Question

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person starts to walk around the perimeter of the disk at a speed of 0.80 m/s relative to the ground. What rotation rate does this motion impart to the disk if
`I_(disk) = 20,000 kg*m^2`. (The person's moment of inertia is
`I = mr^2`)

Answer 1

`\omega = 0.016\,(rad)/(s)`

Step-by-step explanation:

The rotation rate of the man is:

`\omega = (v)/(R)`

`\omega = (0.80\,(m)/(s) )/(5\,m)`

`\omega = 0.16\,(rad)/(s)`

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

`(90\,kg)\cdot (5\,m)^(2)\cdot (0.16\,(rad)/(s) ) = [(90\,kg)\cdot (5\,m)^(2)+20000\,kg\cdot m^(2)]\cdot \omega`

The final angular speed is:

`\omega = 0.016\,(rad)/(s)`