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Old Jun 22nd 2015, 08:50 AM   #1
bes
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Exclamation [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?
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Old Jun 22nd 2015, 02:46 PM   #2
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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.
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Old Jun 24th 2015, 07:06 AM   #3
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Exclamation

----------------------------------------------
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......
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Old Jun 24th 2015, 08:52 AM   #4
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Originally Posted by bes View Post
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 12:10 PM.
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Old Jun 28th 2015, 12:43 PM   #5
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