326_PartUniversity Physics Solution

# 326_PartUniversity - Dynamics of Rotational Motion SET UP 10-21 The force diagram for the cylinder is given in Figure 10.62b EXECUTE z = I z Fl TR

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Dynamics of Rotational Motion 10-21 SET UP: The force diagram for the cylinder is given in Figure 10.62b. EXECUTE: z z I τ α = , z Fl TR I −= where 0.12 m l = and 0.25 m R = aR = so / z = / Fl TR Ia R = + Figure 10.62b 22 0.25 m (2.9 kg m )(0.80 m/s ) 530 N 1200 N 0.12 m (0.25 m)(0.12 m) RI a FT lR l ⎛⎞ =+ = + = ⎜⎟ ⎝⎠ EVALUATE: The tension in the rope is greater than the weight of the crate since the crate accelerates upward. If F were applied to the rim of the cylinder ( l = 0.25 m), it would have the value 567 N. F = This is greater than T because it must accelerate the cylinder as well as the crate. And F is larger than this because it is applied closer to the axis than R so has a smaller moment arm and must be larger to give the same torque. 10.63. IDENTIFY: Apply ext cm m = F a ! ! and cm z z I = to the roll. SET UP: At the point of contact, the wall exerts a friction force f directed downward and a normal force n directed to the right. This is a situation where the net force on the roll is zero, but the net torque is not zero. EXECUTE: (a) Balancing vertical forces, rod cos , Ff w F θ =++ and balancing horizontal forces rod k sin . With , Fn f n θμ == these equations become rod k cos , F w = ++ rod sin . = Eliminating n and solving for rod F gives 2 rod k (16.0 kg) (9.80 m/s ) (40.0 N) 266 N. cos sin cos 30 (0.25)sin30 wF F θμ θ = −° ° (b) With respect to the center of the roll, the rod and the normal force exert zero torque. The magnitude of the net torque is k () , a n d FfR f n μ may be found by insertion of the value found for rod F into either of the above relations; i.e., kr o d sin 33.2 N. fF μθ Then, 2 2 2 (40.0 N 31.54 N)(18.0 10 m) 4.71 rad/s . (0.260 kg m ) I −× = EVALUATE: If the applied force F is increased, rod F increases and this causes n and f to increase. The angle φ changes as the amount of paper unrolls and this affects for a given F . 10.64. IDENTIFY: Apply z z I = to the flywheel and m = F a ! ! to the block. The target variables are the tension in the string and the acceleration of the block. (a) SET UP: Apply z z I = to the rotation of the flywheel about the axis. The free-body diagram for the flywheel is given in Figure 10.64a. EXECUTE: The forces n and Mg act at the axis so have zero torque. z TR = z TR I = Figure 10.64a

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10-22 Chapter 10 SET UP: Apply m = F a ! ! to the translational motion of the block. The free-body diagram for the block is given in Figure 10.64b. EXECUTE: y y Fm a = cos36.9 0 nm g °= cos36.9 g = ° kk k cos36.9 fn m g μμ = Figure 10.64b x x a = k sin36.9 cos36.9 mg T mg ma μ °− − k (sin36.9 cos36.9 ) mg T ma °− ° − = But we also know that block wheel , aR α = so / . = Using this in the z z I τ = equation gives / TR Ia R = and 2 (/ ). TI R a = Use this to replace T in the x x a = equation: 2 k (sin36.9 cos36.9 ) ( / ) mg I R a ma ° − = k 2 (sin36.9 cos36.9 ) / mg a mIR ° = + 2 2 22 (5.00 kg)(9.80 m/s )(sin36.9 (0.25)cos36.9 ) 1.12 m/s 5.00 kg 0.500 kg m /(0.200 m) a ° == +⋅ (b) 2 2 2 0.500 kg m (1.12 m/s ) 14.0 N (0.200 m) T EVALUATE: If the string is cut the block will slide down the incline with 2 k sin36.9 cos36.9 3.92 m/s .
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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326_PartUniversity - Dynamics of Rotational Motion SET UP 10-21 The force diagram for the cylinder is given in Figure 10.62b EXECUTE z = I z Fl TR

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