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Download Chapter 10 Homework and Practice Problems 10.1, 10.10, 10.17
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Chapter 10 Homework and Practice Problems 10.1, 10.10, 10.17, 10.32, 10.37, 10.42, 10.63, 10.82 10.1. Use Eq. (10.2) to calculate the magnitude of the torque and use the right-hand rule illustrated in Figure 10.4 in the textbook to calculate the torque direction. (a) SET UP: Consider Figure 10.1a. EXECUTE: Fl l rsin (400 m)sin90 l 400 m (10.0 N)(4.00 m) 400 N m Figure 10.1a This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector is directed out of the plane of the figure. (b) SET UP: Consider Figure 10.1b. EXECUTE: Fl l rsin (400 m)sin120 l 3464 m (10.0 N)(3.464 m) 346 N m Figure 10.1b This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector is directed out of the plane of the figure. (c) SET UP: Consider Figure 10.1c. EXECUTE: Fl l rsin (400 m)sin30 l 200 m (10.0 N)(2.00 m) 200 N m Figure 10.1c This force tends to produce a counterclockwise rotation about the axis; by the right-hand rule the vector is directed out of the plane of the figure. (d) SET UP: Consider Figure 10.1d. EXECUTE: Fl l rsin (2.00 m)sin60 1.732 m (10.0 N)(1.732 m) 173 N m Figure 10.1d This force tends to produce a clockwise rotation about the axis; by the right-hand rule the vector is directed into the plane of the figure. (e) SET UP: Consider Figure 10.1e. EXECUTE: Fl r 0 so l 0 and 0 Figure 10.1e (f) SET UP: Consider Figure 10.1f. EXECUTE: Fl l rsin, 180, so l 0 and 0 Figure 10.1f The torque is zero in parts (e) and (f) because the moment arm is zero; the line of action of the force passes through the axis. 10.10. The constant force produces a torque which gives a constant angular acceleration to the disk and a linear acceleration to points on the disk. ANALYZE: z I z applies to the disk, z2 02z 2 z ( 0 ) because the angular acceleration is constant. The acceleration components of the rim are atan r and arad r 2 , 2 2 arad . and the magnitude of the acceleration is a atan EXECUTE: (a) z I z gives Fr I z . For a uniform disk, I 1 1 Fr (300 N)(0200 m) 750 rad/s 2. MR 2 (400 kg)(0200 m) 2 0800 kg m 2. z 2 I 2 2 0800 kg m 0 0.200 rev 1.257 rad. 0 z 0, so z2 02z 2 z ( 0 ) gives z 2(750 rad/s2 )(1257 rad) 4342 rad/s. v r (0200 m)(4342 rad/s) 0868 m/s. (b) atan r (0200 m)(750 rad/s2 ) 150 m/s 2. 2 2 arad 406 m/s2. arad r 2 (0200 m)(4342 rad/s)2 3771 m/s2. a atan The net acceleration is neither toward the center nor tangent to the disk. 10.17. Apply F ma to each box and z I z to the pulley. The magnitude a of the acceleration of each box is related to the magnitude of the angular acceleration of the pulley by a R . ANALYZE: The free-body diagrams for each object are shown in Figure 10.17a–c. For the pulley, R 0250 m and I 12 MR2. T1 and T2 are the tensions in the wire on either side of the pulley. m1 120 kg, m2 500 kg and M 200 kg. F is the force that the axle exerts on the pulley. For the pulley, let clockwise rotation be positive. EXECUTE: (a) Fx max weight gives a R and and for the 12.0 kg box gives m2 g T2 m2a. z I z T2 T1 12 Ma. T1 m1a. Fy ma y for the pulley gives for the 5.00 kg (T2 T1) R Adding these three equations gives 12 MR2 . m2 g (m1 m2 12 M )a m2 500 kg 2 2 a g (980 m/s ) 272 m/s . m1 m2 1 M 12 0 kg 5 00 kg 1 00 kg 2 T1 m1a (120 kg)(272 m/s2 ) 326 N. m2 g T2 m2a Then gives T2 m2 ( g a) (500 kg)(980 m/s 272 m/s ) 354 N. The tension to the left of the pulley is 32.6 N and below the pulley it is 35.4 N. 2 (b) a 272 m/s2 2 (c) For the pulley, Fx max gives Fx T1 326 N and Fy ma y gives Fy Mg T2 (200 kg)(980 m/s2 ) 354 N 550 N. The equation three objects. m2 g (m1 m2 12 M )a says that the external force m2 g must accelerate all 10.32. The power output of the motor is related to the torque it produces and to its angular velocity by P zz , where z must be in rad/s. ANALYZE: The work output of the motor in P 60.0 s is 2 (900 kJ) 600 kJ, 3 so 600 kJ 100 W. z 2500 rev/min 262 rad/s. 600 s EXECUTE: z P z 100 W 0382 N m 262 rad/s For a constant power output, the torque developed decreases when the rotation speed of the motor increases. 