ROTATIONAL CHAPTER
9 DYNAMICS
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION The magnitude of the torque produced by a force F is given by the magnitude of the force times the lever arm. When a long pipe is slipped over a wrench handle, the length of the wrench handle is effectively increased. The same force can then be applied with a larger lever arm. Therefore, a greater torque can be used to remove the "frozen" nut. REASONING AND SOLUTION where is the lever arm. The torque produced by a force F is given by F
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2.
a. A large force can be used to produce a small torque by choosing a small lever arm. Then the product is small even though the magnitude of the force, F, is large. If the line of F action passes through the axis of rotation, = 0 m, and the torque is zero regardless of how large F is. b. A small force may be used to produce a large torque by choosing a large lever arm . Then, the product can be large even though the magnitude of the force F is small. F REASONING AND SOLUTION In a cassette deck, the magnetic tape applies a torque to the supply reel. The magnitude of the torque is equal to the product of the tension FT in the tape, assumed to be constant, and the lever arm . As shown below, the lever arm is the radius of the tape remaining on the supply reel. Figure A shows the reel and tape at an earlier time; figure B shows the tape and the reel at a later time.
FT
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3.
F
T
Figure A
Figure B
As the reel becomes empty, the lever arm gets smaller, as suggested in the figures above. Therefore, the torque exerted on the reel by the tape becomes smaller.
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Chapter 9 Conceptual Questions
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4.
REASONING AND SOLUTION A flat rectangular sheet of plywood is fixed so that it can rotate about an axis perpendicular to the sheet through one corner, as shown in the figure. In order to produce the largest possible torque with the force F (acting in the plane of the paper), the force F must be applied with the largest possible lever arm. This can be attained by orienting the force F so that it is perpendicular to the diagonal that passes through the rotation axis as shown below. The lever arm is equal to the length of the diagonal of the plywood.
Axis of rotation
F
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5.
REASONING AND SOLUTION To unscrew either cap, you apply a force tangentially to it and rotate it about an axis perpendicular to the end of the cap at its center. It is the torque created by this force that turns the cap. The greater the torque, the more easily is the cap unscrewed. For a given force, a greater torque is created when the force has a greater lever arm than when it has a smaller lever arm. In the case of the two caps, the lever arms are the radii of the caps, and the "easy-off" cap has a greater radius than the normal cap. Thus, it is the greater radius of the "easy-off" cap that makes it easier to open. REASONING AND SOLUTION A person stands on a train, both feet together, facing a window. The front of the train is to the person's left. When the train starts to move forward, a force of static friction is applied to the person's feet. This force tends to produce a torque about the center of gravity of the person, causing the person to fall backward. If the right foot is slid out toward the rear of the train, the normal force of the floor on the right foot produces a counter torque about the center of gravity of the person. The resultant of these two torques is zero, and the person can maintain his balance. REASONING AND SOLUTION If we assume that the branch is fairly uniform, then in the spring, just when the fruit begins to grow, the center of gravity of the branch is near the center of the branch. As the fruit grows, the mass at the outer end of the branch increases. Therefore, as the fruit grows, the center of gravity of the fruit-growing branch moves along the branch toward the fruit.
6.
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7.
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426 ROTATIONAL DYNAMICS
8.
SSM REASONING AND SOLUTION For a rigid body in equilibrium, there can be no net torque acting on the object with respect to any axis. Consider an axis perpendicular to the plane of the paper at the right end of the rod. The lines of action of both F1 and F3 pass through this axis. Therefore, no torque is created by either of these forces. However, the force F2 does create a torque with respect to this axis. As a result, there is a net torque acting on the object with respect to at least one axis, and the three forces shown in the drawing cannot keep the object in equilibrium. REASONING AND SOLUTION The forces that keep the right section of the ladder in equilibrium are shown in the free-body diagram at the right. They are: the weight of the section, mg, the normal force exerted on the ladder by the floor, FN, the tension in the crossbar, FT, and the reaction force due to the left side of the ladder, FR.
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9.
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10. REASONING AND SOLUTION Treating the wine rack and the bottle as a rigid body, the two external forces that keep it in equilibrium are its weight, mg, located at the center of gravity, and the normal force, FN, exerted on the base by the table. Note that the total weight of the wine rack and the bottle acts at the center of gravity (CG), which must be located exactly above the point where the normal force acts.
CG
mg
N
FN
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11. REASONING AND SOLUTION The torque required to give the triangular sheet an angular acceleration is given by the rotational analog of Newton's second law for a rigid body: I . The greatest net torque will be required about the axis for which the moment of inertia I is the greatest. The greatest value for I occurs when the greatest portion of the mass of the sheet is far from the axis. This will be for axis B. Therefore, for axis B, the greatest net external torque is required to bring the rectangular plate up to 10.0 rad/s in 10.0 s, starting from rest.
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Chapter 9 Conceptual Questions
427
12. REASONING AND SOLUTION The object has an angular velocity and an angular acceleration due to torques that are present. a. If additional torques are applied so as to make the net torque suddenly equal to zero, the object will continue to rotate with constant angular velocity. The angular velocity will remain constant at the value that it had at the instant that the net torque became zero. b. If all the torques are suddenly removed, the effect is the same as when the net torque is zero as in part (a) above. That is, the object will continue to rotate with constant angular velocity. The angular velocity will remain constant at the value that it had at the instant that the torques are removed.
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13. REASONING AND SOLUTION The space probe is initially moving with a constant translational velocity and zero angular velocity through outer space. a. When the two engines are fired, each generates a thrust T in opposite directions; hence, the net force on the space probe is zero. Since the net force on the probe is zero, there is no translational acceleration and the translational velocity of the probe remains the same. b. The thrust of each engine produces a torque about the center of the space probe. If R is the radius of the probe, the thrust of each engine produces a torque of TR, each in the same direction, so that the net torque on the probe is 2TR. Since there is a net torque on the probe, there will be an angular acceleration. The angular velocity of the probe will increase.
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14. SSM REASONING AND SOLUTION When you are lying down with your hands behind your head, your moment of inertia I about your stomach is greater than it is when your hands are on your stomach. Sit-ups are then more difficult because, with a greater I, you must exert a greater torque about your stomach to give your upper torso the same angular acceleration, according to Equation 9.7.
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15. REASONING AND SOLUTION The translational kinetic energy of a rigid body depends only on the mass and the speed of the body. It does not depend on how the mass is distributed. Therefore, for purposes of computing the body's translational kinetic energy, the mass of a rigid body can be considered as concentrated at its center of mass. Unlike the translational kinetic energy, the moment of inertia of a body does depend on the location and orientation of the axis relative to the particles that make up the rigid body. Therefore, for purposes of computing the body's moment of inertia, the mass of a rigid body cannot be considered as concentrated at its center of mass. The distance of any individual mass particle from the axis is important.
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428 ROTATIONAL DYNAMICS
16. REASONING AND SOLUTION A thin sheet of plastic is uniform and has the shape of an equilateral triangle, as shown in the figure at the right. The axes A and B are two rotation axes that are perpendicular to the plane of the triangle. If the triangle rotates about each axis with the same angular speed , the rotational 1 kinetic energy of the sheet is given by KE I 2 .
2
The rotational kinetic energy will be greater if the rotation occurs about the axis for which the triangle has the larger moment of inertia I. The value for I will be greater when more of the mass is located farther away from the axis, as is the case for axis B. Therefore, the triangle has the greater rotational kinetic energy when it rotates about axis B.
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17. REASONING AND SOLUTION With feet flat on the floor and body straight, you can lean over until your center of gravity is above your toes. If you lean any farther, it is no longer possible to balance the torque due to your weight with a torque created by the upward reaction force that pushing downward with your toes produces. Bob's center of gravity must be closer to the ground. The reason is that, with the center of gravity closer to the ground, a greater amount of leaning can occur before it moves beyond his toes.
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18. REASONING AND SOLUTION Let each object have mass M and radius R. The objects are initially at rest. In rolling down the incline, the objects lose gravitational potential energy Mgh , where h is the height of the incline. The objects gain kinetic energy equal to ( 1 / 2 ) I 2 ( 1 / 2 ) Mv 2 , where I is the moment of inertia about the center of the object, v is the linear speed of the center of mass at the bottom of the incline, and is the angular speed about the center of mass at the bottom. From conservation of mechanical energy
1 1 Mgh 2 I 2 2 Mv 2
Since the objects roll down the incline, = v/R. For the hoop, I MR 2 , hence,
1 Mgh 2 ( MR 2 )
v2 1 Mv 2 R2 2
Solving for v gives v gh for the hoop. Repeating the calculation for the solid cylinder, the spherical shell, and the solid sphere gives
v gh
(hoop) (solid cylinder) (spherical shell) (solid sphere)
v 1.15 gh
v 1.10 gh
v 1.20 gh
Chapter 9 Conceptual Questions
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The solid sphere is faster at the bottom than any of the other objects. The solid cylinder is faster at the bottom than the spherical shell or the hoop. The spherical shell is faster at the bottom than the hoop. Thus, the solid sphere reaches the bottom first, followed by the solid cylinder, the spherical shell, and the hoop, in that order.
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19. REASONING AND SOLUTION A woman sits on the spinning seat of a piano stool with her arms folded. Let the woman and the piano stool comprise the system. When the woman extends her arms, the torque that she applies is an internal torque. Since there is no net external torque acting on the system, the angular momentum of the woman and the piano stool, I, must remain constant. a. With her arms extended, the woman's moment of inertia I about the rotation axis is greater than it was when her arms were folded. Since the product of I must remain constant and I increases, must decrease. Therefore, the woman's angular velocity decreases. b. As discussed above, since there is no net external torque acting on the system, the angular momentum of the system remains the same.
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20. REASONING AND SOLUTION Consider the earth to be an isolated system. Note that the earth rotates about an axis that passes through the North and South poles and is perpendicular to the plane of the equator. If the ice cap at the South Pole melted and the water were uniformly distributed over the earth's oceans, the mass at the South Pole would be uniformly distributed and, on average, be farther from the earth's rotational axis. The moment of inertia of the earth would, therefore, increase. Since the earth is an isolated system, any torques involved in the redistribution of the water would be internal torques; therefore, the angular momentum of the earth must remain the same. If the moment of inertia increases, and the angular momentum is to remain constant, the angular velocity of the earth must decrease.
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21. REASONING AND SOLUTION Note that the earth rotates about an axis that passes through the North and South poles and is perpendicular to the plane of the equator. When rivers like the Mississippi carry sediment toward the equator, they redistribute the mass from a more uniform distribution to a distribution with more mass concentrated around the equator. This increases the moment of inertia of the earth. If the earth is considered to be an isolated system, then any torques involved in the redistribution of mass are internal torques. Therefore, the angular momentum of the earth must remain constant. The moment of inertia increases, and the angular momentum must remain the same; therefore, the angular velocity must decrease.
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430 ROTATIONAL DYNAMICS
22. REASONING AND SOLUTION Let the system be the rotating cloud of interstellar gas. The gravitational force that pulls the particles together is internal to the system; hence, the torque resulting from the gravitational force is an internal torque. Since there is no net external torque acting on the collapsing cloud, the angular momentum, I , of the cloud must remain constant. Since the cloud is shrinking, its moment of inertia I decreases. Since the product I remains constant, the angular velocity must increase as I decreases. Therefore, the angular velocity of the formed star would be greater than the angular velocity of the original gas cloud.
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23. SSM REASONING AND SOLUTION A person is hanging motionless from a vertical rope over a swimming pool. She lets go of the rope and drops straight down. Consider the woman to be the system. When she lets go of the rope her angular momentum is zero. It is possible for the woman to curl into a ball; however, since her angular momentum is zero, she cannot change her angular velocity by exerting any internal torques. In order to begin spinning, an external torque must be applied to the woman. Therefore, it is not possible for the woman to curl into a ball and start spinning.
