# Angular Velocity of Rotating Mass

1. Sep 14, 2011

### 20930997

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

A solid sphere of mass m rotates uniformly in a horizontal circle suspended from a fixed
point on a cord of length L and negligible mass. The cord
makes an angle theta from the vertical. What is the angular
velocity?

2. Relevant equations

Torque= (Rotational Inertia)x(angular acceleration)

3. The attempt at a solution

Tried to mess around with a few things but ultimately got no-where so any help would be greatly appreciated.

2. Sep 14, 2011

### 20930997

This is the Diagram for the problem

#### Attached Files:

• ###### Diagram.png
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3. Sep 14, 2011

### Hootenanny

Staff Emeritus
Welcome to Physics Forums.

I would suggest that you draw a free body diagram, identifying all the forces acting on the mass.

4. Sep 14, 2011

### 20930997

thanks for the welcome,

so there would be a tension force through the cord, a gravitational force acting straight down and a centripetal force. am i missing anything?

5. Sep 14, 2011

### Hootenanny

Staff Emeritus
No, that's spot on. So, what can you say about the horizontal and vertical components of the forces acting on the mass?

6. Sep 14, 2011

### 20930997

the vertical components sum to zero.
does the horizontal component of the tension= the centripetal force?

7. Sep 14, 2011

### Hootenanny

Staff Emeritus
Correct!

8. Sep 14, 2011

### 20930997

ok so i got to
tan(theta)=(v^2)/rg where r= Lsin(theta)

i was thinking solve for v and then use Angular velocity=v(r)

is this heading in the right direction?

9. Sep 14, 2011

### Hootenanny

Staff Emeritus
Looks okay to me.

10. Sep 14, 2011

### 20930997

ok so i got angular velocity= g/Lcos(theta)

but the answers tell me that it should be the sqrt of that.
im a bit lost as to where i have made my mistake :S

11. Sep 14, 2011

### 20930997

oh wait i forgot to square both sides of the equation!

12. Sep 14, 2011

### Hootenanny

Staff Emeritus
So you have

$$\frac{v^2}{r} = g\tan\theta$$

And you know that $v = \omega r$. Thus,

$$\omega^2 r = g\tan\theta$$

Do you follow and can you take it from here?

13. Sep 14, 2011

### Hootenanny

Staff Emeritus
Ahh! Never-mind!

No problem, it was a pleasure!