301_PartUniversity Physics Solution

301_PartUniversity Physics Solution - Rotation of Rigid...

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Rotation of Rigid Bodies 9-23 E VALUATE : If the pulley is massless, 2 1 2 98.0 J (4.00 kg 2.00 kg) v = + and 5.72 m/s v = . The moment of inertia of the pulley reduces the final speed of the blocks. 9.87. I DENTIFY and S ET U P : Apply conservation of energy to the motion of the hoop. Use Eq.(9.18) to calculate grav . U Use 2 1 2 K I ω = for the kinetic energy of the hoop. Solve for . ω The center of mass of the hoop is at its geometrical center. Take the origin to be at the original location of the center of the hoop, before it is rotated to one side, as shown in Figure 9.87. Figure 9.87 cm1 cos (1 cos ) y R R R β β = = cm2 0 y = (at equilibrium position hoop is at original position) E XECUTE : 1 1 other 2 2 K U W K U + + = + other 0 W = (only gravity does work) 1 0 K = (released from rest), 2 1 2 2 2 K I ω = For a hoop, 2 cm , I MR = so 2 2 I Md MR = + with d R = and 2 2 , I MR = for an axis at the edge. Thus 2 2 2 2 1 2 2 2 2 (2 ) . K MR MR ω ω = = 1 cm1 (1 cos ), U Mgy MgR β = = 2 cm2 0 U mgy = = Thus 1 1 other 2 2 K U W K U + + = + gives 2 2 2 (1 cos ) MgR MR β ω = and (1 cos )/ g R ω β = E VALUATE : If 0, β = then 2 0. ω = As β increases, 2 ω increases. 9.88. I DENTIFY : 2 1 2 K I ω = , with ω in rad/s. energy P t = S ET U P : For a solid cylinder, 2 1 2 I MR = . 1 rev/min (2 /60) rad/s π = E XECUTE : (a) 3000 rev/min 314 rad/s ω = = . 2 2 1 2 (1000 kg)(0.900 m) 405 kg m I = = 2 2 7 1 2 (405 kg m )(314 rad/s) 2.00 10 J K = = × . (b) 7 3 4 2.00 10 J 1.08 10 s 17.9 min 1.86 10 W K t P × = = = × = × . E VALUATE : In 2 1 2 K I ω = , we must use ω in rad/s. 9.89. I DENTIFY : 1 2 I I I = + . Apply conservation of energy to the system. The calculation is similar to Example 9.9. S ET U P : 1 v R ω = for part (b) and 2 v R ω = for part (c). E XECUTE : (a) 2 2 2 2 2 2 1 1 2 2 1 1 1 ((0.80 kg)(2.50 10 m) (1.60 kg)(5 00 10 m) ) 2 2 2 I M R M R . = + = × + × 3 2 2.25 10 kg m . I = × (b) The method of Example 9.9 yields 2 1 2 1 ( ) gh v I mR = + . 2 3 2 2 2(9.80 m s )(2.00 m) 3.40 m s. (1 ((2.25 10 kg m ) (1.50 kg)(0.025 m) )) v = = + × The same calculation, with 2 R instead of 1 R gives 4.95 m s. v = E VALUATE : The final speed of the block is greater when the string is wrapped around the larger disk. v R ω = , so when 2 R R = the factor that relates v to ω is larger. For 2 R R = a larger fraction of the total kinetic energy resides
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9-24 Chapter 9 with the block. The total kinetic energy is the same in both cases (equal to mgh ), so when 2 R R = the kinetic energy and speed of the block are greater. 9.90. I DENTIFY : Apply conservation of energy to the motion of the mass after it hits the ground. S ET U P : From Example 9.9, the speed of the mass just before it hits the ground is 2 1 /2 gh v M m = + . E XECUTE : (a) In the case that no energy is lost, the rebound height h is related to the speed v by 2 2 v h g ′ = , and with the form for v given in Example 9.9, . 1 2 h h M m ′ = + (b) Considering the system as a whole, some of the initial potential energy of the mass went into the kinetic energy of the cylinder. Considering the mass alone, the tension in the string did work on the mass, so its total energy is not conserved.
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