- #1

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I am trying to find the moment of inertia of a uniform rod, that has a mass added to it at some position along it's length, which is equal to the mass of the rod itself, and the axis of rotation is at one end.

1. Homework Statement

1. Homework Statement

A uniform, [itex]\mathrm{1.00m}[/itex] stick hangs from a horizontal axis at one end and oscillates as a physical pendulum with period [itex]T_{0}[/itex]. A small object of mass equal to that of the stick can be clamped to the stick at a distance [itex]y[/itex] below the axis. The system then has a period [itex]T[/itex].

Find the ratio [itex]\frac{T}{T_{0}}[/itex]

## Homework Equations

## The Attempt at a Solution

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I know that the moment of inertia of a uniform rod with the axis about one end is equal to;

[itex]I = \frac{1}{3}ML^{2}[/itex]

The period of a physical pendulum is given by;

[itex]T = 2 \pi \sqrt{\frac{I}{mgd}}[/itex] where d is the distance from the pivot to the center of gravity. So, for the initial case, we have a period of

[itex]T_{0} = 2 \pi \sqrt{\frac{\frac{1}{3}ML^{2}}{mg\frac{L}{2}}} = 2 \pi \sqrt{\frac{2L}{3g}}[/itex]

I am stuck here though, because I don't know how to find a moment of inertia for the second case. I understand how to calculate moments of inertia of non-uniform rods using integration, but I don't have a function for the linear density...

Thanks for any help you can give!