1. Nov 7, 2005

### HPVic03

A 3.30 m long pole is balanced vertically on its tip. It is given a tiny push. What will be the speed of the upper end of the pole just before it hits the ground? Assume the lower end does not slip.

I know to use MgH = 1/2 MV^2 + 1/2 Iw^2

but when you plug in for w, what do you use? w is rotational velocity and you have to know the time frame, and it doesn't give you that. how do you do this problem??

2. Nov 7, 2005

### cscott

My editing caused double posted for some reason. :\

3. Nov 7, 2005

### cscott

Since there is no slippage, all energy is rotational (PE = RE.) Your moment of inertia in this case is $\frac{1}{3}mL^2$, where "L" is the length of the pole. You can find the potential energy when the pole is standing up (mgh) but remember that the center of mass is at h = L/2. And remember that $v = r\omega$

Oh, and let me guess, Physics - Giancoli :tongue:

Last edited: Nov 7, 2005
4. Nov 7, 2005

### HPVic03

haha yeah its giancoli. thanks so much for your help, but i still have one question though. you you are plugging in v/r for w, what do you use for r?

5. Nov 7, 2005

### cscott

The length of the pole because it asks for the speed of the top end of the pole which is 3.30m from the pivot point.

Last edited: Nov 7, 2005
6. Nov 7, 2005

### HPVic03

hey thanks i got it

7. Nov 7, 2005

No Problem