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A large stick is pivoted about one end and allowed to swing back and forth as a physical penulum. The mass of the stick is 5.90 kg and its center of gravity (found by finding its balance point) is 1.2 m from the pivot. If the period of the swinging stick is 3.90 seconds, what is its moment of inertia, about an axis through the pivot?

I used the parallel-axis theorem (Ip=Icm+Mh^2)....which turns out to be 1/3ML^2 ....so I get 11.328 kgm^2 as the moment of inertia. I am not sure why the period is given. I know T=(2pi)/omega and omega I think is sqrt(MgL/Ip)....so if I solve for Ip given the period, I get a different Ip than my previous calculation. Any enlightenment would be great.

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# Homework Help: Physical pendulum of stick swing

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