10.37. (a) Use L mvr sin (Eq. (10.25)): Consider Figure 10.37. L mvrsin (200 kg)(120 m/s)(800 m)sin1431 EXECUTE: L 115 kg m2 /s Figure 10.37 To find the direction of L apply the right-hand rule by turning r into the direction of v by pushing on it with the fingers of your right hand. Your thumb points into the page, in the direction of L. (b) By Eq. (10.26) the rate of change of the angular momentum of the rock equals the torque of the net force acting on it. EXECUTE: mg (800 m) cos 369 125 kg m2 /s2 To find the direction of and hence of dL/dt , apply the right-hand rule by turning r into the direction of the gravity force by pushing on it with the fingers of your right hand. Your thumb points out of the page, in the direction of dL/dt. L and dL/dt are in opposite directions, so L is decreasing. The gravity force is accelerating the rock downward, toward the axis. Its horizontal velocity is constant but the distance l is decreasing and hence L is decreasing. 10.42. ANALYZE: L is conserved if there is no net external torque. Use conservation of angular momentum to find at the new radius and use K 12 I 2 to find the change in kinetic energy, which is equal to the work done on the block. EXECUTE: (a) Yes, angular momentum is conserved. The moment arm for the tension in the cord is zero so this force exerts no torque and there is no net torque on the block. (b) L1 L2 so I11 I22. Block treated as a point mass, so of the block from the hole. I mr 2, where r is the distance mr121 mr222 2 2 r1 0300 m 1 (175 rad/s) 700 rad/s 0150 m r2 2 (c) K1 12 I112 12 mr1212 12 mv12 v1 r11 (0300 m)(175 rad/s) 0525 m/s K1 12 mv12 12 (00250 kg)(0525 m/s)2 000345 J K2 12 mv22 v2 r22 (0150 m)(700 rad/s) 105 m/s K2 12 mv22 12 (00250 kg)(105 m/s) 2 001378 J K K2 K1 001378 J 000345 J 00103 J (d) But Wtot K Wtot W , the work done by the tension in the cord, so W 00103 J. Smaller r means smaller I. L I is constant so increases and K increases. The work done by the tension is positive since it is directed inward and the block moves inward, toward the hole. 10.63. ANALYZE: Use z I z to find the angular acceleration just after the ball falls off and use conservation of energy to find the angular velocity of the bar as it swings through the vertical position. SET UP: The axis of rotation is at the axle. For this axis the bar has 1 m L2 , I 12 bar where and L 0800 m. Energy conservation gives K1 U1 K2 U2. The gravitational potential energy of the bar doesn’t change. Let y1 0, so y2 L/2. mbar 380 kg EXECUTE: (a) z mball g (L/2) and z z mball g ( L/2) 2 1 m L2 m ball ( L/2) 12 bar 2 2 1 m L2 m I I ball I bar 12 bar ball ( L/2) . z I z 2g mball L mball mbar /3 gives and 2(980 m/s ) 250 kg 2 163 rad/s . 0800 m 250 kg [380 kg]/3 (b) As the bar rotates, the moment arm for the weight of the ball decreases and the angular acceleration of the bar decreases. (c) K1 U1 K2 U 2 . 0 K2 U 2. mball gL mball L2 /4 mbar L2 /12 1 (I 2 bar I ball ) 2 mball g ( L/2). g 4mball 980 m/s2 4[250 kg] L mball mbar /3 0800 m 250 kg [380 kg]/3 570 rad/s. As the bar swings through the vertical, the linear speed of the ball that is still attached to the bar is v (0400 m)(570 rad/s) 228 m/s. A point mass in free-fall acquires a speed of 2.80 m/s after falling 0.400 m; the ball on the bar acquires a speed less than this. 10.82. Apply conservation of energy to the motion of the ball as it rolls up the hill. After the ball leaves the edge of the cliff it moves in projectile motion and constant acceleration equations can be used. (a) SET UP: Use conservation of energy to find the speed v2 of the ball just before it leaves the top of the cliff. Let point 1 be at the bottom of the hill and point 2 be at the top of the hill. Take y 0 at the bottom of the hill, so y1 0 and y2 280 m. K1 U1 K2 U 2 EXECUTE: 1 mv 2 1 2 12 I12 mgy2 12 mv22 12 I22 Rolling without slipping means v/r and 7 mv12 10 7 mgy2 10 mv22 v2 v12 10 gy2 1526 m/s 7 1 I 2 2 12 2 mr 2 5 (v/r) 2 15 mv2. SET UP: Consider the projectile motion of the ball, from just after it leaves the top of the cliff until just before it lands. Take y to be downward. Use the vertical motion to find the time in the air: v0 y 0, a y 980 m/s2 , y y0 280 m, t ? EXECUTE: y y0 v0 yt 12 a yt 2 gives t 239 s During this time the ball travels horizontally x x0 v0 xt (1526 m/s)(239 s) 365 m. Just before it lands, v y v0 y a yt 234 m/s and vx v0 x 153 m/s v vx2 v 2y 280 m/s (b) At the bottom of the hill, v/r (250 m/s)/r. The rotation rate doesn’t change while the ball is in the air, after it leaves the top of the cliff, so just before it lands (153 m/s)/r. The total kinetic energy is the same at the bottom of the hill and just before it lands, but just before it lands less of this energy is rotational kinetic energy, so the translational kinetic energy is greater.