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24. REASONING AND SOLUTION The box that creates the greatest torque is the one that appears shaded in the drawing at the right. All boxes give rise to the same force mg. Each box produces a torque about the axle. The box that produces the greatest torque is the one for which the lever arm is the greatest. By inspection of the drawing, the lever arm relative to the axle is greatest for the shaded box.
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Chapter 9 Problems
431
CHAPTER 9 ROTATIONAL DYNAMICS
PROBLEMS
______________________________________________________________________________ 1. SSM REASONING According to Equation 9.1, we have Magnitude of torque = F where F is the magnitude of the applied force and is the lever arm. From the figure in the text, the lever arm is given by (0.28 m) sin 50.0 . Since both the magnitude of the torque and are known, Equation 9.1 can be solved for F. SOLUTION Solving Equation 9.1 for F, we have
Magnitude of torque 45 N m 2.1102 N (0.28 m) sin 50.0 ______________________________________________________________________________ F
2. REASONING AND SOLUTION According to Equation 9.1, we have
Magnitude of torque = F where F is the magnitude of the applied force and is the lever arm. For an axis passing through the cable car at its center and perpendicular to the ground, the lever arm is one-half the length of the car or 4.60 m. Both people rotate the car in the same direction, so the torques that they create reinforce one another. Therefore, using Equation 9.1 to express the magnitude of each torque, we find that the magnitude of the net torque is
Magnitude of net torque 185 N 4.60 m 185 N 4.60 m 1.70 103 N m ______________________________________________________________________________
3. REASONING The drawing shows the wheel as it rolls to the right, so the torque applied by the engine is assumed to be clockwise about the axis of rotation. The force of static friction that the ground applies to the wheel is labeled as fs. This force
Axis of rotation
produces a counterclockwise torque about the axis of rotation, which is given by Equation 9.1 as = fs , where is the lever arm. Using this relation we can find the magnitude fs of the static frictional force.
r
fs
SOLUTION The countertorque is given as = fs , where fs is the magnitude of the static frictional force and is the lever arm. The lever arm is the distance between the line of
432 ROTATIONAL DYNAMICS
action of the force and the axis of rotation; in this case the lever arm is just the radius r of the tire. Solving for fs gives 295 N m fs 843 N 0.350 m ______________________________________________________________________________ 4. REASONING In both parts of the problem, the magnitude of the torque is given by Equation 9.1 as the magnitude F of the force times the lever arm . In part (a), the lever arm is just the distance of 0.55 m given in the drawing. However, in part (b), the lever arm is less than the given distance and must be expressed using trigonometry as 0.55 m sin . See the drawing at the right. SOLUTION a. Using Equation 9.1, we find that
axis
0.55 m
lever arm 49 N
Magnitude of torque F 49 N 0.55 m 27 N m
b. Again using Equation 9.1, this time with a lever arm of 0.55 m sin , we obtain
Magnitude of torque 15 N m F 49 N 0.55 m sin sin 15 N m 49 N 0.55 m or
sin 1
15 N m 34 49 N 0.55 m
5.
SSM REASONING The torque on either wheel is given by Equation 9.1, F , where F is the magnitude of the force and is the lever arm. Regardless of how the force is applied, the lever arm will be proportional to the radius of the wheel. SOLUTION The ratio of the torque produced by the force in the truck to the torque produced in the car is
truck F truck Frtruck rtruck 0.25 m 1.3 car F car Frcar rcar 0.19 m
Chapter 9 Problems
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6.
REASONING To calculate the torques, we need to determine the lever arms for each of the forces. These lever arms are shown in the following drawings: Axis 2.50 m 32.0 2.50 m T W = (2.50 m) sin 32.0 Axis 32.0 T = (2.50 m) cos 32.0 W
SOLUTION a. Using Equation 9.1, we find that the magnitude of the torque due to the weight W is
Magnitude of torque W W 10 200 N 2.5 m sin 32 13 500 N m
b. Using Equation 9.1, we find that the magnitude of the torque due to the thrust T is
Magnitude of torque T T 62 300 N 2.5 m cos 32 132 000 N m
______________________________________________________________________________ 7. SSM REASONING Since the meter stick does not move, it is in equilibrium. The forces and the torques, therefore, each add to zero. We can determine the location of the 6.00 N force, by using the condition that the sum of the torques must add to zero.
F pinned end
2
30.0
2
1
F
TOP VIE W 1
SOLUTION If we take counterclockwise torques as positive, then the torque of the first force about the pin is 1 F1 1 (2.00 N)(1.00 m) = 2.00 N m The torque due to the second force is
2 F2 (sin 30.0 ) 2 (6.00 N)(sin 30.0 ) 2 = ( 3.00 N) 2
434 ROTATIONAL DYNAMICS
The torque 2 is negative because the force F2 tends to produce a clockwise rotation about the pinned end. Since the net torque is zero, we have
2.00 N m +[ ( 3.00 N) 2 ] 0
Thus,
2
2.00 N m 0.667 m 3.00 N
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8.
REASONING Each of the two forces produces a torque about the axis of rotation, one clockwise and the other counterclockwise. By setting the sum of the torques equal to zero 0 , we will be able to determine the distance x in the drawing
2
Meter Stick x Axis of Rotation
90 F2 = 6.00 N 60.0
F1 = 4.00 N
Table (Top view) SOLUTION
1 = +F1 1 , where the lever arm is 1 1.00 m It is a positive torque, since it tends to produce a counterclockwise rotation. The torque 2 produced by F2 is 2 = F2 2 , where
2 x sin 60.0 It is a negative torque, since it tends to produce a clockwise rotation. Setting the net torque equal to zero, we have
+ F11 + F2 2 = 0
The torque 1 produced by the force F1 is given by Equation 9.1 as
or
+F1 1.00 m F2 x sin 60.0 = 0 1 2
Solving for x gives
0.770 m F2 sin 60.0 ______________________________________________________________________________ x F1 1.00 m
4.00 N 1.00 m 6.00 N sin 60.0
9.
REASONING AND SOLUTION The torque produced by each force of magnitude F is given by Equation 9.1, F , where is the lever arm and the torque is positive since each force causes a counterclockwise rotation. In each case, the torque produced by the couple is equal to the sum of the individual torques produced by each member of the couple.
Chapter 9 Problems
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a. When the axis passes through point A, the torque due to the force at A is zero. The lever arm for the force at C is L. Therefore, taking counterclockwise as the positive direction, we have A C 0 FL FL b. Each force produces a counterclockwise rotation. The magnitude of each force is F and each force has a lever arm of L / 2 . Taking counterclockwise as the positive direction, we have L L A C F F FL 2 2 c. When the axis passes through point C, the torque due to the force at C is zero. The lever arm for the force at A is L. Therefore, taking counterclockwise as the positive direction, we have A C FL 0 FL Note that the value of the torque produced by the couple is the same in all three cases; in other words, when the couple acts on the tire wrench, the couple produces a torque that does not depend on the location of the axis. ______________________________________________________________________________ 10. REASONING AND SOLUTION The net torque about the axis in text drawing (a) is = 1 + 2 = F1b F2a = 0 Considering that F2 = 3F1, we have b 3a = 0. The net torque in drawing (b) is then = F1(1.00 m a) F2b = 0 or 1.00 m a 3b = 0
Solving the first equation for b, substituting into the second equation and rearranging, gives a = 0.100 m 11. REASONING The drawing shows the forces acting on the person. It also shows the lever arms for a rotational axis perpendicular to the plane of the paper at the place where the person's toes touch the floor. Since the person is in equilibrium, the sum of the forces must be zero. Likewise, we know that the sum of the torques must be zero. and b = 0.300 m
FFEET
FHANDS
Axis W W = 0.840 m HANDS = 1.250 m
436 ROTATIONAL DYNAMICS
SOLUTION Taking upward to be the positive direction, we have
FFEET FHANDS W 0
Remembering that counterclockwise torques are positive and using the axis and the lever arms shown in the drawing, we find
W W FHANDS HANDS 0 FHANDS W W HANDS 584 b N g0.840 m g 392 N b 1.250 m
Substituting this value into the balance-of-forces equation, we find
FFEET W FHANDS 584 N 392 N = 192 N
The force on each hand is half the value calculated above, or 196 N . Likewise, the force on each foot is half the value calculated above, or 96 N . ______________________________________________________________________________ 12. REASONING Although this arrangement of body parts is vertical, we can apply Equation 9.3 to locate the overall center of gravity by simply replacing the horizontal position x by the vertical position y, as measured relative to the floor. SOLUTION Using Equation 9.3, we have
y cg
W1 y1 W2 y 2 W3 y 3 W1 W2 W3 438 1 87 b b N g.28 mgb N g0.760 mgb N g0.250 mg b 144 b
1.03 m 438 N 144 N 87 N ______________________________________________________________________________
13. SSM REASONING The drawing shows the bridge and the four forces that act on it: the upward force F1 exerted on the left end by the support, the force due to the weight Wh of the hiker, the weight Wb of the bridge, and the upward force F2 exerted on the right side by the support. Since the bridge is in equilibrium, the sum of the torques about any axis of rotation must be zero 0 , and the sum of the forces in the vertical direction must be zero Fy 0 . These two conditions will allow us to determine the magnitudes of F1 and F2.
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F1 Wh
1 5
F2 Axis
+y + +x
L
Wb
1 2
L
SOLUTION a. We will begin by taking the axis of rotation about the right end of the bridge. The torque produced by F2 is zero, since its lever arm is zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, F1, in it. Setting the sum of the torques produced by the three forces equal to zero gives
F1L Wh
4 5 L Wb 1 L 0 2
Algebraically eliminating the length L of the bridge from this equation and solving for F1 gives 4 4 F1 5 Wh 1 Wb 5 985 N 1 3610 N 2590 N 2 2 b. Since the bridge is in equilibrium, the sum of the forces in the vertical direction must be zero: Fy F1 Wh Wb F2 0 Solving for F2 gives
F2 F1 Wh Wb 2590 N + 985 N + 3610 N = 2010 N ______________________________________________________________________________
14. REASONING Multiple-Concept Example 8 discusses the static stability factor (SSF) and rollover. In that example, it is determined that the maximum speed v at which a vehicle can negotiate a curve of radius r is related to the SSF according to v rg SSF . No value is given for the radius of the turn. However, by applying this result separately to the sport utility vehicle (SUV) and to the sports car (SC), we will be able to eliminate r algebraically and determine the maximum speed at which the sports car can negotiate the curve without rolling over. SOLUTION Applying v rg SSF to each vehicle, we obtain
vSUV rg SSFSUV
and
vSC rg SSF SC
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Dividing these two expressions gives
vSC vSUV
rg SSF SC rg SSF SUV
or
vSC vSUV
SSFSC 18 m/s SSFSUV
1.4 24 m/s 0.80
15. REASONING Since the forearm is in equilibrium, the sum of the torques about any axis of rotation must be zero 0 . For convenience, we will take the elbow joint to be the axis of rotation. SOLUTION Let M and F be the magnitudes of the forces that the flexor muscle and test apparatus, respectively, exert on the forearm, and let M and F be the respective lever arms about the elbow joint. Setting the sum of the torques about the elbow joint equal to zero (with counterclockwise torques being taken as positive), we have
M M F F 0
Solving for M yields
M F F 190 N 0.34 m 1200 N M 0.054 m
The direction of the force is to the left . ______________________________________________________________________________ 16. REASONING AND SOLUTION between the tray and the thumb is The net torque about an axis through the contact point
= F(0.0400 m) (0.250 kg)(9.80 m/s2)(0.320 m) (1.00 kg)(9.80 m/s2)(0.180 m) (0.200 kg)(9.80 m/s2)(0.140 m) = 0
F 70.6 N, up
Similarly, the net torque about an axis through the point of contact between the tray and the finger is = T (0.0400 m) (0.250 kg)(9.80 m/s2)(0.280 m) (1.00 kg)(9.80 m/s2)(0.140 m) (0.200 kg)(9.80 m/s2)(0.100 m) = 0
T 56.4 N, down
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Chapter 9 Problems
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17. SSM REASONING The figure at the right shows the door and the forces that act upon it. Since the door is uniform, the center of gravity, and, thus, the location of the weight W, is at the geometric center of the door. Let U represent the force applied to the door by the upper hinge, and L the force applied to the door by the lower hinge. Taking forces that point to the right and forces that point up as positive, we have
Fx Lx U x 0 or Lx U x
(1) (2)
Fy Ly W 0
or
Ly W 140 N
Taking torques about an axis perpendicular to the plane of the door and through the lower hinge, with counterclockwise torques being positive, gives
d W U x h 0 2
(3)
Equations (1), (2), and (3) can be used to find the force applied to the door by the hinges; Newton's third law can then be used to find the force applied by the door to the hinges. SOLUTION a. Solving Equation (3) for Ux gives
Ux
Wd (140 N)(0.81 m) 27 N 2h 2(2.1 m)
From Equation (1), Lx = 27 N. Therefore, the upper hinge exerts on the door a horizontal force of 27 N to the left . b. From Equation (1), Lx = 27 N, we conclude that the bottom hinge exerts on the door a horizontal force of 27 N to the right . c. From Newton's third law, the door applies a force of 27 N on the upper hinge. This force points horizontally to the right , according to Newton's third law. d. Let D represent the force applied by the door to the lower hinge. Then, from Newton's third law, the force exerted on the lower hinge by the door has components Dx = 27 N and
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Dy = 140 N. From the Pythagorean theorem,
D
2 2 Dx D y ( 27N) 2 ( 140N) 2
143 N
The angle is given by
tan 1
F N I 79 140 G NJ 27 H K
The force is directed 79 below the horizontal .
