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

given
wo = 0
w = 15 rpm = 15*2*pi/60 = 1.57 rad/s

t = 10.0 s
M = 560 kg
r = 2.5 m
m = 25 kg

a) angular acceleration of the merry-go-round,

alfa = (w - wo)/t

= (1.57 - 0)/10

= 0.157 rad/s^2

moment of inertia of the system,

I = 0.5*M*R^2 + 2*m*R^2

= 0.5*560*2.5^2 + 2*25*2.5^2

= 2062.5 kg.m^2

let F is the applied force

use, Net torque = I*alfa

F*R = I*alfa

F = I*alfa/R

= 2062.5*0.157/2.5

= 129.5 N <<<<<<<<<-------------Answer

b)
when one child gets down the moment of inertia of the system,

I' = 0.5*M*R^2 + m*R^2

= 0.5*560*2.5^2 + 25*2.5^2

= 1906.25 kg.m^2

let w' is the new angular speed.

now apply conservation of angular momentum

I'*w' = I*w

w' = I*w/I'

= 2062.5*15/1906.25

= 16.2 rpm <<<<<<<<<<------------------Answer