Question

What is the angular momentum of a merry go round if we approximate it as a 2kg solid disk with radius 0.1m? The merry go round makes a complete revolution once each second

Answer 1

I = M* R^2 / 2

= 2 * (0.1^2) / 2

= 0.01 kg -m^2

omega = 1 rev / s

= 2pi rads

= 6.28 rad/s

angular momentum = I * omega

= 0.01 * 6.28

= 0.628

Answer 2

f=w/2 $\pi$

=> w=2 $\pi$ rad/s

M=Iw=2*(0.1)²*2 $\pi$ /2=0.628 kg.m².s^-1