circular motion q from patel (1 Viewer)

[totallybord]

hi,
can anybody explain (esp the part where M moved up O) 9Fq9)
A particle of mass m is attached to a fixed point O by a string of length 1 metre and by another string of the same length to a small ring of mass M which can slide on a smooth verticle wire through O. If m describes a horizontal circle with a constant angular velocity w, prove htat its depth below O is {(m+2M)/mw^2}g
thank you!

 

[Yip]

My mechanics are a bit rusty, but this is my attempt.

Denote the tension in the string connected to O as T1, and the tension connected to the sliding ring as T2.

Since the particle is moving in circular motion with constant angular velocity w, horizontal resolution of forces yields
T1sinx+T2sinx=mrw^2=mw^2sinx (where x is the angle between the vertical wire and the string)
T1+T2=mw^2
x is also the angle between the string connected to the ring and the vertical wire, as the 2 strings form an isosceles triangle.
In the string connected to the ring, since the ring is free to move,
T2cosx=Mg
Resolving forces vertically,
T1cosx=mg+T2cosx=g(m+M)

T1+T2=g(m+M)secx+Mgsecx=mw^2
secx=1/y (where y is the depth below O)
Thus, y={(m+2M)/mw^2}g as required