Circular motion: net work after one period

  • Thread starter nx01
  • Start date
  • #1
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Homework Statement



"A 40 kg child sitting 6.0 m from the center of a merry-go-round has a constant speed of 6.0 m/s. The net work (in Joules) done on the child after making one complete circle on the merry-go-round is?"


Homework Equations



a=V^2/r
F=ma
C=2pi*r

The Attempt at a Solution



C ≈ 37.7 m
a = 6 m/s
F at a point = (40)(6) = 240 N
Net work ≈ (240)(37.7) ≈ 9048 J

My question is, is the net work actually 0 J after one full rotation on the merry-go-round, since force is a vector quantity, and the force in one-half of the circular motion negates the force in the other half for each point?

Thanks in advance!
 

Answers and Replies

  • #2
116
2
You are right and you are wrong... the work is indeed zero, but actually not only on a full rounf, every time... indeed the work is the scalar product of the force times the displacement vector... and in an uniform circular motion the acceleration (and hence the force) is the centripetal one only, and is perpendicular ALWAYS to the displacement, so that the scalar product is identically zero
 
  • #3
16
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Thank you :)
 

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