# Homework Help: Angular frequency of a pendulum

1. Mar 31, 2012

### dlp211

1. The problem statement, all variables and given/known data

A physical pendulum consists of a small 1.5kg mass at the bottom end of a uniform 1.00m long 1.5kg stick swinging about its upper end. The moment of inertia of the pendulum about its upper end is 2.00kg*m^2. What is the angular frequency

2. Relevant equations

sqrt(mgd/I)=w

I = I(cm)+md^2 = I = (1/3)(1.5)(1^2)+ (1.5)(d^2) = 2

3. The attempt at a solution

m = 1.5+1.5
g = 9.81
d = 1
I = 2

w = sqrt([3*9.81*1]/2) = 3.83 rad/s

According to my solutions manual this is wrong and the correct answer is 3.32 rad/s and somehow d = .75. I don't know how they calculated that, can anyone help?

2. Apr 1, 2012

### ehild

Do you know what d means in the formula for angular frequency?

ehild

3. Apr 1, 2012

### dlp211

I thought d was distance, but I am guessing that it isn't?

4. Apr 1, 2012

### ehild

Distance of what?

ehild

5. Apr 1, 2012

### dlp211

Wait, is supposed to be distance to center mass? The lecture notes don't say this, but I think this is right.

[1.5(.5)+1.5(1.0)]/3 = .75

6. Apr 1, 2012

### ehild

Yes, it is the distance of the CM from the pivot.

ehild

7. Apr 1, 2012

Thanks

8. Apr 2, 2012

### jtc143

Can someone explain to me how this was done? Is that the equation for the center of mass? Where did the number 0.5 come from? Thanks!

9. Apr 2, 2012

### dlp211

Yes that is the center of mass equation.

The 0.5 comes from the uniform rod of 1m.