Question

A 210-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 0.600 rev/s in 2.00 s

Answer 1

The value of the constant force is
`\bf{296.88~N}`.

Step-by-step explanation:

Given:

Mass of the merry-go-round,
`m = 210~Kg`

Radius of the horizontal disk,
`r = 1.5~m`

Time required,
`t = 2.00~s`

Angular speed,
`\omega = 0.600~rev/s`

Torque on an object is given by

`\tau = F.r = I.\alpha~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)`

where
`I` is the moment of inertia of the object,
`\alpha` is the angular acceleration and
`F` is the force on the disk.

The moment of inertia of the horizontal disk is given by

`I = (1)/(2)mr^(2)`

and the angular acceleration is given by

`\alpha = (2\pi \omega)/(t)`

Substituting all these values in equation (1), we have

`F &=& (I\alpha)/(r)\\&=& (\pi m r \omega)/(t)\\&=& \dfarc{\pi(210~Kg)(1.5~m)(0.600~rev/s)}{2.00~s}\\&=& 296.88~N`