18. REASONING AND SOLUTION The net torque about an axis through the elbow joint is = (111 N)(0.300 m) (22.0 N)(0.150 m) (0.0250 m)M = 0 or M 1.20 103 N ______________________________________________________________________________ 19. REASONING The jet is in equilibrium, so the sum of the external forces is zero, and the sum of the external torques is zero. We can use these two conditions to evaluate the forces exerted on the wheels. SOLUTION a. Let Ff be the magnitude of the normal force that the ground exerts on the front wheel. Since the net torque acting on the plane is zero, we have (using an axis through the points of contact between the rear wheels and the ground) = W w + Ff f = 0 where W is the weight of the plane, and w and f are the lever arms for the forces W and Ff , respectively. Thus, = N)(15.0 m 12.6 m) + Ff (15.0 m) = 0 Solving for Ff gives Ff = 1.60 10 5 N . b. Setting the sum of the vertical forces equal to zero yields Fy = Ff + 2Fr W = 0
Chapter 9 Problems
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where the factor of 2 arises because there are two rear wheels. Substituting the data gives Fy = 1.60 105 N + 2Fr 1.00 106 N = 0 Fr = 4.20 105 N ______________________________________________________________________________ 20. REASONING The drawing shows the beam and the five forces that act on it: the horizontal and vertical components Sx and Sy that the wall exerts on the left end of the beam, the weight Wb of the beam, the force due to the weight Wc of the crate, and the tension T in the cable. The beam is uniform, so its center of gravity is at the center of the beam, which is where its weight can be assumed to act. Since the beam is in equilibrium, the sum of the torques about any axis of rotation must be zero 0 , and the sum of the forces in the horizontal and vertical directions must be zero Fx 0, Fy 0 . These three conditions will allow us to determine the magnitudes of Sx, Sy, and T.
T +y 50.0 30.0 Wc Wb 30.0 Sx + +x
Sy Axis
SOLUTION a. We will begin by taking the axis of rotation to be at the left end of the beam. Then the torques produced by Sx and Sy are zero, since their lever arms are zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, T, in it. Setting the sum of the torques produced by the three forces equal to zero gives (with L equal to the length of the beam)
Wb
1 L cos30.0 Wc L cos30.0 T L sin 80.0 0 2
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Algebraically eliminating L from this equation and solving for T gives
T Wb
1 cos 30.0 Wc cos 30.0 2
sin 80.0 sin 80.0 2260 N
1220 N 1 cos 30.0 1960 N cos 30.0 2
b. Since the beam is in equilibrium, the sum of the forces in the vertical direction must be zero: Fy S y Wb Wc T sin 50.0 0 Solving for Sy gives
S y Wb Wc T sin 50.0 N + 1960 N 2260 N sin 50.0 1450 N
The sum of the forces in the horizontal direction must also be zero:
Fx S x T cos50.0 0
S x T cos50.0 2260 N cos50.0 N ______________________________________________________________________________
21. SSM WWW REASONING AND SOLUTION The figure below shows the massless board and the forces that act on the board due to the person and the scales.
so that
a. Applying Newton's second law to the vertical direction gives 315 N + 425 N W = 0 or W = 7.40 10 2 N, downward
b. Let x be the position of the center of gravity relative to the scale at the person's head. Taking torques about an axis through the contact point between the scale at the person's head and the board gives (315 N)(2.00 m) (7.40 102 N)x = 0 or x = 0.851 m
Chapter 9 Problems
443
______________________________________________________________________________
0.61 m 22. REASONING The arm, being stationary, is in equilibrium, since it has no translational or angular acceleration. Therefore, the net 0.28 m external force and the net external torque 98 N acting on the arm are zero. Using the fact 0.069 m that the net external torque is zero will allow us to determine the magnitude of the force cg Axis M. The drawing at the right shows three 29 forces: M, the 47-N weight of the arm acting 47 N at the arm's center of gravity (cg), and the 98-N force that acts upward on the right end M of the arm. The 98-N force is applied to the (0.070 m) sin 29 arm by the ring. It is the reaction force that arises in response to the arm pulling downward on the ring. Its magnitude is 98 N, because it supports the 98-N weight hanging from the pulley system. Other forces also act on the arm at the shoulder joint, but we can ignore them. The reason is that their lines of action pass directly through the axis at the shoulder joint, which is the axis that we will use to determine torques. Thus, these forces have zero lever arms and contribute no torque.
SOLUTION The magnitude of each individual torque is the magnitude of the force times the corresponding lever arm. The forces and their lever arms are as follows: Force 98 N 47 N M Lever Arm 0.61 m 0.28 (0.069 m) sin 29
Each torque is positive if it causes a counterclockwise rotation and negative if it causes a clockwise rotation about the axis. Thus, since the net torque must be zero, we see that
98 N 0.61 m 47 N 0.28 m M 0.069 m sin 29 0
Solving for M gives
M
98 N 0.61 m 47 N 0.28 m 1400 N 0.069 m sin 29
23. SSM REASONING Since the man holds the ball motionless, the ball and the arm are in equilibrium. Therefore, the net force, as well as the net torque about any axis, must be zero. SOLUTION Using Equation 9.1, the net torque about an axis through the elbow joint is = M(0.0510 m) (22.0 N)(0.140 m) (178 N)(0.330 m) = 0
444 ROTATIONAL DYNAMICS
Solving this expression for M gives M = 1.21103 N . The net torque about an axis through the center of gravity is = (1210 N)(0.0890 m) + F(0.140 m) (178 N)(0.190 m) = 0 Solving this expression for F gives F = 1.01103 N . Since the forces must add to give a net force of zero, we know that the direction of F is downward . ______________________________________________________________________________ 24. REASONING AND SOLUTION a. The net torque about an axis through the point of contact between the floor and her shoes is = (5.00 102 N)(1.10 m)sin 30.0 + FN(cos 30.0)(1.50 m) = 0
FN 212 N
b. Newton's second law applied in the horizontal direction gives Fh FN = 0, so Fh 212 N . c. Newton's second law in the vertical direction gives Fv W = 0, so Fv 5.00 102 N . ______________________________________________________________________________ 25. REASONING AND SOLUTION Taking torques on the right leg about the apex of the "A" gives = (120 N) L cos 60.0 FL cos 30.0 + T (2L) cos 60.0 = 0 where F is the horizontal component of the force exerted by the crossbar and T is the tension in the right string. Due to symmetry, the tension in the left string is also T. Newton's second law applied to the vertical gives 2T 2W = 0 so T = 120 N
Substituting T = 120 N into the first equation yields F = 69 N . ______________________________________________________________________________ 26. REASONING AND SOLUTION The weight W of the left side of the ladder, the normal force FN of the floor on the left leg of the ladder, the tension T in the crossbar, and the reaction force R due to the right-hand side of the ladder, are shown in the following figure. In the vertical direction W + FN = 0, so that
Chapter 9 Problems
445
FN = W = mg = (10.0 kg)(9.80 m/s2) = 98.0 N In the horizontal direction it is clear that R = T. The net torque about the base of the ladder is = T [(1.00 m) sin 75.0o] W [(2.00 m) cos 75.0o] + R [(4.00 m) sin 75.0o] = 0 Substituting for W and using R = T, we obtain FN
R 75.0
o
W T
T
9.80 N 2.00 m cos 75.0 17.5 N 3.00 m sin 75.0
______________________________________________________________________________ 27. REASONING The drawing shows the forces acting on the board, which has a length L. Wall 2 exerts a normal force P2 on the lower end of the board. The maximum force of static friction that wall 2 can apply to the lower end of the board is sP2 and is directed upward in the drawing. The weight W acts downward at the board's center. Wall 1 applies a normal force P1 to the upper end of the board. We take upward and to the right as our positive directions.
