Mechanics - conical pendulum,tension/mass?

Question

No luck with this one...please help if you can!

A string of length 3L is threaded through a ring and carries a particle at each end. One particle A, of mass m, is at rest at a distance L below the ring while the other particle B, of mass M, is rotating in a horizontal circle whose centre is A. Find m in terms of M.

Note: there is no friction, and the string is weightless...the only forces are gravity and centripetal force.

Help is greatly appreciated, thank you!

Answer 1

m=2M

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here goes

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if the angle between m and M is x

then

2Lcosx=L

cosx=1/2

if the tention is T

Tcosx =Mg

T=mg

mg cosx=Mg

m*1/2=M

m=2M

sorry it is taking a lot of time to download through what u asked

yeah the link works

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As it is the same string and you say it doesn't have a mass the tention should be same everywhere

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apart from the M s motion the system is  still (esp in the direction of gravity)

Answer 2

Let's call C the point where the thread rests on the ring. C divides the 3L-long thread into CA = 1L and CB = (3L - L) = 2L. It's easy to see that triangle ABC is right-angled at A and cosα = CA/CB = L/2L = 1/2,  ÃƒÂŽÃ‚±= 60° (α is the angle CB makes with the vertical)

The forces acting on particle B are (1) its weight Mg oriented vertically and (2) the tension T of the portion of thread CB, whose direction makes an angle α= 60° with the vertical. The resultant of these 2 forces is the centripetal force oriented on BA, that is horizontal. It follows that Mg/|T|= sinα or |T| = Mg/sinα

The tension that portion CA acts on particle A is |T'| = mg.

For the thead not to slide on the ring it requires that |T'| = |T|, thus

mg = Mg/sinα, or

m = M/sin60°

m = (2√3) M/3