Physical pendulum of stick swing

1. Sep 2, 2006

scienceman2k9

Problem:

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.

2. Sep 2, 2006

Päällikkö

You assumed that the stick could be approximated with a thin rod of uniform mass distribution. Apparently this assumption is wrong, the stick doesn't have to be of uniform material nor does it have to resemble a rod.
The following equation applies to a physical pendulum:

$$T = 2 \pi \sqrt{\frac{I_A}{mgl}}$$

where IA is the moment of inertia about the axis through the pivot and l the distance between the pivot and centre of mass.

Last edited: Sep 2, 2006
3. Sep 3, 2006

Thanks!