A child of mass 20 kg sits at the edge of a merry-go-round of radius 2 m and mass 100 kg...

Question

The merry go round spins freely (w/out friction) at 10 rpm. What is the total angular momentum of the system? Assume that the child is a point mass.

Answer 1

I₀ = moment of inertia of the merry-go-round = MR²/2 = (100)(2)²/2 kg·m² = 200 kg·m²

 

I₁ = moment of inertia of the child = mR² = (20)(2)² kg·m² = 80 kg·m²

 

ω = angular speed of rotating system = 2π(10)/60 rad/s = π/3 rad/s

 

L = total angular momentum = (I₀ + I₁)ω

 

Now just plug in the numbers for I₀, I₁, and ω, and finish from here.