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Conservation of angular momemtum

  • Thread starter nautica
  • Start date

nautica

Seems easy enough - but could I get a check on this.

Thanks
Nautica

A 1.8 m radius merry go round has a mass of 120 kg and is rotating with an angular velocity of .5 rev/s.

What is its angular velocity after a 22 kg childs gets on its edge which was initially at rest???

First, I determined that this was an Inelastic collision with conservation of momentum

Formula

Iw1 + Iw2 = Iw (final)

I for a disc was determined to be (1/2)MR^2

I converted the .5rev/s to 3.14rad/s

So the work is as follows:

(1/2)(120 kg * 1.8m^2)(3.14 rad/s) + 0 = ((22kg + 120 kg) * 1.8m^2)W(final)

so w(final) = 1.33 rad/s

Thanks again
Nautica
 
512
0
I think the equation should read
(1/2)*120 kg*(1.8m)^2*(3.14 rad/s) + 0 = ((1/2)*120 kg*(1.8m)^2 + 22 kg*(1.8m)^2)W(final)

Remember, your final I is a sum of the merry-go-round's (whose mass is distributed over a disk), and the child's (whose mass isn't.)
 

nautica

Thanks
 

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