Question

An 85.0 kg child runs in a straight line towards the edge of a stationary merry-go-round at 2.50 m/s. The merry-go-round is in the shape of a disk and has a diameter of 4.50 m and a mass of 235 kg. The child jumps onto and stays on the merry-go- round. What is the angular momentum of the child before this collision, in kg m2/s? (A) What is the moment of inertia of the merry-go-round, in kg m27 [B) What is the final angular speed, in radians/s, of the child-merry-go-round system after the collision? IC]

Answer 1

Mass of the merry-go-round = M = 235 kg

Diameter of the merry-go-round = D = 4.5 m

Radius of the merry-go-round = R = D/2 = 4.5/2 = 2.25 m

Moment of inertia of the merry-go-round = I

I = MR²/2

I = (235)(2.25)²/2

I = 594.84 kg.m²

Mass of the child = m = 85 kg

Velocity of the child at which he is running towards the edge of the merry-go-round = V = 2.5 m/s

Angular momentum of the child before the collision = L₁

L₁ = mVR

L₁ = (85)(2.5)(2.25)

L₁ = 478.125 kg.m²/s

Moment of inertia of the child-merry-go-round after the child jumps on it = I₂

I₂ = I + mR²

I₂ = 594.84 + (85)(2.25)²

I₂ = 1025.15 kg.m²

Angular speed of the merry-go-round system after the collision = $\omega$

Angular momentum of the system after the collision = L₂

L₂ = I₂$\omega$

By conservation of angular momentum,

L₁ = L₂

L₁ = I₂$\omega$

478.125 = (1025.15)$\omega$

$\omega$ = 0.466 rad/s

A) Angular momentum of the child before the collision = 478.125 kg.m²/s

B) Moment of inertia of the merry-go-round = 594.84 kg.m²

C) Final angular speed of the child-merry-go-round system after the collision = 0.466 rad/s