# Light inextensible string problem

1. Mar 9, 2008

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

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.

2. Relevant equations

accel = aw^2

3. 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?

Last edited by a moderator: May 3, 2017
2. Mar 9, 2008

### Staff: Mentor

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

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.

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

3. Mar 9, 2008

### Shooting Star

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.)

Last edited: Mar 9, 2008
4. Mar 9, 2008

### Staff: Mentor

This is perfectly correct.

5. Mar 9, 2008

### Shooting Star

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

Last edited: Mar 9, 2008
6. Mar 9, 2008

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

7. Mar 9, 2008

Looks good!

8. Mar 9, 2008