be1. the same.2. larger because she’s rotating faster.3. smaller because her rotational inertia is smaller.Answer: 2. Rotational kinetic energy is 1⁄2 Iω2. Substituting L, the angularmomentum, for Iω, we find Krot = 1⁄2 Lω. We know (a) that L is constantbecause the angular momentum is conserved in this isolated systemand (b) that ω increases. Therefore Krot must increase. The “extra” energycomes from the work she does on her arms: She has to pull them inagainst their tendency to continue in a straight line, thereby doing workon them. If you prefer to consider the skater and her arms as one system,the extra energy comes from potential energy.
18the center. Which reaches the bottom of the incline first? 1. A 2. B Two cylinders of the same size and mass roll down an incline. Cylinder A has most of its weight concentrated at the rim, while cylinder B has most of its weight concentrated atthe center. Which reaches the bottom of the incline first?1. A2. B3. Both reach the bottom at the same time.Answer: 2.When the cylinders roll down the incline, gravitational potentialenergy gets converted to translational and rotational kinetic energy:mgh = 1⁄2 mv2 + 1⁄2 Iω2. We can substitute v/r for ω and write mgh =1⁄2 (m + I/r2)v2. The values for m and r are the same for both cylinders,and so the cylinder having the smaller rotational inertia has the largerspeed. The smaller rotational inertia is obtained when more of the massis concentrated near the center.
191. mring= mdisk, where m is the inertial mass. A solid disk and a ring roll down an incline. The ring is slower than the disk if1. mring= mdisk, where m is the inertial mass.2. rring = rdisk, where r is the radius.3. mring = mdisk and rring = rdisk.4. The ring is always slower regardless of the relative valuesof m and r.Answer: 4. The ring has more rotational inertia per unit mass than thedisk. Therefore as it starts rolling, it has a relatively larger fraction of itstotal kinetic energy in rotational form and so its translational kinetic energyis lower than that of the disk.As the disk and ring roll down the incline, the potential energy of each is reducedby an amount mgh, where h is the difference in height between thebottom and top of the incline.This energy is converted to kinetic energy (translationaland rotational): mgh = 1⁄2 mv2 + 1⁄2 Iω2.We can write the rotationalinertia as I= cmR2,where c is a constant equal to 1 for the ring and 1⁄2 for thedisk.Because ω= v/R,we have:mgh= 1⁄2mv2+ 1⁄2 cmR2(v/R)2= 1⁄2 (1 + c)mv2.The larger c, therefore, the smaller v.Thus the ring, which has the larger rotationalinertia, takes longer to go down the incline, regardless of inertiaand radius.
20Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertiais I = mR2. In order to impart identical angular accelerations, how large must F2 be?NN3. 1 N4. 2 N5. 4 NAnswer: 4. For each wheel, the moment of inertia times the angular accelerationmust equal the force on the wheel rim times the distance fromthe rim to the center of the wheel.
213. The kinetic energies are the same. 4. need more information Two wheels initially at rest roll the same distance without slipping down identical inclined planes starting from rest. Wheel B has twice the radius but the same mass aswheel A. All the mass is concentrated in their rims, so that the rotational inertias are I = mR2. Which has more translational kinetic energy when it gets to the bottom?1. Wheel A2. Wheel B3. The kinetic energies are the same.4. need more informationAnswer: 3. For each wheel, the gain in total kinetic energy (translationalplus rotational) equals the loss in gravitational potential energy. Becausethe two wheels have identical mass and roll down the same distance theyboth lose the same amount of potential energy. Both wheels also have thesame ratio of translational to rotational kinetic energy, so their translationalkinetic energies are the same.
224. yes, some other direction 5. no, the choice is really arbitrary Consider the uniformly rotating object shown below. If the object’s angular velocity is a vector (in other words, it points in a certain direction in space) is there a particular direction we should associate with the angular velocity?1. yes, ±x2. yes, ±y3. yes, ±z4. yes, some other direction5. no, the choice is really arbitraryAnswer: 3. Only the z direction—the direction perpendicular to the planeof rotation—is unique; the x and y directions represent only the instantaneousdirection in which the object is moving (i.e., at the instant representedin the diagram). The z direction has the same relationship to theinstantaneous velocity at all times during the motion.
23This causes a change in angular momentum DL in the A person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow in the forward direction.This causes a change in angular momentum DL in the1. x direction2. y direction3. z directionAnswer: 3.The force F makes the ball go around the circle faster than before,and so the ball’s angular momentum increases in magnitude withoutchanging direction. According to the right-hand rule, the angularvelocity is in the +z direction, and so the change in angular momentummust also be in the +z direction.
24which direction does the axis of rotation tilt after the blow? A person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow vertically downward. Inwhich direction does the axis of rotation tilt after the blow?1. +x direction2. –x direction3. +y direction4. –y direction5. It stays the same (but the magnitude of the angularmomentum changes).6. The ball starts wobbling in all directions.Answer: 1. As it receives the blow, the ball accelerates downward and itsvelocity gains a downward component. Still constrained by the string to goaround in a circle, the ball moves in a new circular path the plane of whichis determined by the instantaneous velocity v the ball has immediatelyafter the blow. This velocity v′ defines a plane that is tilted downward inthe forward direction (+x direction). The angular momentum thereforealso tilts forward in the +x direction. The change in angular momentum isthe dashed vector connecting L and L′.Explanation using torque: According to the right-hand rule, the torqueexerted by the force is in the forward direction (+x), and so the change inangular momentum must also be in this direction.
25ExplanationAnswer: 1. As it receives the blow, the ball accelerates downward and itsvelocity gains a downward component. Still constrained by the string to goaround in a circle, the ball moves in a new circular path the plane of whichis determined by the instantaneous velocity v the ball has immediatelyafter the blow. This velocity v′ defines a plane that is tilted downward inthe forward direction (+x direction). The angular momentum thereforealso tilts forward in the +x direction. The change in angular momentum isthe dashed vector connecting L and L′.Explanation using torque: According to the right-hand rule, the torqueexerted by the force is in the forward direction (+x), and so the change inangular momentum must also be in this direction.
26A suitcase containing a spinning flywheel is rotated about the vertical axis as shown in (a). As it rotates, the bottom of the suitcase moves out and up, as in (b). From this, we can conclude that the flywheel, as seen from the side of the suitcase as in (a), rotates1. clockwise.2. counterclockwise.Answer: 1. To rotate the suitcase in the direction indicated, a downwardtorque must be exerted on it. This downward torque gives a downwardcomponent to the angular momentum. Because the angularmomentum initially points into the page, this downward componentcauses the suitcase to tilt as indicated. The flywheel must therefore bespinning clockwise.