Question

A merry-go-round (radius = 4 m) with a perfect frictionless bearing is pushed with a force of 24 N by a young girl. She pushes with a constant force that is oriented tangentially to the edge of the merry-go-round. After she pushes the merry-go-round through 14 full rotations (at which point she lets go) it is spinning with an angular speed of 3 rad/s.

a) What is the moment of inertia of the merry-go-round?

b) After the girl lets go, a 20 kg boy jumps onto the merry-go-round and sticks halfway between the center and the edge. How fast are the merry-go-round and the boy spinning after he lands?

Answer 1

here,

the force applied , Ft = 24 N

radius , r = 4 m

theta = 14 * 2pi = 87.92 rad

final angular speed , w = 3 rad/s

the angular accelration , alpha = w^2 /( 2*theta)

alpha = 3^2 /( 2*87.92) = 5.12 * 10^-2 rad/s^2

a)

let the moment of inertia be I

equating the torque

I * alpha = Ft * r

I * 0.0512 = 24 * 4

solving for I

I = 1875.6 kg.m^2

b)

mass of boy , m2 = 20 kg

let the final angular speed be w'

using conservation of angular momentum

(I * w) = ( I + m2 * (r/2)^2) * w'

1875.6 * 3 = ( 1875.6 + 20 * ( 4/2)^2) * w'

solving for w'

w' = 2.88 rad/s

the final angular speed is 2.88 rad/s