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?

w=?

(b) How much work did the child do on the merry-go-round?

W=?

(c) What is the average power supplied by the child?

P=?

PHYSICS

### 2 Answers

Torque τ = I α where α is angular acceleration

Torque τ = F*r

F*r = I α = I (ωf – ωi)/t

=============================

A)

Given ωi = 0

F*r = I ωf/t

ωf = F*rt/ I = 23*2.6*21/2300 = 0.546 rad/s

————————————–…

B)

W =0.5*I* ωf² = 1150*0.546 ² = 342.8 J

—————–

C)

Power Total work/time =342.8334/21 =16.3W

=================

torque applied

= 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

answer

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

answer

average power supplied by the child

= work the child did on the merry-go-round / time

= 342.83 / 21

= 16.33 W

answer

Source(s): my brain (Prof TBT)