# Moment of Inertia of a Pendulum

1. Jul 3, 2009

### MyNewPony

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

The 20 cm-long wrench in the figure swings on its hook with a period of 0.92s. When the wrench hangs from a spring of spring constant 350 N/m, it stretches the spring 3.5 cm.

What is the wrench's moment of inertia about the hook?

http://session.masteringphysics.com/problemAsset/1070606/9/14.EX25.jpg

2. Relevant equations

I = m*g*L*T^2/2pi

3. The attempt at a solution

Fsp = Fg
kx = mg
m = kx/g
m = (350)(0.035)/9.8 = 1.25

I = (1.25)(9.8)(0.14)(0.92)^2/2pi = 0.23 kg*m^2

This isn't the correct answer however. Can someone explain what I did wrong?

2. Jul 3, 2009

### alphysicist

Hi MyNewPony,

This formula does not look quite right to me. Do you see what it needs to be?

3. Jul 3, 2009

### MyNewPony

Ah. I forgot to square the 2pi.

So the equation becomes:

I = mgLT^2/4pi^2

Is that correct?

4. Jul 3, 2009

### Staff: Mentor

This equation isn't quite right. It's off by a factor of 2pi.

5. Jul 3, 2009

### alphysicist

That looks right to me.

Last edited: Jul 3, 2009
6. Jul 3, 2009

Yes.

7. Jul 3, 2009

### MyNewPony

Thanks a bunch!

8. Jul 3, 2009

### alphysicist

9. Dec 13, 2010

### ElTaco

Sorry for bumping up an old thread, but thought it'd be better than making a new one about the same problem. Can someone tell me where equation for I is derived from? From my knowledge I know that I = cMR^2 (as an estimated value), but what exactly do you plug in to get to that point? (I = mgLT^2/4pi^2 )

10. Dec 14, 2010

### alphysicist

Hi ElTaco,

It's a standard form for the period of a physical pendulum. The derivation should be in your book, and you might also look at:

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

11. Dec 14, 2010