Wall 1
P1
1
Wall 2
s P2
W
P2
SOLUTION Then, since the horizontal forces balance to zero, we have P1 P2 = 0 The vertical forces also balance to zero: (1)
sP2 W = 0
Using an axis through the lower end of the board, we now balance the torques to zero:
(2)
L W cos P L sin 0 1 2
Rearranging Equation (3) gives
tan W 2P 1
(3)
(4)
446 ROTATIONAL DYNAMICS
But W = sP2 according to Equation (2), and P2 = P1 according to Equation (1). Therefore, W = sP1, which can be substituted in Equation (4) to show that
tan
or From the drawing at the right,
s P 1
2P 1
s
2
0.98 2
= tan1(0.49) = 26
cos
d L
L
Therefore, the longest board that can be propped between the two walls is
d d 1.5 m 1.7 m cos cos 26 ______________________________________________________________________________
L
28. REASONING The moment of inertia of the stool is the sum of the individual moments of inertia of its parts. According to Table 9.1, a circular disk of radius R has a moment of inertia of Idisk 1 M disk R2 with respect to an axis perpendicular to the disk center. Each thin rod is 2 attached perpendicular to the disk at its outer edge. Therefore, each particle in a rod is located at a perpendicular distance from the axis that is equal to the radius of the disk. This means that each of the rods has a moment of inertia of Irod = Mrod R2. SOLUTION Remembering that the stool has three legs, we find that the its moment of inertia is 1 I stool I disk 3 I rod 2 M disk R 2 3 M rod R 2
1 2
1 0 b.2 kg g0.16 mg 3b.15 kg g0.16 mg b b
2 2
0.027 kg m 2
______________________________________________________________________________ 29. REASONING According to Newton's second law for rotational motion, = I, the angular acceleration of the blades is equal to the net torque applied to the blades divided by their total moment of inertia I, both of which are known. SOLUTION The angular acceleration of the fan blades is
1.8 N m 8.2 rad/s 2 2 I 0.22 kg m ______________________________________________________________________________
Chapter 9 Problems
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30. REASONING According to Table 9.1 in the text, the moment of inertia I of the disk is
I MR 2 , where M is the disk's mass and R is its radius. Thus, we can determine the mass
1 2
from this expression, provided we can obtain a value for I. To obtain the value for I, we will use Newton's second law for rotational motion, I (Equation 9.7), where is the net torque and is the angular acceleration. SOLUTION From the moment of inertia of the disk, we have
I MR 2
1 2
or
M
2I R2
(1)
Using Newton's second law for rotational motion, we find for I that
I
or
I
Substituting this expression for I into Equation (1) gives
M 2 I 2 R 2 R 2
(2)
The net torque is due to the 45-N force alone. According to Equation 9.1, the magnitude of the torque that a force of magnitude F produces is F , where is the lever arm of the force. In this case, we have R , since the force is applied tangentially to the disk and perpendicular to the radius. Substituting FR into Equation (2) gives
M 2 45 N 2 2 FR 2 F 2 5.0 kg R 2 R R 0.15 m 120 rad/s 2
31. REASONING AND SOLUTION a. The net torque on the disk about the axle is = F1R F2R = (0.314 m)(90.0 N 125 N) = 11 N m b. The angular acceleration is given by = /I. From Table 9.1, the moment of inertia of the disk is I = (1/2) MR2 = (1/2)(24.3 kg)(0.314 m)2 = 1.20 kgm2
= ( Nm)/(1.20 kgm2) = 9.2 rad / s 2
______________________________________________________________________________
448 ROTATIONAL DYNAMICS
32. REASONING The rotational analog of Newton's second law is given by Equation 9.7, I . Since the person pushes on the outer edge of one section of the door with a force F that is directed perpendicular to the section, the torque exerted on the door has a magnitude of FL, where the lever arm L is equal to the width of one section. Once the moment of inertia is known, Equation 9.7 can be solved for the angular acceleration . The moment of inertia of the door relative to the rotation axis is I = 4IP, where IP is the moment of inertia for one section. According to Table 9.1, we find I P 1 ML2 , so that the 3 rotational inertia of the door is I 4 ML2 . 3 SOLUTION Solving Equation 9.7 for , and using the expression for I determined above, we have
4 3
FL ML2
4 3
F 68 N 4 0.50 rad/s 2 ML 3 (85 kg)(1.2 m)
______________________________________________________________________________ 33. SSM REASONING AND SOLUTION a. The rim of the bicycle wheel can be treated as a hoop. Using the expression given in Table 9.1 in the text, we have
I hoop MR2 (1.20 kg)(0.330 m)2 0.131 kg m2
b. Any one of the spokes may be treated as a long, thin rod that can rotate about one end. The expression in Table 9.1 gives
I rod 1 ML2 1 (0.010 kg)(0.330 m)2 3.6 104 kg m 2 3 3
c. The total moment of inertia of the bicycle wheel is the sum of the moments of inertia of each constituent part. Therefore, we have
I I hoop 50I rod 0.131 kg m2 50(3.6 104 kg m2 ) 0.149 kg m2
______________________________________________________________________________ 34. REASONING The angular acceleration that results from the application of a net external torque to a rigid object with a moment of inertia I is given by Newton's second law for rotational motion: (Equation 9.7). By applying this equation to each of the two I situations described in the problem statement, we will be able to obtain the unknown angular acceleration.
Chapter 9 Problems
449
SOLUTION The drawings at the right illustrate the two types of rotational motion of the object. In applying Newton's second law for rotational motion, we need to keep in mind that the moment of inertia depends on where the axis is. For axis 1 (see top drawing), piece B is rotating about one of its ends, so according to Table 9.1, the moment of inertia is
I for axis 1 M B L2 , where MB and LB are the mass B
1 3
B Axis 1
Axis 2
and length of piece B, respectively. For axis 2 (see bottom drawing), piece A is rotating about an axis through its midpoint. According to Table 9.1 the moment of inertia is I for axis 2
I for axis 1
1 12
M A L2 . Applying A
A
the second law for each of the axes, we obtain
for axis 1
and
for axis 2
I for axis 2
As given, the net torque is the same in both expressions. Dividing the expression for axis 2 by the expression for axis 1 gives for axis 2 I for axis 2 I for axis 1 for axis 1 I for axis 2 I for axis 1 Using the expressions from Table 9.1 for the moments of inertia, this result becomes
for axis 2 I for axis 1 for axis 1 I for axis 2
1 M B L2 B 3 1 M A L2 A 12
4M B L2 B M A L2 A
Noting that M A 2M B and LA 2LB , we find that
for axis 2 4M B L2 4M B L2 1 B B 2 for axis 1 M A L2 2 2M B 2 LB A for axis 2 for axis 1
1 2 4.6 rad/s 2 2.3 rad/s 2 2
450 ROTATIONAL DYNAMICS
35. REASONING a. The angular acceleration is defined as the change, 0, in the angular velocity divided by the elapsed time t (see Equation 8.4). Since all these variables are known, we can determine the angular acceleration directly from this definition. b The magnitude of the torque is defined by Equation 9.1 as the product of the magnitude F of the force and the lever arm . The lever arm is the radius of the cylinder, which is known. Since there is only one torque acting on the cylinder, the magnitude of the force can be obtained by using Newton's second law for rotational motion, F I. SOLUTION a. From Equation 8.4 we have that = ( 0)/t. We are given that 0 = 76.0 rad/s,
1 0 38.0 rad/s , and t = 6.40 s, so 2
=
0
t
38.0 rad/s 76.0 rad/s 5.94 rad/s 2 6.40 s
The magnitude of the angular acceleration is 5.94 rad/s2 . b. Using Newton's second law for rotational motion, we have that F I. Thus, the magnitude of the force is
0.615 kg m 5.94 rad/s I F 44.0 N 0.0830 m ______________________________________________________________________________
2 2
36. REASONING The drawing shows the two identical sheets and the axis of rotation for each.
Axis of Rotation
L1 = 0.40 m
L2 = 0.20 m
The time t it takes for each sheet to reach its final angular velocity depends on the angular acceleration of the sheet. This relation is given by Equation 8.4 as t 0 / , where and 0 are the final and initial angular velocities, respectively. We know that 0 = 0 rad/s in each case and that the final angular velocities are the same. The angular acceleration can be determined by using Newton's second law for rotational motion, Equation 9.7, as / I , where is the torque applied to a sheet and I is its moment of inertia.
Chapter 9 Problems
451
SOLUTION Substituting the relation / I into t 0 / gives
t
0 0 I 0
I
The time it takes for each sheet to reach its final angular velocity is:
tLeft
I Left 0
and
tRight
I Right 0
The moments of inertia I of the left and right sheets about the axes of rotation are given by the following relations, where M is the mass of each sheet (see Table 9.1 and the drawings 2 above): I Left 1 ML1 and I Right 1 ML2 . Note that the variables M, 0, and are the 2 3 3 same for both sheets. Dividing the time-expression for the right sheet by that for the left sheet gives I Right 0 tRight I Right 1 ML2 L2 2 2 3 2 1 ML2 tLeft I Left I Left 0 L1 1 3
Solving this expression for tRight yields
L2 0.20 m 2 2.0 s 2 tRight tLeft 2 8.0 s L 0.40 m 2 1 ______________________________________________________________________________
37. SSM REASONING AND SOLUTION a. From Table 9.1 in the text, the moment of inertia of the rod relative to an axis that is perpendicular to the rod at one end is given by
I rod 1 ML2 1 (2.00 kg)(2.00 m)2 2.67 kg m2 3 3
b. The moment of inertia of a point particle of mass m relative to a rotation axis located a perpendicular distance r from the particle is I mr 2 . Suppose that all the mass of the rod were located at a single point located a perpendicular distance R from the axis in part (a). If this point particle has the same moment of inertia as the rod in part (a), then the distance R, the radius of gyration, is given by
2.67 kg m 2 I 1.16 m m 2.00 kg ______________________________________________________________________________ R
452 ROTATIONAL DYNAMICS
38. REASONING AND SOLUTION
r = 0.28 m. The angular acceleration needed to produce this angular speed is = (0t. The net torque required is = I. This torque is due solely to the force M, so that = ML. Thus, 0 I t M L L Setting 0 = 0 rad/s and = vT/r, the force becomes
The final angular speed of the arm is = vT/r, where
v I T 2 r t I vT 0.065 kg m 5.0 m/s 460 N M L Lrt 0.025 m 0.28 m 0.10 s ______________________________________________________________________________
39. SSM REASONING The angular acceleration of the bicycle wheel can be calculated from Equation 8.4. Once the angular acceleration is known, Equation 9.7 can be used to find the net torque caused by the brake pads. The normal force can be calculated from the torque using Equation 9.1. SOLUTION The angular acceleration of the wheel is, according to Equation 8.4,
0
t
3.7 rad/s 13.1 rad/s 3.1 rad/s 2 3.0 s
If we assume that all the mass of the wheel is concentrated in the rim, we may treat the wheel as a hollow cylinder. From Table 9.1, we know that the moment of inertia of a hollow cylinder of mass m and radius r about an axis through its center is I mr 2 . The net torque that acts on the wheel due to the brake pads is, therefore,
I (mr 2 )
(1)
From Equation 9.1, the net torque that acts on the wheel due to the action of the two brake pads is (2) 2 fk where fk is the kinetic frictional force applied to the wheel by each brake pad, and 0.33 m is the lever arm between the axle of the wheel and the brake pad (see the drawing in the text). The factor of 2 accounts for the fact that there are two brake pads. minus The sign arises because the net torque must have the same sign as the angular acceleration. The kinetic frictional force can be written as (see Equation 4.8)
Chapter 9 Problems
453
f k k FN
(3)
where k is the coefficient of kinetic friction and FN is the magnitude of the normal force applied to the wheel by each brake pad. Combining Equations (1), (2), and (3) gives
2(k FN ) (mr 2 )
mr 2 (1.3 kg)(0.33 m)2 (3.1 rad/s 2 ) FN 0.78 N 2 k 2(0.85)(0.33 m) ______________________________________________________________________________
40. REASONING To determine the angular acceleration of the pulley and the tension in the cord attached to the block, we will apply Newton's second law to the pulley and the block separately. Only one external forces acts on the pulley, as its free-body diagram at the right shows. This force T is due to the tension in the cord. The torque that results from this force is the net torque acting on the pulley and obeys Newton's second law for rotational motion (Equation 9.7). Two external forces act on the block, as its free-body diagram at the right indicates. These are (1) the force T due to the tension in the cord and (2) the weight mg of the block. The net force that results from these forces obeys Newton's second law for translational motion (Equation 4.2b).
