hw12solution - Homework 12 Solution CHANG 1 From the...

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Homework 12 Solution CHANG 1 From the right-hand rule F 1 produces a counterclockwise torques while F 2 produces a clockwise torque. So the total torque is   0 1 2 F 1 l 1 F 2 l 2 Here the counterclockwise rotation is taken to be positive. For the vertical force F 1 the lever arm is the horizontal distance from the pivot: l 1 2.5 1 cos . The lever arm corresponding to the horizontal force F 2 is the vertical distance from the pivot: l 2 2.5sin . Using cos 0.8 and sin 0.6, we get the net torque to be:   0 F 1 l 1 F 2 l 2 100 1.5 0.8 120 0.6 120 180 60  0 60 Nm, CW CHANG 2 The total moment of inertia of the entire assembly is the sum of the moments of inertia from all the parts. Using the subscripts r for the rod and s for the sphere, we have I I r I s 1 3 m r L 2 m s r 2 1 3 5 2 10 0.2 2 1Kgm 2 Note that for a point mass r is the distance from the pivot to the particle. CHANG 3 The moment of inertia for the entire assembly is the sum from the disk and the rod: I I d I r . For the disk its moment of inertia is I d 1/2 m d R 2 . Since the pivot is neither at the center nor at the end of the slender rod, we apply the parallel axis theorem to calculate the rod’s moment of inertia: I r I cm m r d 2 ,inwhich d is the distance from the center of mass to the pivot and can be found from geometry. So the total moment of inertia is:
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This note was uploaded on 03/05/2009 for the course PHYS 2305 taught by Professor Tschang during the Fall '08 term at Virginia Tech.

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hw12solution - Homework 12 Solution CHANG 1 From the...

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