Question

. A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from

rest to 15 rpm in 10.0 seconds. The merry-go-round has a radius of 2.5m and mass of 560 kg and his

two kids (each 25 kg) sit on opposite sides of each other on the edge.

(a) Calculate the magnitude of the dad’s pushing force on the edge of the merry-go-round.

(b) After these 10.0 seconds, the dad stops pushing. One kid quickly steps off the merry-go-round.

Assuming that the pivot of the merry-go-round is fairly friction-free, what is the angular velocity

in rpm of the merry-go-round now?

Answer 1

here,

the final angular speed , w = 15 rpm = 1.57 rad/s

time taken , t = 10 s

let the angular acceleration be alpha

equating the torques

w = w0 + alpha * t

1.57 = 0 + alpha * 10

alpha = 0.157 rad/s^2

mass of merry-go-round , m1 = 560 kg

radius , r = 2.5 m

mass of each kid , m2 = 25 kg

A)

let the force applied bv man be F

equating the torques

I * alpha * F * r

(0.5 * m1 * r^2 + 2 *m2 * r^2) * alpha = F * r

(0.5 * 560 + 2 * 25) * 2.5 * 0.157 = F

F= 129.5 N

the force applied by the man is 129.5 N

b)

when on child steps out

let the new angular speed be w'

using conservation of angular momentum

Li = Lf

(0.5 * m1 * r^2 + 2 *m2 * r^2) * w = (0.5 * m1 * r^2 + m2 * r^2) * w'

(0.5 * 560 + 2 * 25) * 1.57 = ( 0.5 * 560 + 25) * w'

w' = 1.7 rad/s

the new angular velocity is 1.7 rad/s