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326_PartUniversity Physics Solution

326_PartUniversity Physics Solution - Dynamics of...

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Dynamics of Rotational Motion 10-21 S ET U P : The force diagram for the cylinder is given in Figure 10.62b. E XECUTE : z z I τ α = , z Fl TR I α = where 0.12 m l = and 0.25 m R = a R α = so / z a R α = / Fl TR Ia R = + Figure 10.62b 2 2 0.25 m (2.9 kg m )(0.80 m/s ) 530 N 1200 N 0.12 m (0.25 m)(0.12 m) R Ia F T l Rl = + = + = E VALUATE : 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. I DENTIFY : Apply ext cm m = F a ! ! and cm z z I τ α = to the roll. S ET U P : 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. E XECUTE : (a) Balancing vertical forces, rod cos , F f w F θ = + + and balancing horizontal forces rod k sin . With , F n f n θ μ = = these equations become rod k cos , F n F w θ μ = + + rod sin . F n θ = 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 w F F θ μ 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 ( ) , and F f R f n μ = may be found by insertion of the value found for rod F into either of the above relations; i.e., k rod sin 33.2 N. f F μ θ = = Then, 2 2 2 (40.0 N 31.54 N)(18.0 10 m) 4.71 rad/s . (0.260 kg m ) I τ α × = = = E VALUATE : 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 α for a given F . 10.64. I DENTIFY : Apply z z I τ α = 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) S ET U P : Apply z z I τ α = to the rotation of the flywheel about the axis. The free-body diagram for the flywheel is given in Figure 10.64a. E XECUTE : The forces n and Mg act at the axis so have zero torque. z TR τ = z TR I α = Figure 10.64a
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10-22 Chapter 10 S ET U P : Apply m = F a ! ! to the translational motion of the block. The free-body diagram for the block is given in Figure 10.64b. E XECUTE : y y F ma = cos36.9 0 n mg ° = cos36.9 n mg = ° k k k cos36.9 f n mg μ μ = = ° Figure 10.64b x x F ma = k sin36.9 cos36.9 mg T mg ma μ ° − ° = k (sin36.9 cos36.9 ) mg T ma μ ° − ° − = But we also know that block wheel , a R α = so / . a R α = Using this in the z z I τ α = equation gives / TR Ia R = and 2 ( / ) . T I R a = Use this to replace T in the x x F ma = equation: 2 k (sin36.9 cos36.9 ) ( / ) mg I R a ma μ ° − ° − = k 2 (sin36.9 cos36.9 ) / mg a m I R μ ° − ° = + 2 2 2 2 (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 = = E VALUATE : If the string is cut the block will slide down the incline with 2 k sin36.9 cos36.9 3.92 m/s . a
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