hw11solution

# hw11solution - Homework 11 Solution CHANG 1 The total...

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Homework 11 Solution CHANG 1 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 The sphere’s moment of inertia is based on a point mass: I s m s r s 2 10 0.6 0.2 2 1.6 Kgm 2 Note that for a point mass r is the distance from the pivot to the particle. We then apply parallel-axis theorem to calculate the moment of inertia for the rod: I r I cm m r d 2 1 12 m r L 2 m r d 2 1 12 5 2 5 0.3 2 0.2 Kgm 2 Note that the distance d in parallel-axis theorem is from the pivot point to the center of mass of the rigid body, and it can be measured from geometry. So the total moment of inertia is simply I I r I s 1.6 1.8 Kgm 2 CHANG 2 The moment of inertia for the entire assembly is the sum of those from the disk and the rod: I I d I r For the disk its moment of inertia is found by applying the parallel-axis theorem: I d I cm m d d 2 1 2 m d R 2 m d R 2 1 2 8 0.5 2 8 2 3Kgm 2 Note that for the disk the distance d equals the radius of the disk since the pivot is at the rim of the disk. We also apply the parallel-axis theorem to calculate the rod’s moment of inertia: I r I cm m r d 2 1 12 m r L 2 m r d 2 1 12 5 1.2 2 5 1 2 1.4 Kgm 2 So the total moment of inertia is: I I d I s 3 1.4 4.4 Kgm 2 CHANG 3 To find the angular acceleration using Newton’s 2nd law for rotation, I ,

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## This note was uploaded on 10/17/2009 for the course PHYS 2305 taught by Professor Tschang during the Spring '08 term at Virginia Tech.

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hw11solution - Homework 11 Solution CHANG 1 The total...

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