Question

A 150 kg merry go round in the shape of a uniform solid horizontal disc of radius 2 m is set in motion by wrapping a rope around the rim of disc 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 $0.5$ rev/s in 2 s ?

• A. 574 N
• B. 157 N
• C. 314 N
• D. 81 N

Answer 1

The correct option is C 314 N

Moment of inertia of merry go round will be
$I=\frac{M{R}^2}{2}=\frac{\left(150\right)\times {\left(2\right)}^2}{2}=300kg.m^2$
To find the angular acceleration, we use
$\alpha =\frac{\Delta \omega }{\Delta t}=\frac{\omega f-\omega i}{\Delta t}=\left(\frac{0.5-0}{2}\right)\left(\frac{2\pi }{1}\right)=\frac{\pi }{2}\left(rad\right)s^{-2}$
From the definition of torque, $\tau =r\times F=I\alpha$ we get,
$F=\frac{I}{r}\alpha =\left(\frac{300}{1.5}\right)\times \left(\frac{\pi }{2}\right)=314N.$