A child of mass 50 kg is standing on the edge of a merry go round of mass 250 kg and radius 3.0 m which is rotating with an angular velocity of 3.0 rad s⁻¹. The child then starts walking towards the centre of the merry go round. What will be the final angular velocity of the merry go round when the child reaches the centre?

Introduction

This problem is based on the principle of conservation of angular momentum. When no external torque acts on a system, the total angular momentum remains constant. As the child moves toward the center, the system’s moment of inertia changes, which affects the angular velocity.

Given Data

• Mass of child (m) = 50 kg
• Mass of merry-go-round (M) = 250 kg
• Radius of merry-go-round (R) = 3.0 m
• Initial angular velocity (ω₁) = 3.0 rad/s
• Final position of child = center (r = 0)

Step 1: Moment of Inertia

We assume the merry-go-round is a uniform disc, so its moment of inertia is:

I_disc = (1/2)MR² = 0.5 × 250 × (3)² = 0.5 × 250 × 9 = 1125 kg·m²

Initially, the child is at the edge (r = 3.0 m), treated as a point mass:

I_{child_initial} = m × r² = 50 × 3² = 450 kg·m²

Total I_initial = 1125 + 450 = 1575 kg·m²

When the child reaches the center (r = 0):

I_{child_final} = 0

Total I_final = 1125 kg·m²

Step 2: Conservation of Angular Momentum

L_initial = I_initial × ω₁ = 1575 × 3 = 4725 kg·m²/s

L_final = I_final × ω₂

Set L_initial = L_final:

4725 = 1125 × ω₂ → ω₂ = 4725 / 1125 ≈ 4.2 rad/s

Final Answer

• Final angular velocity of the merry-go-round: 4.2 rad/s

Conclusion

As the child moves toward the center of the merry-go-round, the moment of inertia of the system decreases. Due to the conservation of angular momentum, this results in an increase in the angular velocity. Thus, the system spins faster when the mass distribution is closer to the center.