# Light inextensible string problem

## 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.

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 [Broken]

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?

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Doc Al
Mentor
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
Homework Helper
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.)

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Doc Al
Mentor
T2 - 2mg = 0 by resolving.

So T2 = 2mg.
This is perfectly correct.

Shooting Star
Homework Helper
This is perfectly correct.
So it is. I thought the static mass was also m; overlooked the 2mg in the diagram...Thanks Doc!

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