+y + +x Free-body diagram for pulley
R T
T
mg Free-body diagram for block
SOLUTION Applying Newton's second law for rotational motion to the pulley gives
TR I
(1)
where we have written the torque as the magnitude of the tension force times the lever arm (the radius) as specified by Equation 9.1. In addition, we have assigned this torque a positive value, since it causes a counterclockwise rotation of the pulley. To obtain a value for T, we note that the tension has the same magnitude everywhere in the massless cord, so that T T . Thus, by applying Newton's second law for translation (Equation 4.2b), we obtain
F T mg ma
or
T T mg ma
454 ROTATIONAL DYNAMICS
where we have used a to denote the magnitude of the vertical acceleration of the block and included the minus sign to account for the fact that the block is accelerating downward. Substituting this result for T into Equation (1) gives
mg ma R I
(2)
To proceed further, we must deal with a. Note that the pulley rolls without slipping against the cord, so a and are related according to a R (Equation 8.13). With this substitution, Equation (2) becomes
mg m R R I
Solving for , we find that
or
mgR mR 2 I
2.0 kg 9.80 m/s2 0.040 m mgR 180 rad/s 2 2 2 3 2 mR I 2.0 kg 0.040 m 1.110 kg m
The value for the tension can be now obtained by substituting this value for into Equation (1):
TR I
or
1.1103 kg m2 180 rad/s2 I T 5.0 N R 0.040 m
41. REASONING AND SOLUTION The moment of inertia of a solid cylinder about an axis coinciding with the cylinder axis which contains the center of mass is Icm = (1/2) MR2. The parallel axis theorem applied to an axis on the surface (h = R) gives I = Icm + MR2, so that
I
3 2
MR 2
______________________________________________________________________________
Chapter 9 Problems
455
42. REASONING The drawing shows the drum, pulley, and the crate, as well as the tensions in the cord
r2 T1 Pulley m2 = 130 kg T2
T1 r1 Drum m1 = 150 kg r1 = 0.76 m
T2
Crate m3 = 180 kg
Let T1 represent the magnitude of the tension in the cord between the drum and the pulley. Then, the net torque exerted on the drum must be, according to Equation 9.7, = I1, where I1 is the moment of inertia of the drum, and 1 is its angular acceleration. If we assume that the cable does not slip, then Equation 9.7 can be written as
R EAS ONI
a T1 r1 m1 r12 r1 I1
c hF I GJ HK
1
(1)
where is the counterclockwise torque provided by the motor, and a is the acceleration of the cord (a = 1.2 m/s2). This equation cannot be solved for directly, because the tension T1 is not known. We next apply Newton's second law for rotational motion to the pulley in the drawing:
a 1 T1 r2 T2 r2 2 m2 r22 r2 I2
c
hF I GJ HK
2
(2)
where T2 is the magnitude of the tension in the cord between the pulley and the crate, and I2 is the moment of inertial of the pulley. Finally, Newton's second law for translational motion (Fy = m a) is applied to the crate, yielding
456 ROTATIONAL DYNAMICS
T2 m3 g m3 a Fy
(3)
SOLUTION Solving Equation (1) for T1 and substituting the result into Equation (2), then solving Equation (2) for T2 and substituting the result into Equation (3), results in the following value for the torque
r1 a m1 1 m2 m3 m3 g 2
(0.76 m) (1.2 m/s 2 ) 150 kg + 1 130 kg + 180 kg (180 kg)(9.80 m/s 2 ) 1700 N m 2 ______________________________________________________________________________
43. SSM REASONING The kinetic energy of the flywheel is given by Equation 9.9. The moment of inertia of the flywheel is the same as that of a solid disk, and, according to Table 1 9.1 in the text, is given by I 2 MR 2 . Once the moment of inertia of the flywheel is known, Equation 9.9 can be solved for the angular speed in rad/s. This quantity can then be converted to rev/min. SOLUTION Solving Equation 9.9 for , we obtain,
2 ( KE R ) I
2 ( KE R )
1 2
MR 2
4(1.2 10 9 J) 6.4 10 4 rad / s (13kg)(0.30m) 2
Converting this answer into rev/min, we find that
6.4 10 4 rad / s
c
G rad JGmin J hF1rev I F60 s I H K1 K H 2
6.1 10 5 rev / min
______________________________________________________________________________ 44. REASONING 1 a. The kinetic energy is given by Equation 9.9 as KE R 2 I 2 . Assuming the earth to be a uniform solid sphere, we find from Table 9.1 that the moment of inertia is I 2 MR 2 . The 5 mass and radius of the earth is M = 5.98 1024 kg and R = 6.38 106 m (see the inside of the text's front cover). The angular speed must be expressed in rad/s, and we note that the earth turns once around its axis each day, which corresponds to 2 rad/day. b. The kinetic energy for the earth's motion around the sun can be obtained from 1 Equation 9.9 as KE R 2 I 2 . Since the earth's radius is small compared to the radius of the earth's orbit (Rorbit = 1.50 1011 m, see the inside of the text's front cover), the moment
Chapter 9 Problems
457
2 of inertia in this case is just I MRorbit . The angular speed of the earth as it goes around the sun can be obtained from the fact that it makes one revolution each year, which corresponds to 2 rad/year.
SOLUTION a. According to Equation 9.9, we have
1 KE R 2 I 2 1 2
c MR h
2 5 2 24
2
1 2
2 5
5 c.98 10
kg 6.38 10
c h
6
L2 rad IF day IF1 h IO F J1 J J mh M G G G P H H H Q N1 day K24 h K3600 s K
2 2
2 .57 10 29 J
b. According to Equation 9.9, we have
1 KE R 2 I 2 1 2
MR c h
2 orbit 24
2
1 2
5 c.98 10
kg 1.50 10
c h
11
L2 rad IF 1 yr IF day IF1 h IO F J m hM G G P G yr K365 day J 124 h J 3600 s J H H Q H H N1 G K K K
2 2
2 .67 10 33 J ______________________________________________________________________________
45. SSM REASONING AND SOLUTION a. The tangential speed of each object is given by Equation 8.9, v T r . Therefore, For object 1: For object 2: For object 3: vT1 = (2.00 m)(6.00 rad/s) = vT2 = (1.50 m)(6.00 rad/s) = vT3 = (3.00 m)(6.00 rad/s) =
12.0 m / s
9.00 m/ s
18.0 m / s
b. The total kinetic energy of this system can be calculated by computing the sum of the kinetic energies of each object in the system. Therefore,
2 2 2 1 1 1 KE 2 m1 v1 2 m2 v 2 2 m3 v 3
KE
1 2
( 6.00 kg)(12.0 m / s) 2 ( 4 .00 kg)(9.00 m / s) 2 ( 3.00 kg)(18.0 m / s) 2 1.08 10 3 J
458 ROTATIONAL DYNAMICS
c. The total moment of inertia of this system can be calculated by computing the sum of the moments of inertia of each object in the system. Therefore,
I mr 2 m1 r12 m2 r22 m3 r32
I ( 6.00 kg)(2.00 m) 2 ( 4 .00 kg)(1.50 m) 2 ( 3.00 kg)(3.00 m) 2 60.0 kg m 2
d. The rotational kinetic energy of the system is, according to Equation 9.9,
1 1 KE R 2 I 2 2 (60.0 kg m 2 )(6.00 rad / s) 2 1.08 10 3 J
This agrees, as it should, with the result for part (b). ______________________________________________________________________________ 46. REASONING Each blade can be approximated as a thin rod rotating about an axis perpendicular to the rod and passing through one end. The moment of inertia of a blade is given in Table 9.1 as 1 ML2 , where M is the mass of the blade and L is its length. The total 3 moment of inertia I of the two blades is just twice that of a single blade. The rotational kinetic energy KER of the blades is given by Equation 9.9 as KER 1 I 2 , where is the 2 angular speed of the blades. SOLUTION a. The total moment of inertia of the two blades is
I 1 ML2 1 ML2 2 ML2 3 3 3
2 3
240 kg 6.7 m 2
7200 kg m 2
b. The rotational kinetic energy is
KER 1 I 2 1 7200 kg m2 44 rad/s 7.0 106 J 2 2
2
______________________________________________________________________________ 47. REASONING The rotational kinetic energy of a solid sphere is given by Equation 9.9 as KER 1 I 2 , where I is its moment of inertia and its angular speed. The sphere has 2 translational motion in addition to rotational motion, and its translational kinetic energy is KET 1 mv2 (Equation 6.2), where m is the mass of the sphere and v is the speed of its 2 center of mass. The fraction of the sphere's total kinetic energy that is in the form of rotational kinetic energy is KER/(KER + KET).
Chapter 9 Problems
459
2 SOLUTION The moment of inertia of a solid sphere about its center of mass is I 5 mR 2 ,
where R is the radius of the sphere (see Table 9.1). The fraction of the sphere's total kinetic energy that is in the form of rotational kinetic energy is
KE R KET
KE R
=
1 I 2 2 1 I 2 1 mv 2 2 2
1 2 mR 2 2 5 1 2 mR 2 2 2 5
2
1 mv 2 2
2 R 2 2 5 2 R 2 2 v 2 5
Since the sphere is rolling without slipping on the surface, the translational speed v of the center of mass is related to the angular speed about the center of mass by v = R (see Equation 8.12). Substituting v = R into the equation above gives
KE R KE R KE T
2 R 2 2 5 2 R 2 2 v 2 5
2 5
R 2 2 R
2 5
R 2 2
2
2 7
______________________________________________________________________________ 48. REASONING The drawing shows the rod in its initial (dashed lines) and final (solid lines) orientations. Since both friction and air resistance are absent, the total mechanical energy is conserved. In this case, the total mechanical energy is
Center of mass
E
1 I 2 2 Rotational kinetic energy
mgh
Gravitational potential energy
Pivot hf
In this expression, m is the rod's mass, I is its moment of inertia, is its angular speed, and h is the height of its center of mass above a reference level that we take to be the ground. Since the rod is uniform, its center of mass is located at the center of the rod. The energy-conservation principle will guide our solution.
