Question

A 180-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 would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s

Answer 1

The force acting on the rope is 211.95 N.

Step-by-step explanation:

Given that,

Mass of merry-go-round, m = 180 kg

Radius of merry-go-round, r = 1.5 m

Initial angular speed is 0 it was at rest

Final angular speed,
`\omega_f=0.4\ rev/s=3.14\ rad/s`

Time, t = 2 s

When it moves in circular path, the torque acting on it is given by :

`\tau=F* r=I\alpha \\\\F=(I\alpha)/(r)\\\\F=(mr^2\omega_f)/(2rt)\\\\F=(180* (1.5)^2* 3.14)/(2* 1.5* 2)\\\\F=211.95\ N`

So, the force acting on the rope is 211.95 N.