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anonymous

  • 4 years ago

A 4.20 m diameter merry-go-round is rotating freely with angular speed of 0.800 rad/s. Its total moment of inertia is 1260 kg m squared. How much work will bring the merry-go-round to rest?

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  1. JamesJ
    • 4 years ago
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    The rotational kinetic energy of the merry-go-round is \[ KE_r = \frac{1}{2}I\omega^2 \] Now, you want to reduce that to zero. Hence by the work-energy theorem, you need to do exactly an equal amount of work.

  2. JamesJ
    • 4 years ago
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    make sense? It's analogous to asking "How much work is required to stop a 1000 kg car moving at 10 m/s?" We need to do \[ KE = \frac{1}{2}mv^2 \] of work.

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