# 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!