Physics Help Forum [exam problem] angular momentum problem

 Jun 22nd 2015, 07:50 AM #1 Junior Member   Join Date: Jun 2015 Posts: 4 [exam problem] angular momentum problem A particle of mass m travels in a horizontal circle of radius R on a frictionless table. The centripetal force is provided by a string passing through a hole in the table attached to two blocks of equal mass M, as shown in the figure. If one hanging block is removed, what will the radius of the particle motion become? My answer is 3/2R, and the correct answer is 2^(1/3)*R......How to count?
 Jun 22nd 2015, 01:46 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,339 You can set up two requations: one for conservation of angular momentum and one for the cetripetal acceleration being cut in half. Then solve for r2/r1. Show us your attempt at solving this and we'll be happy to comment or help you along if you are still stuck.
 Jun 24th 2015, 06:06 AM #3 Junior Member   Join Date: Jun 2015 Posts: 4 ---------------------------------------------- A particle of mass m travels in a horizontal circle of radius R on a frictionless table. The centripetal force is provided by a string passing through a hole in the table attached to two blocks of equal mass M, as shown in the figure. If one hanging block is removed, what will the radius of the particle motion become? ----------------------------------------------- for initial r, w and final R W (ww=w^2) condition 2Mg=mrww Mg=mRWW mrrw=mRRW my derivation mrww=2mRWW-->wRRW/r=2RWW wR/r=2W for E=K+U mrrww/2=mRRWW/2 +Mg(R-r) w=W + Mg(R-r)/mrrw = wR/2r + (R-r)w/2r = wR/2r + Rw/2r -w/2 3w/2=Rw/r 3/2=R/r And there's no 3r/2 in choices......
Jun 24th 2015, 07:52 AM   #4
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 Originally Posted by bes my derivation mrww=2mRWW-->wRRW/r=2RWW
How did you go from the first equation to the second? Try this: you have wr^2=WR^2 from conservation of momentum. Rearrange: w=W(R/r)^2. Now substitute this value for w into the force equation and rearrange to get R/r.

Last edited by ChipB; Jun 24th 2015 at 11:10 AM.

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