A playground merry-go-round has radius 2.60 m and moment of inertia 2300 kg*m^2 about a vertical axle through its center, and it turns with negligible friction.
(a) A child applies an 23.0 N force tangentially to the edge of the merry-go-round for 21.0 . If the merry-go-round is initially at rest, what is its angular speed after this 21.0 s interval?
(b) How much work did the child do on the merry-go-round?
(c) What is the average power supplied by the child?
Torque τ = I α where α is angular acceleration
Torque τ = F*r
F*r = I α = I (ωf – ωi)/t
Given ωi = 0
F*r = I ωf/t
ωf = F*rt/ I = 23*2.6*21/2300 = 0.546 rad/s
W =0.5*I* ωf² = 1150*0.546 ² = 342.8 J
Power Total work/time =342.8334/21 =16.3W
= 23 * 2.6
= 59.8 Nm
= moment of inertia * angular acc
= 2300 *a
a = 0.026 rad/s^2
t =21; u=0
v = u +at
v = 0 + 0.026 * 21
= 0.546 rad /s
angular speed after this 21.0 s interval
work the child did on the merry-go-round
= rotational KE gained
= 0.5 * MoI * w^2
= 0.5 * 2300 * 0.546^2
= 342.83 J
average power supplied by the child
= work the child did on the merry-go-round / time
= 342.83 / 21
= 16.33 W
answerSource(s): my brain (Prof TBT)