Question

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m2. The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.
(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 18.2 rpm.
(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest.
revolutions
(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in three revolutions.

Answer 1

Given: I = 84.4 kg·m^2, r = 1.50 m, τ = 375 N·m, ω = 18.2 rpm.

First convert ω to rad/s:
ω = 18.2 (rev/min) × (2π rad/rev) / 60 s/min = 1.9048 rad/s.

(a) Rotational kinetic energy:
K = (1/2) I ω^2 = 0.5(84.4)(1.9048^2) ≈ 153 J.

(b) Work by the torque to reach this energy: τθ = K ⇒ θ = K/τ.
θ = 153 J / 375 N·m = 0.4083 rad = 0.4083/(2π) ≈ 0.0650 revolutions.

(c) To stop in 3 revolutions (θ = 3 × 2π = 18.8496 rad), braking work τ_brake θ = K, τ_brake = K/θ. Force F = τ_brake / r = K/(r θ):
F = 153 J / (1.50 m × 18.8496 rad) ≈ 5.41 N.

Answers:
(a) 1.53 × 10^2 J
(b) 6.50 × 10^-2 revolutions
(c) 5.41 N