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A thin 0.100-kg rod that is 370 mm long has a small hole drilled through it...

A thin 0.100-kg rod that is 370 mm long has a small hole drilled through it 92.5 mm from one end. A metal wire is strung through the hole, and the rod is free to rotate about the wire.

part a Determine the rod's rotational inertia I about this axis.

part b Determine the period of the rod's oscillation.

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Answer #1

a) I = Icm + m r2 = m L2 / 12 + m r2

= [1/12 * 0.100 * 0.3702] + [0.100 * 0.09252]

= 1.99 * 10-3 kg.m2

b) T = 2 pi * sqrt [I / m g d]

= 2 pi * sqrt [ 1.99 * 10-3 / (0.100 * 9.8 * 0.0925)]

= 0.932 s

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