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University Physics I - Classical Mechanics, 2019

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11.5. EXAMPLES 271<br />

AsshowninSection11.3.2, wehavethen<br />

√<br />

Mgw<br />

ω =<br />

2I<br />

(11.30)<br />

Squaring this, and solving for I/M,<br />

I<br />

M = gw<br />

2ω 2 = 9.8m/s2 × 0.025 m<br />

2 × (2π/5s) 2 =0.0776 m 2 (11.31)<br />

The moment of inertia is to be calculated around the point O, that is to say, the point of suspension<br />

(where the knot is in the figure). For reference, the moment of inertia of a thin rod of length l<br />

around its midpoint is Ml 2 /12 = 0.083l 2 . The length of the meter stick is, of course, 1 m, so the<br />

result I/M ∼ 0.08 m 2 obtained above seems reasonable.

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