Center of mass h0 v0
SOLUTION Conservation of the total mechanical energy dictates that
1 I f2 mghf 2 Final total mechanical energy
1 2 I 0 mgh0 2 Initial total mechanical energy
(1)
Since the rod is momentarily at rest in its final orientation, we know that f = 0 rad/s, so that Equation (1) becomes
2 mghf I 0 mgh0 1 2
or
1 2
2 I 0 mg hf h0
(2)
460 ROTATIONAL DYNAMICS
We can relate the initial angular speed 0 to the initial linear speed v0 by using Equation 8.9: v 0 0 . With this substitution, the fact that hf h0 L (see the drawing), and the fact that L
I mL2 (see Table 9.1 in the text), Equation (2) indicates that
1 2 2 I 0
1 3
mg hf h0
or
1 1 mL2 2 3
v0 mgL L
2
or
1 2 v 6 0
gL
Solving for v0 gives
v0 6 gL 6 9.80 m/s 2 0.80 m 6.9 m/s
49. SSM REASONING AND SOLUTION The only force that does work on the cylinders as they move up the incline is the conservative force of gravity; hence, the total mechanical energy is conserved as the cylinders ascend the incline. We will let h = 0 on the horizontal plane at the bottom of the incline. Applying the principle of conservation of mechanical energy to the solid cylinder, we have
2 2 mghs 1 mv0 1 Is0 2 2
(1)
where, from Table 9.1, Is 1 mr 2 . In this expression, v0 and 0 are the initial translational 2 and rotational speeds, and hs is the final height attained by the solid cylinder. Since the cylinder rolls without slipping, the rotational speed 0 and the translational speed v0 are related according to Equation 8.12, 0 v0 / r . Then, solving Equation (1) for hs , we obtain
hs
2 3v0
4g
Repeating the above for the hollow cylinder and using I h mr 2 we have
hh
2 v0
g
The height h attained by each cylinder is related to the distance s traveled along the incline and the angle of the incline by h h ss s and sh h sin sin Dividing these gives ss 3/4 sh ______________________________________________________________________________
Chapter 9 Problems
461
50. REASONING AND SOLUTION The conservation of energy gives for the cube 2 (1/2) mvc = mgh, so vc = 2gh . The conservation of energy gives for the marble (1/2) mvm2 + (1/2) I2 = mgh. Since the marble rolls without slipping: = v/R. For a solid sphere we know I = (2/5) mR2. Combining the last three equations gives
vm 10 gh 7
Thus,
7 1.18 5 vm ______________________________________________________________________________
51. REASONING We first find the speed v0 of the ball when it becomes airborne using the
vc
conservation of mechanical energy. Once v0 is known, we can use the equations of kinematics to find its range x. SOLUTION When the tennis ball starts from rest, its total mechanical energy is in the form of gravitational potential energy. The gravitational potential energy is equal to mgh if we take h = 0 m at the height where the ball becomes airborne. Just before the ball becomes airborne, its mechanical energy is in the form of rotational kinetic energy and translational kinetic 2 energy. At this instant its total energy is 1 mv0 1 I 2 . If we treat the tennis ball as a thin2 2 walled spherical shell of mass m and radius r, and take into account that the ball rolls down the hill without slipping, its total kinetic energy can be written as
1 2 mv0 2
v 1 1 5 2 1 2 2 I 2 mv0 ( mr 2 ) 0 mv0 2 2 2 3 6 r
2
Therefore, from conservation of mechanical energy, he have
2 mgh mv0 5 6
or
v0
6 gh 5
The range of the tennis ball is given by x v0 xt v0 (cos ) t , where t is the flight time of the ball. From Equation 3.3b, we find that the flight time t is given by
t v v0 y ay (v0 y ) v0 y ay 2v0 sin ay
Therefore, the range of the tennis ball is
462 ROTATIONAL DYNAMICS
2v sin x v0 xt v0 (cos ) 0 ay
If we take upward as the positive direction, then using the fact that a y g and the expression for v0 given above, we find
2cos sin 2 2cos sin 6 gh 12 x h cos sin v0 g g 5 5 12 (1.8 m) (cos 35)(sin 35)= 2.0 m 5 ______________________________________________________________________________
52. REASONING Before any sand strikes the disk, only the disk is rotating. After the sand has landed on the disk, both the sand and the disk are rotating. If the sand and disk are taken to be the system of objects under consideration, we note that there are no external torques acting on the system. As the sand strikes the disk, each exerts a torque on the other. However, these torques are exerted by members of the system, and, as such, are internal torques. The conservation of angular momentum states that the total angular momentum of a system remains constant (is conserved) if the net average external torque acting on the system is zero. We will use this principle to find the final angular velocity of the system. SOLUTION The angular momentum of the system (sand plus disk) is given by Equation 9.10 as the product of the system's moment of inertia I and angular velocity , or L = IThe conservation of angular momentum can be written as
I
Final angular momentum
2
I 00
Initial angular momentum
where and 0 are the final and initial angular velocities, respectively, and I and I0 are final and initial moments of inertia. The initial moment of inertia is given, while the final moment of inertia is the sum of the values for the rotating sand and disk, I = Isand + I0. We note that the sand forms a thin ring, so its moment of inertia is given by (see Table 2 9.1) Isand = M sand Rsand , where Msand is the mass of the sand and Rsand is the radius of the ring. Thus, the final angular velocity of the system is, then,
Chapter 9 Problems
463
0
I0 I0 I0 0 0 2 I M I sand I 0 sand Rsand I 0
0.10 kg m 2 0.037 rad/s 0.067 rad/s 0.50 kg 0.40 m 2 + 0.10 kg m 2 ______________________________________________________________________________
53. SSM WWW REASONING Let the two disks constitute the system. Since there are no external torques acting on the system, the principle of conservation of angular momentum applies. Therefore we have Linitial Lfinal , or
I A A I B B ( I A I B ) final
This expression can be solved for the moment of inertia of disk B. SOLUTION Solving the above expression for I B , we obtain
IB IA
F G H
final B
A
final
I ( 3.4 kg m ) L 2.4 rad / s 7.2 rad / s O 4.4 kg m M rad / s (2.4 rad / s) P J 9.8 K N Q
2
2
______________________________________________________________________________ 54. REASONING AND SOLUTION The conservation of angular momentum applies about the indicated axis Idd + MpRp2p = 0 where
p = vp/Rp
Now Id = (1/2) MdRd2. Combining and solving yields
d = 2(Mp/Md)(Rp/Rd2)vp = 2
F 40.0 kg IL1.25 m O.00 m / sg= 0.500 rad/s 2 M b G 10 kg J b mgP 1.00 H K 2.00 P M N Q
2 2
The magnitude of d is 0.500 rad/s
. The negative sign indicates that the disk rotates in a
direction opposite to the motion of the person. ______________________________________________________________________________ 55. REASONING The rod and bug are taken to be the system of objects under consideration, and we note that there are no external torques acting on the system. As the bug crawls out to the end of the rod, each exerts a torque on the other. However, these torques are internal torques. The conservation of angular momentum states that the total angular momentum of a system remains constant (is conserved) if the net average external torque acting on the system is zero. We will use this principle to find the final angular velocity of the system.
464 ROTATIONAL DYNAMICS
SOLUTION The angular momentum L of the system (rod plus bug) is given by Equation 9.10 as the product of the system's moment of inertia I and angular velocity , or L = IThe conservation of angular momentum can be written as
I
Final angular momentum
I 00
Initial angular momentum
where and 0 are the final and initial angular velocities, respectively, and I and I0 are the final and initial moments of inertia. The initial moment of inertia is given. The initial moment of inertia of the bug is zero, because it is located at the axis of rotation. The final moment of inertia is the sum of the moment of inertia of the bug and that of the rod; I = Ibug + I0. When the bug has reached the end of the rod, its moment of inertia is Ibug = mL2, where m is its mass and L is the length of the rod. The final angular velocity of the system is, then,
0
I0 I I0 0 2 0 0 mL I I bug I 0 I 0
1.1103 kg m 2 0.26 rad/s 0.32 rad/s 4.2 103 kg 0.25 m 2 + 1.1 103 kg m 2 ______________________________________________________________________________
56. REASONING Since friction and air resistance play no role in this problem, no net external torque acts on the system comprised of the carousel and the person on it. Therefore, the total angular momentum of the system is conserved, and it is via this conservation principle that we will determine the person's mass. In applying this principle, we note that the moment of inertia of the carousel itself remains constant as the person moves toward the center, but the moment of inertia of the person changes. SOLUTION Since the total angular momentum of the system is conserved, we have
I Cf I Pf f I C0 I P00
Final total angular momentum Initial total angular momentum
(1)
where IC is the moment of inertia of the carousel, IPf and IP0 are, respectively, the final and initial moments of inertia of the person, and f and 0 are, respectively, the final and initial angular velocities. As the person (mass = m) moves, his distance from the center changes to a final value of rf from an initial value of r0. Correspondingly, his moment of inertia changes to a final value of mrf2 from an initial value of mr02 , according to Equation 9.6. Substituting these values for IPf and IP0 into Equation (1) gives
Chapter 9 Problems
465
ICf mrf2 f IC0 mr02 0
Solving for m gives
or ICf IC0 mr02 0 mrf2 f m r020 rf2f
m
I C f 0 r020 rf2f
125 kg m2 0.800 rad/s 0.600 rad/s
1.50 m 2 0.600 rad/s 0.750 m 2 0.800 rad/s
28 kg
57. SSM REASONING When the modules pull together, they do so by means of forces that are internal. These pulling forces, therefore, do not create a net external torque, and the angular momentum of the system is conserved. In other words, it remains constant. We will use the conservation of angular momentum to obtain a relationship between the initial and final angular speeds. Then, we will use Equation 8.9 (v = r) to relate the angular speeds 0 and f to the tangential speeds v0 and vf.
SOLUTION Let L be the initial length of the cable between the modules and 0 be the initial angular speed. Relative to the center-of-mass axis, the initial momentum of inertia of the two-module system is I0 = 2M(L/2)2, according to Equation 9.6. After the modules pull together, the length of the cable is L/2, the final angular speed is f, and the momentum of inertia is If = 2M(L/4)2. The conservation of angular momentum indicates that
I f f I 0 0
Final angular momentum 2 Initial angular momentum 2
LM F I O LM F I O L L 2 2 M GJ P M GJ P 4 Q N 2 Q M HKP M HKP N
f
0
f 4 0
According to Equation 8.9, f = vf/(L/4) and 0 = v0/(L/2). With these substitutions, the result that f = 40 becomes
v f 2 v 0 2 17 m / s 34 m / s L/4 ______________________________________________________________________________
0
vf
4
Fv I or G/ 2 J L H K
b
g
58. REASONING AND SOLUTION Since the change occurs without the aid of external torques, the angular momentum of the system is conserved: Iff = I00. Solving for f gives
466 ROTATIONAL DYNAMICS
f 0
I0 If
1 12
where for a rod of mass M and length L, I 0
ML2 . To determine If , we will treat the
arms of the "u" as point masses with mass M/4 a distance L/4 from the rotation axis. Thus,
1 M If 12 2
L 2
2
M L 2 1 ML2 2 4 4 24
and
1 ML2 (7.0 rad/s)(2) 14 rad/s f 0 12 1 ML2 24 ______________________________________________________________________________
59. REASONING AND SOLUTION The block will just start to move when the centripetal force on the block just exceeds f smax . Thus, if rf is the smallest distance from the axis at
which the block stays at rest when the angular speed of the block is f, then
s FN = mrf f2, or s mg = mrf f2. Thus, s g = rf f2
(1)
Since there are no external torques acting on the system, angular momentum will be conserved. I0 0 = If f where I0 = mr02, and If = mrf2. Making these substitutions yields r02 0 = rf2 f Solving Equation (2) for f and substituting into Equation (1) yields:
2 s g rf 0
(2)
r04 rf4
1/3
Solving for rf gives
2 0 r04 (2.2 rad/s) 2 (0.30 m) 4 rf 0.17 m 2 g (0.75)(9.80 m/s ) s ______________________________________________________________________________ 1/3
60. REASONING AND SOLUTION After the mass has moved inward to its final path the centripetal force acting on it is T = 105 N.
Chapter 9 Problems
467
r
105 N
Its centripetal acceleration is ac = v2/R = T/m Now v = R so R = T/(2m)
The centripetal force is parallel to the line of action (the string), so the force produces no torque on the object. Hence, angular momentum is conserved. I = Ioo Substituting and simplifying R3 = (mRo4 o2)/T, so that R =
0.573 m
so that
= (Io/I)o = (Ro2/R2)o
______________________________________________________________________________ 61. SSM WWW REASONING The maximum torque will occur when the force is applied perpendicular to the diagonal of the square as shown. The lever arm is half the length of the diagonal. From the Pythagorean theorem, the lever arm is, therefore,
1 2
(0.40 m ) (0.40 m ) 0.28 m
2
2
Since the lever arm is now known, we can use Equation 9.1 to obtain the desired result directly. SOLUTION Equation 9.1 gives
F (15 N)(0.28 m) = 4.2 N m ______________________________________________________________________________
468 ROTATIONAL DYNAMICS
62. REASONING The net torque acting on the CD is given by Newton's second law for rotational motion (Equation 9.7) as where I is the moment of inertia of the CD and is its angular acceleration. The moment of inertia can be obtained directly from Table 9.1, and the angular acceleration can be found from its definition (Equation 8.4) as the change in the CD's angular velocity divided by the elapsed time. SOLUTION The net torque is Assuming that the CD is a solid disk, its moment of inertia can be found from Table 9.1 as I 1 MR2 , where M and R are the mass and radius of 2 the CD. Thus, the net torque is
I
1 MR2 2
The angular acceleration is given by Equation 8.4 as 0 / t , where and 0 are the final and initial angular velocities, respectively, and t is the elapsed time. Substituting this expression for into Newton's second law yields
1 MR2 1 MR2 2 2
0
t
2 21 rad/s 0 rad/s 4 1 17 103 kg 6.0 102 m 8.0 10 N m 2 0.80 s ______________________________________________________________________________
63. REASONING AND SOLUTION a. Angular momentum is conserved, so that I = Ioo, where I = Io + 10mbRb2 = 2100 kgm2
= (Io/I)o = [(1500 kgm2)/(2100 kgm2)](0.20 rad/s) = 0.14 rad/s
b. A net external torque must be applied in a direction that is opposite to the angular deceleration caused by the baggage dropping onto the carousel. ______________________________________________________________________________ 64. REASONING When the board just begins to tip, three forces act on the board. They are the weight W of the board, the weight WP of the person, and the force F exerted by the right support. At the moment tipping begins, the left support exerts no force on the board and, therefore, is omitted from the following drawing.
