A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in
motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force
must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s
in 2.00 s?
mass, m = 150 kg
radius, r = 1.5 m
initial angular velocity, wo = 0
final angular velocity, w = 0.5 rps
time, t = 2 s
mass, m = 150 kg
radius, r = 1.5 m
initial angular velocity, wo = 0
final angular velocity, w = 0.5 rps
time, t = 2 s
Moment of inertia of the disk, I = 0.5 mr^{2} = 0.5 x 150 x 1.5 x 1.5 = 168.75 kgm2
The angular acceleration is given by
$w=wo+\alpha t\phantom{\rule{0ex}{0ex}}2x3.14x0.5=0+\alpha x2\phantom{\rule{0ex}{0ex}}\alpha =1.57rad/{s}^{2}$
According to the equation of torque, let the force is F.
$\tau =I\alpha =rF$
168.75 x 1.57 = 1.5 x F
F = 176.625 N