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

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

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

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$