Chapter 9 Problems
469
F 1.4 m x W = 225 N WP 450 N
Since the board will rotate around the right support, the lever arm for this force is zero, and the torque exerted by the right support is zero. The lever arm for the weight of the board is equal to one-half the length of the board minus the overhang length: 2.5 m 1.1 m = 1.4 m. The lever arm for the weight of the person is x. Therefore, taking counterclockwise torques as positive, we have WP x W (1.4 m) 0 This expression can be solved for x. SOLUTION Solving the expression above for x, we obtain
W (1.4 m) (225 N)(1.4 m) 0.70 m WP 450 N ______________________________________________________________________________ x
65. SSM REASONING The figure below shows eight particles, each one located at a different corner of an imaginary cube. As shown, if we consider an axis that lies along one edge of the cube, two of the particles lie on the axis, and for these particles r = 0. The next four particles closest to the axis have r , where is the length of one edge of the cube. The remaining two particles have r d , where d is the length of the diagonal along any one of the faces. From the Pythagorean theorem, d 2 2 2 .
470 ROTATIONAL DYNAMICS
According to Equation 9.6, the moment of inertia of a system of particles is given by I mr 2 . SOLUTION Direct application of Equation 9.6 gives
I mr 2 4(m2 ) 2(md 2 ) 4(m2 ) 2(2m2 ) 8m2
or
I 8(0.12 kg)(0.25 m) 2 0.060 kg m 2
______________________________________________________________________________ 66. REASONING When the wheel is resting on the ground it is in equilibrium, so the sum of the torques about any axis of rotation is zero 0 . This equilibrium condition will provide us with a relation between the magnitude of F and the normal force that the ground exerts on the wheel. When F is large enough, the wheel will rise up off the ground, and the normal force will become zero. From our relation, we can determine the magnitude of F when this happens. SOLUTION The free body diagram shows the forces acting on the wheel: its weight W, the normal force FN, the horizontal force F, and the force FE that the edge of the step exerts on the wheel. We select the axis of rotation to be at the edge of the step, so that the torque produced by FE is zero. Letting
r FN FE h W F Axis of Rotation
N , W , and F represent the lever arms for the forces FN, W, and F, the sum of the torques is
FN r2 r h W r 2 r h F r h 0
2 2
N
W
F
Solving this equation for F gives
W FN F
r2 r h r h
2
When the bicycle wheel just begins to lift off the ground the normal force becomes zero (FN = 0 N). When this happens, the magnitude of F is
29 N r h 0.340 m 0.120 m __________________________________________________________________________
W FN F
r2 r h
2
25.0 N 0 N 0.340 m 2 0.340 m 0.120 m 2
Chapter 9 Problems
471
67. SSM REASONING The drawing shows the forces acting on the board, which has a length L. The ground exerts the vertical normal force V on the lower end of the board. The maximum force of static friction has a magnitude of sV and acts horizontally on the lower end of the board. The weight W acts downward at the board's center. The vertical wall applies a force P to the upper end of the board, this force being perpendicular to the wall since the wall is smooth (i.e., there is no friction along the wall). We take upward and to the right as our positive directions. Then, since the horizontal forces balance to zero, we have sV P = 0 (1)
The vertical forces also balance to zero giving V W = 0 (2)
Using an axis through the lower end of the board, we express the fact that the torques balance to zero as L (3) PL sin W cos = 0 2 Equations (1), (2), and (3) may then be combined to yield an expression for . SOLUTION Rearranging Equation (3) gives
tan =
W 2P
(4)
But, P = sV according to Equation (1), and W = V according to Equation (2). Substituting these results into Equation (4) gives V 1 tan = 2 sV 2 s Therefore, 1 1 tan 1 tan 1 37.6 2 s 2(0.650) ______________________________________________________________________________ 68. REASONING If we assume that the system is in equilibrium, we know that the vector sum of all the forces, as well as the vector sum of all the torques, that act on the system must be zero. The figure below shows a free body diagram for the boom. Since the boom is assumed to be uniform, its weight WB is located at its center of gravity, which coincides with its geometrical center. There is a tension T in the cable that acts at an angle to the horizontal, as shown. At the hinge pin P, there are two forces acting. The vertical force V that acts on
472 ROTATIONAL DYNAMICS
the end of the boom prevents the boom from falling down. The horizontal force H that also acts at the hinge pin prevents the boom from sliding to the left. The weight WL of the wrecking ball (the "load") acts at the end of the boom.
By applying the equilibrium conditions to the boom, we can determine the desired forces. SOLUTION The directions upward and to the right will be taken as the positive directions. In the x direction we have (1) Fx H T cos 0 while in the y direction we have
Fy V T sin WL WB 0
(2)
Equations (1) and (2) give us two equations in three unknown. We must, therefore, find a third equation that can be used to determine one of the unknowns. We can get the third equation from the torque equation. In order to write the torque equation, we must first pick an axis of rotation and determine the lever arms for the forces involved. Since both V and H are unknown, we can eliminate them from the torque equation by picking the rotation axis through the point P (then both V and H have zero lever arms). If we let the boom have a length L, then the lever arm for WL is L cos , while the lever arm for WB is ( L / 2)cos . From the figure, we see that the lever arm for T is L sin( ) . If we take counterclockwise torques as positive, then the torque equation is L cos WB WL L cos TL sin 0 2 Solving for T, we have 1 W W L T 2 B cos (3) sin( ) a. From Equation (3) the tension in the support cable is
Chapter 9 Problems
473
T
1 (3600 2
N) 4800 N
sin(48 32)
cos 48 1.6 104 N
b. The force exerted on the lower end of the hinge at the point P is the vector sum of the forces H and V. According to Equation (1),
H T cos 1.6 104 N cos 32 1.4 104 N
and, from Equation (2)
V WL WB T sin 4800 N 3600 N 1.6 104 N sin 32 1.7 104 N
Since the forces H and V are at right angles to each other, the magnitude of their vector sum can be found from the Pythagorean theorem:
FP H 2 V 2 (1.4 104 N)2 (1.7 104 N) 2 2.2 104 N
______________________________________________________________________________ 69. SSM REASONING Let the space station and the people within it constitute the system. Then as the people move radially from the outer surface of the cylinder toward the axis, any torques that occur are internal torques. Since there are no external torques acting on the system, the principle of conservation of angular momentum can be employed. SOLUTION Since angular momentum is conserved,
I final final I 0 0
Before the people move from the outer rim, the moment of inertia is
2 I 0 I station 500mperson rperson
or
I 0 3.00 10 9 kg m 2 ( 500)(70.0 kg)(82.5 m) 2 3.24 10 9 kg m 2
If the people all move to the center of the space station, the total moment of inertia is
I final I station 3.00 10 9 kg m 2
Therefore,
final I0 3.24 10 9 kg m 2 1.08 0 I final 3.00 10 9 kg m 2
This fraction represents a percentage increase of 8 percent . ______________________________________________________________________________
474 ROTATIONAL DYNAMICS
70. REASONING AND SOLUTION The conservation of energy gives mgh + (1/2) mv2 + (1/2) I2 = (1/2) mvo2 + (1/2) Io2 If the ball rolls without slipping, = v/R and o = vo/R. We also know I = (2/5) mR2. Substitution of the last two equations into the first and rearrangement gives
2 v v0 10 gh 7
3.50 m/s 2 10 9.80 m/s2 0.760 m 7
1.3 m/s
______________________________________________________________________________ 71. SSM WWW REASONING AND SOLUTION Consider the left board, which has a length L and a weight of (356 N)/2 = 178 N. Let Fv be the upward normal force exerted by the ground on the board. This force balances the weight, so Fv = 178 N. Let fs be the force of static friction, which acts horizontally on the end of the board in contact with the ground. fs points to the right. Since the board is in equilibrium, the net torque acting on the board through any axis must be zero. Measuring the torques with respect to an axis through the apex of the triangle formed by the boards, we have
L + (178 N)(sin 30.0) + fs(L cos 30.0) Fv (L sin 30.0) = 0 2
or
44.5 N + fs cos 30.0 FV sin 30.0 = 0
(178 N)(sin 30.0) 44.5 N 51.4 N cos 30.0 ______________________________________________________________________________ fs
72. REASONING AND SOLUTION Newton's law applied to the 11.0-kg object gives T2 (11.0 kg)(9.80 m/s2) = (11.0 kg)(4.90 m/s2) A similar treatment for the 44.0-kg object yields T1 (44.0 kg)(9.80 m/s2) = (44.0 kg)(4.90 m/s2) For an axis about the center of the pulley T2r T1r = I( = (1/2) Mr2 (a/r) Solving for the mass M we obtain or T1 = 216 N or T2 = 162 N so that
Chapter 9 Problems
475
M = (2/a)(T2 T1) = [2/(4.90 m/s2)](162 N 216 N) = 22.0 kg ______________________________________________________________________________ 73. CONCEPT QUESTIONS a. The magnitude of a torque is the magnitude of the force times the lever arm of the force, according to Equation 9.1. The lever arm is the perpendicular distance between the line of action of the force and the axis. For the force at corner B, the lever arm is half the length of the short side of the rectangle. For the force at corner D, the lever arm is half the length of the long side of the rectangle. Therefore, the lever arm for the force at corner D is greater. Since the magnitudes of the two forces are the same, this means that the force at corner D creates the greater torque. b. As we have seen, the torque at corner D (clockwise) has a greater magnitude than the torque at corner B (counterclockwise). Therefore, the net torque from these two contributions is clockwise. This means that the force at corner A must produce a counterclockwise torque, if the net torque produced by the three forces is to be zero. The force at corner A, then, must point toward corner B. SOLUTION Let L be the length of the short side of the rectangle, so that the length of the long side is 2L. We can use Equation 9.1 for the magnitude of each torque. Remembering that counterclockwise torques are positive and that the torque at corner A is counterclockwise, we can write the zero net torque as follows:
FA A FB B FD D 0 FA L 12 N
1 L 12 N L 0 2
6.0 N (pointing toward corner B)
The length L can be eliminated algebraically from this result, which can then be solved for FA:
FA 12 N
1 12 N 2
______________________________________________________________________________ 74. CONCEPT QUESTIONS a. In the left design, both the 60.0 and the 525-N forces create clockwise torques with respect to the rotational axis at the point where the tire contacts the ground. In the right design, only the 60.0-N force creates a torque, because the line of action of the 525-N force passes through the axis. Thus, the total torque from the two forces is greater for the left design. b. The man is supporting the wheelbarrow in equilibrium. Therefore, his force must create a torque that balances the total torque from the other two forces. To do this, his torque must be counterclockwise and have the same magnitude as the total torque from the other two forces. Since the two forces create more torque in the left design, the magnitude of the man's torque must be greater for that design. c. The magnitude of a torque is the magnitude of the force times the lever arm of the force, according to Equation 9.1. The drawing shows that the lever arms of the man's forces in the
476 ROTATIONAL DYNAMICS
two designs is the same (1.300 m). Thus, to create the greater torque necessitated by the left design, the man must apply a greater force for that design. SOLUTION The lever arms for the forces can be obtained from the distances shown in the drawing for each design. We can use Equation 9.1 for the magnitude of each torque. Remembering that counterclockwise torques are positive, we can write an expression for the zero net torque for each design. These expressions can be solved for the man's force F in each case:
Left design
525 N 0.400 m 60.0 N 0.600 m F 1.300 m 0
b g b gb g b g b g 525 60.0 b b N g0.400 mgb N g0.600 mg 189 N b F
1.300 m
Right design
60.0 N 0.600 m F 1.300 m 0
1.300 m ______________________________________________________________________________
75. CONCEPT QUESTIONS a. The forces in part b of the drawing cannot keep the beam in equilibrium (no acceleration), because neither V nor W, both being vertical, can balance the horizontal component of P. This unbalanced component would accelerate the beam to the right. b. The forces in part c of the drawing cannot keep the beam in equilibrium (no acceleration). To see why, consider a rotational axis perpendicular to the plane of the paper and passing through the pin. Neither P nor H create torques relative to this axis, because their lines of action pass directly through it. However, W does create a torque, which is unbalanced. This torque would cause the beam to have a counterclockwise angular acceleration. c. The forces in part d are sufficient to eliminate the accelerations discussed above, so they can keep the beam in equilibrium. SOLUTION Assuming that upward and to the right are the positive directions, we obtain the following expressions by setting the sum of the vertical and the sum of the horizontal forces equal to zero:
Horizontal forces Vertical forces P cos H 0 P sin V W 0 ( 1) (2)
b g b g b g 60.0 b b N g0.600 mg 27.7 N F
Using a rotational axis perpendicular to the plane of the paper and passing through the pin and remembering that counterclockwise torques are positive, we also set the sum of the torques equal to zero. In doing so, we use L to denote the length of the beam and note that the lever arms for W and V are L/2 and L, respectively. The forces P and H create no torques relative to this axis, because their lines of action pass directly through it.
