Question

A child weighing 22kg is standing on a frictionless merry-go-round at rest. The child begins walking at 1.1m/s around the merry-go-round at a radius of 0.8m. What is the child's speed relative to the ground? (The merry-go-round is a 100kg disk with a radius of 1.2m.)

Answer 1

m = 22 kg, M = 100 kg, r = 0.8 m, R = 1.2 m, v = 1.1 m/s, V= ?

we have to apply conservation of angular momentum;

I_1*w₁ = I_2*w_{2; .........................................................................................................................}(1)

moment of inertia of a clild I₁ = m*r^2 = 14.08 kg-m^2;

moment of inertia of a disk I₂ = 0.5*M*R^2 = 72 kg-m^2;

I_1*w₁ = I_2*w_2;

I_1*v/r = I_2*V/R;      (as we know the relationship v = r*w)

(14.08*1.1) /0.8 = (72*V) /1.2;

linear speed of disk relative to the ground (V) = 0.322 m/s.

therefore speed of child relative to the disk = 1.1 - 0.322 = 0.778 m/s