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

Answer:

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

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

$I = \dfrac{1}{2}mr^{2}$

and the angular acceleration is given by

$\alpha = \dfrac{2\pi \omega}{t}$

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

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