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 radius to achieve an angular acceleration of 4.44 rad/s2.
(a)
How long (in s) does it take the father to give the merry-go-round an angular velocity of 2.31 rad/s? (Assume the merry-go-round is initially at rest.)
(b)How many revolutions must he go through to generate this velocity?
revolutions
(c)If he exerts a slowing force of 350 N at a radius of 1.30 m, how long (in s) would it take him to stop them?

Answer 1

Given I = 84.4 kg·m^2, α = 4.44 rad/s^2, ωf = 2.31 rad/s.

(a) t = ωf/α = 2.31 / 4.44 = 0.520 s

(b) Angular displacement θ = 1/2 α t^2 = ωf^2/(2α) = (2.31^2)/(2·4.44) = 0.6007 rad.
Revolutions = θ/(2π) = 0.6007 / (2π) = 0.0957 rev ≈ 0.096 rev

(c) Torque from the slowing force: τ = F r = 350 N · 1.30 m = 455 N·m.
Angular deceleration magnitude = τ / I = 455 / 84.4 = 5.39 rad/s^2.
Stopping time t_stop = ωf / (τ/I) = 2.31 / 5.39 = 0.429 s

Results (3 s.f.):
(a) 0.520 s
(b) 0.096 revolutions
(c) 0.429 s