# Revolving Ball on String Problem

1. Apr 22, 2005

### PhysicsinCalifornia

Another physics question for all you smart people out there

A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane with the cord making a certain degree with the vertical. If the ball is moving at 4.0 m/s, how can I find the angle that the cord makes with the vertical?

This one is a toughie!

I worked on this a couple of hours, but I keep on running into the same dead ends, could anyone give us some pointers?

(This is the direction I've been trying to go with no luck)

Fx = Tsin(theta) = m((v^2)/r)
Fy = Tcos(theta) - Fg = 0

2. Apr 22, 2005

### Berislav

What does $r$ stand for in your equation?

3. Apr 22, 2005

### PhysicsinCalifornia

4. Apr 22, 2005

### Berislav

Well, of course. :tongue2: I meant what value does $r$ take in your equation.

EDIT: Because, you know, it's not 1.5 m.

Last edited: Apr 22, 2005
5. Apr 23, 2005

### Staff: Mentor

So far, so good. As I think Berislav was trying to point out, r depends on $\theta$: rewrite r in terms of the length and the angle. Also, realize that $F_g = mg$.

Then just eliminate T and solve for the angle. Hint: You may need to use a trig identity.

(Note: We can't tell where you got stuck, since you only showed the first step in your solution.)

6. Apr 27, 2005

### PhysicsinCalifornia

Well,

I got up to where

tan(theta)sin(theta) = 1.088

I just need to find (theta) now.

Any help????

Thanks

7. Apr 27, 2005

### Staff: Mentor

Maybe this will help:
$$\sin^2\theta / \cos\theta = (1 - \cos^2\theta)/ \cos\theta$$

8. Apr 27, 2005

### PhysicsinCalifornia

Yea, that was my next step.

**New info**

I was informed by my friend that I should use the half-angle identity for this!

I don't know how to answer this problem yet, but I'm getting there

9. Apr 28, 2005

### Staff: Mentor

Use what I gave you in post #7 and solve for $\cos\theta$. (Rewrite the expression as a quadratic equation in $\cos\theta$, then solve it.)