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, [tex]\omega_i[/tex] = 0.75 rad/s
The angular velocity after the children jump off = [tex]\omega_f[/tex]
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 = [tex]I_i \times \omega_i = m_i \times R^2 \times \omega_i[/tex]
The total angular momentum after the children jump off = [tex]I_f \times \omega_f = m_f \times R^2 \times \omega_f[/tex]
The initial mass, [tex]m_i[/tex] = m₁ + m₂ + m₃ = 60 kg + 60 kg + 140 kg = 260 kg
The final mass, [tex]m_f[/tex] = m₃ = 140 kg
According to the principle of conservation of linear momentum, we have;
[tex]I_i \times \omega_i[/tex] = [tex]I_f \times \omega_f[/tex]
Therefore;
260 kg × R² × 0.75 rad/s = 140 kg × R² × [tex]\omega_f[/tex]
∴ [tex]\omega _f[/tex] = (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, [tex]\omega _f[/tex] ≈ 1.4 rad/s.
