What is the Speed of the Upper End of a Falling Pole?

[HPVic03]

Please help me get this, it's due tonigh tonight on webassign. The question is:

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??

 

[cscott]

My editing caused double posted for some reason. :\

 

[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

 

[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?

 

[cscott]

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?

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.

 

[HPVic03]

hey thanks i got it

 

[cscott]

No Problem