Angular frequency of a pendulum

[dlp211]

Homework Statement

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

Homework Equations

sqrt(mgd/I)=w

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

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?

 

[ehild]

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

ehild

 

[dlp211]

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

 

[ehild]

Distance of what?

ehild

 

[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

 

[ehild]

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

ehild

 

[dlp211]

Thanks

 

[jtc143]

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

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!

 

[dlp211]

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!

Yes that is the center of mass equation.

The 0.5 comes from the uniform rod of 1m.