Light inextensible string problem

[mattgad]

Homework Statement

Two particles of masses m and 2m are connected by a light inextensible string, which is
threaded through a fixed smooth ring. If the lighter particle moves uniformly round a
horizontal circle of radius a, while the other particle remains stationary, find the lighter
particle's speed.

Homework Equations

accel = aw^2

The Attempt at a Solution

Here is my diagram of the problem, I would appreciate it if someone could tell me if this is correct. p is the angle between the 2 ends of string.

http://img166.imageshack.us/img166/7560/question3diagram122lo7.jpg

Particle is moving uniformly with constant angular speed, w.

T2 - 2mg = 0 by resolving.

So T2 = 2mg.

T1 cos p = mg by resolving.

Using F = m(accel), F = T1 sin p, accel = aw^2

T1 sin p = maw^2

Question 1. Are the tensions equal because if they werent, the string would break?
Question 2. Have I now got to go on and find w? How?

Thanks in advance.

 

[Doc Al]

Question 1. Are the tensions equal because if they werent, the string would break?

Yes, the tensions are equal. There's no friction from the ring.

Question 2. Have I now got to go on and find w? How?

Using the vertical force equation you can solve for the angle. Then use it in the horizontal force equation to solve for \omega or v.

a_c = r\omega^2 = v^2/r

 

[Shooting Star]

Particle is moving uniformly with constant angular speed, w.

T2 - 2mg = 0 by resolving.

So T2 = 2mg.

The tension is the same throughout the string, since there is no friction involved anywhere and the string is massless.

For the static mass, T=2mg.

For the revolving mass, Tcosp=mg and Tsinp=centripetal force, as you have written.

Now you can find v.

(EDIT: Overlooked 2mg in the diagram. Mistake pointed out by Doc Al.)

 

[Doc Al]

T2 - 2mg = 0 by resolving.

So T2 = 2mg.

This is perfectly correct.

 

[Shooting Star]

This is perfectly correct.

So it is. I thought the static mass was also m; overlooked the 2mg in the diagram...Thanks Doc!

 

[mattgad]

Thanks for your replies guys.

I end up with v = root(a*g*root(3)).

 

[Doc Al]

Looks good!

 

[mattgad]

So that's all there is to it? For some reason I thought it would be much more complex than this. My knowledge of physics/mechanics is very very limited. I'm actually not sure I even understand the concept of angular speed.