Conservation of Angular Momentum

  • Thread starter icf927
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  • #1
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Homework Statement


A solid cylinder merry-go-round of mass 250kg and radius of 1.9m spinning at 1 revolution every 5 seconds has a 40kg child sitting at 1.1m from the axis. A 50kg child, running tangentially at 3m/s, jumps on the merry go round at the outer edge. What is the new rotation rate in rev/s?


Homework Equations


I1ω1=I2ω2


The Attempt at a Solution


(1/2MR^2 +m1r^2)ω1 + (m2r^2)(v/r)=(1/2MR^2 +m1r^2 +m2r^2)ω2
I plugged in all the variables and solved for ω2 and got 0.214 rev/s. Did I set up the equation correctly and come about the right answer?
 

Answers and Replies

  • #2
ehild
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The radii are different for the children.

ehild
 
  • #3
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yes I plugged in the different radii. I just forgot to put a 1 and 2 behind them in the equations.
 
  • #4
ehild
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Could you please show your calculations in detail?

ehild
 
  • #5
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(1/2(250kg)(1.9m)^2)(1.256637061rad/s)+((50kg)(1.9m)^2(3m/s/1.9m)=(1/2(250kg)(1.9m)^2+(40kg)(1.1m)^2+(50kg)(1.9m)^2)w2.

solved for angular velocity(w2) and got 0.214rev/s. is this the right set up?
 
  • #6
ehild
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(1/2(250kg)(1.9m)^2)(1.256637061rad/s)+((50kg)(1.9m)^2(3m/s/1.9m)=(1/2(250kg)(1.9m)^2+(40kg)(1.1m)^2+(50kg)(1.9m)^2)w2.

solved for angular velocity(w2) and got 0.214rev/s. is this the right set up?

You left out the first child on the left hand side.

ehild
 
  • #7
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I did it on paper just forgot to type it on here. When I include the first child on the left hand side, is that the right setup?
 
  • #8
ehild
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It is, but I got different result. Show calculations in detail.

ehild
 
  • #9
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Did you change the initial angular velocity from revolution/sec to rad/sec and then change the final answer back to revolutions/sec
 
  • #10
ehild
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Ops! You are right. I calculated w. So your result is correct.

ehild
 

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