Circular motion question

Mar 2017
I'm having trouble with this question and I was wondering if you could help.

A light, inelastic string of length 2a is attached to fixed points A and B where A is vertically above B and the distance AB < 2a. A small smooth ring, P, of mass m slides on the string and is moving in a horizontal circle at a constant angular speed ω. The string sections AP and PB are straight and there is the same tension, T, in each section. The distance AP is x and AP and PB make angles α and β respectively with the vertical.

i) Show that xsinα = (2a-x)sinβ. [Done]
ii) By considering the vertical components of the forces on the ring, explain why x >a.

I've drawn a diagram to see what's going on and tried using F=ma and got xcosα = (2a-x)cosβ + mg and then rearranged the equation to get x(cosα+cosβ)=2acosβ+mg but I don't know where to go from there.

UPDATE: x(cosα+cosβ)=2acosβ+mg is wrong, it's Tcosα = Tcosβ + mg. Then since mg is positive that implies that cosβ < cosα which implies that β > α (when 0°<α,β<180°). Then, since angle β is larger, ring P is closer to point B leading to the inequality 2a-x<x which can then be rearranged to show that x>a. Is this correct?
Last edited:


PHF Helper
Jun 2010
Morristown, NJ USA
Yes, your approach is a good one.