Angular momentum of spinning mass on wheel

  • Thread starter elemis
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  • #1
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Homework Statement



An object of mass 2 kg is held in circular motion by a string of negligible mass.
Initially the radius is 1 m and the tangential velocity is 4 m/s. The string is
subsequently let out gradually until the radius doubles to 2 m. There are no
frictional or resistive forces present. What is the final tangential velocity?


Homework Equations


v=wr
L=Iw


The Attempt at a Solution



Initially the angular momentum is : 2*12*[itex]\frac{4}{1}[/itex] = L = 8

By conservation of angular momentum we should have L = 8 when the string becomes 2m long :

2*22*[itex]\frac{v^2}{2}[/itex] = 8 v=20.5

Is this correct ?
 

Answers and Replies

  • #2
TSny
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In your equation for the final angular momentum, you squared the velocity. Is that correct?
 
  • #3
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In your equation for the final angular momentum, you squared the velocity. Is that correct?
Yes, is that a problem ?
 
  • #4
TSny
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Did you square the velocity when finding the initial angular momentum? Why should it be squared in the final angular momentum?
 
  • #5
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Did you square the velocity when finding the initial angular momentum? Why should it be squared in the final angular momentum?
Ooops !!! I have absolutely no idea where that came in from. YIKES !

So final tangential velocity is 2, correct ?
 
  • #6
TSny
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Yes, that's the correct answer.
 

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