SolutionsSet09

# SolutionsSet09 - Homework Solutions Set 09 Physics 121...

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Homework Solutions Physics 121 Set 09 Spring 2008 - 1 - Problem 1 The moment of inertia of a stick of length L and mass M with respect to the center-of-mass is equal to (1/12) M ( L 2 + W 2 ). The moment of inertia of the stick with respect to the pivot point can be found using the parallel-axis theorem: I O = 1 12 M L 2 + W 2 ( ) + Mx 2 In order to determine the period of the pendulum we need to use the equation of the period of a physical pendulum: T = 2 ! I 0 Mgx = 2 1 12 M L 2 + W 2 ( ) + Mx 2 Mgx = 2 g 1 12 L 2 + W 2 ( ) x + x The period has a minimum or maximum value when dT/dx = 0. This requires that d dx 2 g 1 12 L 2 + W 2 ( ) x + x " # \$ \$ % & = 2 g 1 2 ( 1 12 L 2 + W 2 ( ) x 2 + 1 " # \$ % & 1 12 L 2 + W 2 ( ) x + x = ( 1 12 L 2 + W 2 ( ) x 2 + 1 " # \$ % & 1 12 L 2 + W 2 ( ) x + x = 0 This condition is satisfied when ! 1 12 L 2 + W 2 ( ) x 2 + 1 = 0 or x = 1 12 L 2 + W 2 ( ) = 1 2 3 L 2 + W 2 ( ) The period at this distance is equal to T = 2 g 1 12 L 2 + W 2 ( ) x + x = 2 g x 1 12 L 2 + W 2 ( ) x 2 + 1 = 2 g x = 2 g L 2 + W 2 ( ) 3

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Homework Solutions Physics 121 Set 09 Spring 2008 - 2 - Problem 2 When the cylinder is released, the total mechanical energy of the system is equal to the potential energy of the spring, which is stretched by a distance d : E i = 1 2 kd 2
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SolutionsSet09 - Homework Solutions Set 09 Physics 121...

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