Finding angular speed of a thin rod.

1. Mar 9, 2008

Seraph404

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

A thin 1.0-meter-long rod pivoted at one end falls (rotates) from a horizantal position, starting from rest and with no friction. What is the angular speed of the rod when it is vertical?

2. Relevant equations

I (of the thin rod) = (1/3)ML^2

3. The attempt at a solution

I tried solving this problem using conservation of energy, but I'm not getting the right answer (5.4 rad/s).

For kinetic energy, I was using (1/2)I$$\omega$$^2; and for potential energy, I was using mgL. Then I solved for $$\omega$$. What's wrong with that?

2. Mar 9, 2008

Staff: Mentor

In figuring out the change in PE of the rod, consider what happens to its center of mass.

3. Mar 9, 2008

Seraph404

It accelerates?

4. Mar 9, 2008

Staff: Mentor

How does its position change? That determines the change in PE of the rod.

5. Mar 9, 2008

Seraph404

The CoM moves from a height of L to a height of L/2 ?

6. Mar 9, 2008

Staff: Mentor

Good. So what's the change in PE of the rod?

7. Mar 9, 2008

Seraph404

MgL - Mg(L/2)

I got the answer now, thanks!