# Math Help - Circular Motion - Mechanics Question

1. ## Circular Motion - Mechanics Question

This question has been bugging me for ages.... I can't seem to draw the diagram correctly.

A light inextensible string of length 3 metres is threaded through a smooth vertical ring which is free to turn. The spring carries a particle at each end. One particle, P, of mass m is at rest at a distance of 1 metre below the ring, while the other particle B of mass M is rotating in a horizontal circle whose centre is P.

Show that the velocity, v, of B in terms of m, and M is given by:

$\sqrt {\frac{{(M+m)g }} {M}}$

2. Originally Posted by shinn
This question has been bugging me for ages.... I can't seem to draw the diagram correctly.

A light inextensible string of length 3 metres is threaded through a smooth vertical ring which is free to turn. The spring carries a particle at each end. One particle, P, of mass m is at rest at a distance of 1 metre below the ring, while the other particle B of mass M is rotating in a horizontal circle whose centre is P.

Show that the velocity, v, of B in terms of m, and M is given by:

$\sqrt {\frac{{(M+m)g }} {M}}$
I apologize, my ability to visualize can be rather bad at times. Can you post a diagram of some kind? I'm just not seeing the physical setup here.

-Dan

3. Originally Posted by shinn
This question has been bugging me for ages.... I can't seem to draw the diagram correctly.

A light inextensible string of length 3 metres is threaded through a smooth vertical ring which is free to turn. The spring carries a particle at each end. One particle, P, of mass m is at rest at a distance of 1 metre below the ring, while the other particle B of mass M is rotating in a horizontal circle whose centre is P.

Show that the velocity, v, of B in terms of m, and M is given by:

$\sqrt {\frac{{(M+m)g }} {M}}$
Perhaps I still have the physical situation wrong. I'm getting
$v = \sqrt{2g \left ( \frac{m^2 - M^2}{Mm} \right ) }$

-Dan