A merry go around problem

  • Thread starter phyhelpme
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  • #1
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What do I need to do with this problem?

Homework Statement



A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kgm2 and is rotating at 15.0 rev/min about a frictionless vertical axle. Facing the axle, a 35.0 kg child hopes onto the merry-go-round. What is the new angular speed of the merry-go-round?
 

Answers and Replies

  • #2
DaveC426913
Gold Member
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Again, show your work.
 
  • #3
10
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How does this look?

Ok so moment of intertia =35kg(2^2)=140
15rev/m*pi/30=1.57 rad/s

Linitial=Lfinal
Linitial=I*angular speed
250(1.57)=263.12w
392.5=263.12w
1.49 rad/s=w

Converting back to rev/min:
(1.49*60)/2pi=14.23 rev/min
 
  • #5
10
0
I think this now is the correct answer.
Final moment of inertial (if)= 250kgm^2+(35kg)(2m)^2=390kgm^2

IiWi=IfWf
(250)(15)=390wf
3750=390wf
= 9.62 rev/min
 

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