# Pendulum without oscillation

1. Oct 20, 2005

### amcavoy

If a mass (m) at the end of a length (L) on a pendulum starts at an angle of θ from the vertical, what is the minimum inital velocity v0 it must have to just barely make it over the top and not oscillate?

This is what I did:

$$\Delta K=mgh\implies v_0=\sqrt{v^2-2gh}$$

but at the top, v is zero so it can be written as:

$$v_0=\sqrt{-2gh}$$

and h in this case is L+Lcosθ, so using g=-10m/s2 I get:

$$v_0=\sqrt{20L\left(1+\cos{\theta}\right)}$$

but whenever I plug in values for θ and L, I get the wrong answer. I can't see what I did wrong.

Any ideas?

Thanks a lot.

2. Oct 20, 2005

### JeremyM

What School do you go to?

3. Oct 20, 2005

### amcavoy

I'm in high school.

4. Oct 21, 2005

### HallsofIvy

Staff Emeritus
At the top the pendulum has 0 velocity (actually, it should have just a tiny non-zero velocity but you can take 0 as "boundary" between going over the top and not making it to the top) and so has no kinetic energy but has potential energy, -mgL (relative to the height of the pivot- the center of the circle the pendulum moves in). Initially, it has velocity v0 and so kinetic energy (1/2)mv0[/sub]2 AND some potential energy. Have you taken into account the fact that the potential energy has to be measured with respect to the center of the circle?