Question

Two children (each having a mass of 60 kg) are standing on the edge a merry-go-round (mass of 140 kg) as it spins with an angular velocity of 0.75 rad/s. The two children jump off the merry-go-round. What is the angular velocity of the merry-go-round after the children have jumped off

Answer 1

Answer:

The angular velocity after the children jump off is approximately 1.4 rad/s

Explanation:

The given parameters are;

The masses of each child, m₁, and m₂ = 60 kg

The mass of the merry-go-round, m₃ = 140 kg

The initial angular velocity, $\omega_i$ = 0.75 rad/s

The angular velocity after the children jump off = $\omega_f$  

According to the principle of conservation of angular momentum

The angular momentum = I × ω

The moment of inertia, I = m × R²

The total initial angular momentum = $I_i \times \omega_i = m_i \times R^2 \times \omega_i$

The total angular momentum after the children jump off = $I_f \times \omega_f = m_f \times R^2 \times \omega_f$

The initial mass, $m_i$ = m₁ + m₂ + m₃ = 60 kg + 60 kg + 140 kg = 260 kg

The final mass, $m_f$ = m₃ = 140 kg

According to the principle of conservation of linear momentum, we have;

$I_i \times \omega_i$ = $I_f \times \omega_f$

Therefore;

260 kg × R² × 0.75 rad/s = 140 kg × R² × $\omega_f$

∴ $\omega _f$ = (260 kg × R² × 0.75 rad/s)/(140 kg × R²) = 1.39285714 rad/s. ≈ 1.4 rad/s

The angular velocity after the children jump off, $\omega _f$ ≈ 1.4 rad/s.