Chapter 9 Problems
477
Torques
W
c L hVL 0
1 2
( 3)
Since L can be eliminated algebraically, Equation (3) may be solved immediately for V:
1 1 V 2 W 2 340 N 170 N
b g
Substituting this result into Equation (2) gives
1 P sin 2 W W 0
P
W 340 N 270 N 2 sin 2 sin 39
Substituting this result into Equation (1) gives
F W Icos H 0 Gsin J H K 2
W 340 N H 210 N 2 tan 2 tan 39 ______________________________________________________________________________
76. CONCEPT QUESTIONS a. Equation 8.7 from the equations of rotational kinematics 1 indicates that the angular displacement is 0 t 2 t 2 , where 0 is the initial angular
velocity, t is the time, and is the angular acceleration. Since both wheels start from rest, 0 = 0 rad/s for each. Furthermore, each wheel makes the same number of revolutions in the same time, so and t are also the same for each. Therefore, the angular acceleration must be the same for each.
b. For the hoop, all of the mass is located at a distance from the axis equal to the radius. For the disk, some of the mass is located closer to the axis. According to Equation 9.6, the moment of inertia is greater when the mass is located farther from the axis. Therefore, the moment of inertia of the hoop is greater. Table 9.1 indicates that the moment of inertia of a 1 hoop is I hoop MR 2 , while the moment of inertia of a disk is I disk 2 MR 2 . c. Newton's second law for rotational motion indicates that the net external torque is equal to the moment of inertia times the angular acceleration. Both disks have the same angular acceleration, but the moment of inertia of the hoop is greater. Thus, the net external torque must be greater for the hoop. SOLUTION Using Equation 8.7, we can express the angular acceleration as follows:
1 0 t 2 t 2
or
2 0t t2
c
h
478 ROTATIONAL DYNAMICS
This expression for the acceleration can now be used in Newton's second law for rotational motion. Accordingly, the net external torque is
2 0t Hoop I hoop MR 2 t2 2 2 13 rad 2 0 rad/s 8.0 s 0.20 N m 4.0 kg 0.35 m 2 8.0 s
2 0t Disk I disk 1 MR 2 2 t2 2 13 rad 2 0 rad/s 8.0 s 0.10 N m 8.0 s 2 ______________________________________________________________________________
1 2
4.0 kg 0.35 m 2
77. CONCEPT QUESTIONS a. According to Equation 9.6, the moment of inertia for rod A is just that of the attached particle, since the rod itself is massless. For rod A with its attached particle, then, the moment of inertia is I A ML2 . According to Table 9.1, the moment of inertia for rod B is I B 1 ML2 . The moment of inertia for rod A with its attached particle is 3 greater. b. Since the moment of inertia for rod A is greater, its kinetic energy is also greater, 1 according to Equation 9.9 KE R 2 I 2 . SOLUTION Using Equation 9.9 to calculate the kinetic energy, we find
Rod A
1 KE R 2 I A 2
Rod B KE R
1 2
1 2
1 1.1 J 6 ______________________________________________________________________________
2 2
ML c h 0 4 b.66 kg g0.75 mgb.2 rad / sg b I c ML h 0 4 b.66 kg g0.75 mgb.2 rad / sg b
1 2 2 2 2 2 2 B 1 2 1 3 2 2
3.3 J
78. CONCEPT QUESTIONS a. The force is applied to the person in the counterclockwise direction by the bar that he grabs when climbing aboard.
Chapter 9 Problems
479
b. According to Newton's action-reaction law, the person must apply a force of equal magnitude and opposite direction to the bar and the carousel. This force acts on the carousel in the clockwise direction and, therefore, creates a clockwise torque. c. Since the carousel is moving counterclockwise, the clockwise torque applied to it when the person climbs aboard decreases its angular speed. SOLUTION We consider a system consisting of the person and the carousel. Since the carousel rotates on frictionless bearings, no net external torque acts on this system. Therefore, angular momentum is conserved, and we can set the total angular momentum of the system before the person hops on equal to the total angular momentum afterwards. Afterwards, the angular momentum includes a contribution from the person. According to Equation 9.6, his moment of inertia is I person MR 2 , since he is at the outer edge of the carousel.
I carousel f I person I carousel 0 f
Final total angular momentum Initial total angular momentum
f
I carousel 0 I carousel I person
2
I carousel 0 I carousel MR 2 1.83 rad / s
125 b c kg m h3.14 rad / sg 125 kg m b .0 kg g .50 m g 40 1 b
2 2
______________________________________________________________________________ 79. CONCEPT QUESTIONS a. Two identical systems of objects may have different moments of inertia, depending on where the axis of rotation is located. The moment of inertia depends on the mass of each object and its distance from the axis of rotation. According to 2 Equation 9.6, the moment of inertia for the three-ball system is I m1r12 m2r2 m3r32 , where m1, m2, and m3 are the masses of the balls and r1, r2, and r3 are the distances from the axis. In system A, the ball whose mass is m1 does not contribute to the moment of inertia, because the ball is located on the axis and r1 = 0 m. In B, the ball whose mass is m3 does not contribute to the moment of inertia, because it is located on the axis and r3 = 0 m. b. Even though the force acting on both systems is the same, the torque is not the same. The magnitude of the torque is equal to the magnitude F of the force times the lever arm (Equation 9.1). In system A the lever arm is 3.00 m. In B the lever arm is 0 m, since the line of action of the force passes through the axis of rotation. c. The systems have different angular accelerations. According to Newton's second law for rotational motion, Equation 9.7, the angular acceleration is given by / I , where is the net torque and I is the moment of inertia. Since the net torque and moment of
480 ROTATIONAL DYNAMICS
inertia are different for the two systems, the angular accelerations are different. The angular velocity is given by Equation 8.4 as 0 + t, where 0 is the initial angular velocity and t is the time. Since the angular accelerations are different for the two systems, the angular velocities will also be different at the same later time. SOLUTION a. The moment of inertia for each system is System A I m1r12 m2 r22 m3r32
9.00 kg 0 m + 6.00 kg 3.00 m + 7.00 kg 5.00 m = 229 kg m
2 2 2
2
System B
I m1r12 m2 r22 m3r32 9.00 kg 5.00 m + 6.00 kg 4.00 m + 7.00 kg 0 m = 321 kg m
2 2 2 2
As anticipated, the two systems have different moments of inertia. b. The torque produced by the force has a magnitude that is equal to the product of the force magnitude and the lever arm: System A
F 424 N 3.00 m 1270 N m
The torque is negative because it produces a clockwise rotation about the axis. System B
F 424 N 0 m 0 N m
As anticipated, the torques in the two cases are not the same. c. The final angular velocity is related to the initial angular velocity 0, the angular acceleration , and the time t by 0 t (Equation 8.4). The angular acceleration is given by Newton's second law for rotational motion as / I , where is the net torque and I is the moment of inertia. Since there is only one torque acting on each system, it is the net torque, so . Substituting this expression for into Equation 8.4 yields
0 t 0 t I
In both cases the initial angular velocity is 0 = 0 rad/s, since the systems start from rest. The final angular velocities after 5.00 s are:
Chapter 9 Problems
481
System A
1270 N m 0 t 0 rad/s + 5.00 s 27.7 rad/s 2 I 229 kg m 0 Nm 0 t 0 rad/s + 5.00 s 0 rad/s 2 I
321 kg m ______________________________________________________________________________
80. CONCEPT QUESTIONS a. The rolling wheel has the greater total kinetic energy. Its kinetic energy is the sum of its translational 1 mv 2 and rotational 1 I 2 kinetic energies. The sliding wheel only has 2 2
System B
translational kinetic energy, since it does not rotate. b. The rolling wheel has the greater potential energy when it comes to a halt. As the wheels move up the incline plane, the total mechanical energy is conserved, since only the conservative force of gravity does work on each wheel. Thus, the initial kinetic energy at the bottom of the incline is converted entirely into potential energy when the wheels come to a momentary halt. Since the rolling wheel has the greater kinetic energy to begin with, it also has the greater potential energy. c. The rolling wheel rises to the greater height. The potential energy PE is given by Equation 6.5 as PE = mgh, where h is the height of the wheel above an arbitrary zero level. Since the rolling wheel has the greater potential energy when it comes to a halt, it also rises to a greater height. SOLUTION a. The total kinetic energy KE of the rolling wheel is the sum of its translational and rotational kinetic energies KE = 1 mv2 + 1 I 2 2 2 where is the angular speed of the wheel. Since the wheel is a disk, its moment of inertia is I = 1 mR2 (see Table 9.1), where R is the radius of the disk. Furthermore, the angular speed 2 Rolling Wheel
of the rolling wheel is related to the linear speed v of its center of mass by Equation 8.12 as = v/R. Thus, the total kinetic energy of the rolling wheel is
KE =
1 mv 2 2 3 mv 2 4
+
3 4
1 2
I
2
1 mv 2 2
+
1 1 mR 2 2 2 2
v R
2
2.0 kg 6.0 m/s
54 J
The kinetic energy of the sliding wheel is
482 ROTATIONAL DYNAMICS
Sliding Wheel
KE =
1 2
mv 2
1 2
2.0 kg 6.0 m/s 2
36 J
As expected, the rolling wheel has the greater total kinetic energy. b. As each wheel rolls up the incline, its total mechanical energy is conserved. The initial kinetic energy KE at the bottom of the incline is converted entirely into potential energy PE when the wheels come to a momentary halt. Thus, the potential energies of the wheels are: Rolling Wheel Sliding Wheel
PE = 54 J
PE = 36 J
As anticipated, the rolling wheel has the greater potential energy. The potential energy of a wheel is given by Equation 6.5 as PE = mgh, where g is the acceleration due to gravity and h is the height relative to an arbitrary zero level. Therefore, the height reached by each wheel is as follows: Rolling Wheel Sliding Wheel
h= PE 54 J = = 2.8 m mg 2.0 kg 9.80 m/s 2
h=
PE 36 J = = 1.8 m mg 2.0 kg 9.80 m/s 2
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