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 Mar 30th 2017, 05:29 AM #1 Junior Member   Join Date: Mar 2017 Posts: 5 Circular motion question 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-xa. Is this correct? Last edited by punintentional; Mar 30th 2017 at 07:45 AM. Reason: made progess
 Mar 30th 2017, 07:51 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 Yes, your approach is a good one.

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