Question

A merry-go-round of radius 2.74 m and a moment of inertia of 340 kgm2 rotates without friction. It makes 1 revolution every 4.00 s. A child of mass 25.0 kg sitting at the center crawls out to the rim. Find (a) the new angular speed of the merry-go-round, and (b) the kinetic energy change during this process.

Answer 1

a)
`\omega_(f) = 1.012\,(rad)/(s)`, b)
`\Delta K = - 149.352\,J`

Step-by-step explanation:

a) The initial angular speed of the merry-go-round is:

`\omega_(o) = \left((1\,rev)/(4\,s)\right)\cdot \left((2\pi\,rad)/(1\,rev) \right)`

`\omega_(o) \approx 1.571\,(rad)/(s)`

The final angular speed of the merry-go-round is computed with the help of the Principle of Angular Momentum:

`\left(340\,(kg)/(m^(2))\right)\cdot \left(1.571\,(rad)/(s) \right) = \left[340\,(kg)/(m^(2))+(25\,kg)\cdot (2.74\,m)^(2) \right]\cdot \omega_(f)`

`\omega_(f) = 1.012\,(rad)/(s)`

b) The change in kinetic energy is:

`\Delta K = (1)/(2)\cdot \left\{\left[340\,(kg)/(m^(2)) + \left(25\,kg\right)\cdot \left(2.74\,m\right)^(2)\right]\cdot \left(1.012\,(rad)/(s) \right)^(2) - \left(340\,(kg)/(m^(2)) \right)\cdot \left(1.571\,(rad)/(s) \right)^(2) \right\}`
`\Delta K = - 149.352\,J`