# Physical pendulum hole

1. Apr 6, 2009

### nns91

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

You are given a meter stick and asked to drill a hole in it so that when pivoted about the hole the period of the pendulum will be a minimum. Where should you drill the hole

2. Relevant equations

T=2*pi*sqrt(I/mgd)

3. The attempt at a solution

So I use parallel axis theorem for I and get I=I(com)+mh^2=(1/12)mL^2+mh^2 = m*((1/12)+h^2) because L=1 as in a meter stick.

Plug in the formula I have T= 2*pi * sqrt( (1/12)+h^2 / gh) because m cancels out and d=h because h is the distance from center of mass.

How should I move one to get the final answer ???

2. Apr 6, 2009

### rl.bhat

Take differentiation of T with respect to h and equate to zero.

3. Apr 6, 2009

### nns91

Ok, thanks, I got like 0.238, don't know if it sounds reasonable. Can you check that for me please ??

4. Apr 6, 2009

### rl.bhat

Check it again. I am getting different answer.

5. Apr 6, 2009

### nns91

I punched the whole expression in my calculator and set it equal 0 and use the solver. That's what it gave me. What did you get ?

6. Apr 6, 2009

0.288675 m