# Calculating the period of a pendulum

1. Jul 14, 2009

### dummybbmm

Let the moment of inertia of a pendulum be I and the distance from pivot to the center of gravity of the pendulum, D. Use the conversation of energy principle to show that the period of this pendulum is 2 * pi * sqrt(I / [M*g*D]).

My attempt is as follows,

I, displace the pendulum by small angle theta. Let, h, be the vertical displacement of the center of mass. The potential energy of the displaced pendulum equates to kinetic energy at the bottom. (w is omega)

Hence, m*g*h = 1/2 * I * w^2

When I rearrange the terms of this equation, I obtain, w = sqrt(2*m*g*h/I)

Substituting into the period T = 2*pi / w, I obtain T = 2*pi* sqrt(I/[2*m*g*h])

I am stuck here because I do not know how to eliminate h from the equation. Can anybody help me here? It would be very much appreciated. My apologies for the lack of latex as I am not familiar with that syntax.

Last edited: Jul 14, 2009
2. Jul 14, 2009

### tiny-tim

Welcome to PF!

Hi dummybbmm!

(have a pi: π and an theta: θ and an omega: ω and a square-root: √ and try using the X2 tag just above the Reply box )
Isn't h the same as D?

But I don't think you have the correct formula … your ω is the angular velocity at the bottom, not the ω in the θ = θ0sinωt formula.

3. Jul 14, 2009

### dummybbmm

Thanks tiny-tim, you're a lifesaver.

As you said, the ω is not the same ω in the θ = θ0sinωt formula.