Question

A $150-kg$ merry-go-round in the shape of a uniform, solid, horizontal disk of radius $2\cdot 0m$ 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 $0\cdot 500rev/s$ in $2\cdot 00$ s?

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

Answer 1

The correct option is D $314N$

Step 1:Determine the moment of inertia of the merry-go-round.
$l=\frac{1}{2}M{R}^2$
By putting the values, we get,
$I=\frac{1}{2}\left(150kg\right)(2\cdot 0m{)}^2$
$I=300kg{m}^2$
Step 2:Find the angular acceleration of the merry-go-round.
As we know,
$\alpha =\frac{\Delta \omega }{\Delta t}=\frac{{\omega }_f-{\omega }_i}{\Delta t}$
By putting the values, we get,
$\alpha =\left(\frac{.500rev/s-0}{2.00s}\right)\left(\frac{2\pi rad)}{1rev)}\right)$
Therefore,
$\alpha =\frac{\pi }{2}rad/s^2$
Step 3:Find the torque of the merry-go-round.
As torque, $\tau =F.r=I\alpha$ ,
Hence,
$F=I\alpha /r$
By putting the values, we get,
$F=\frac{\left(300kg{m}^2\right)\left(\frac{\frac{\pi }{2}rad)}{s^2}\right)}{(1\cdot 50m)}$
$F=314N$

Final Answer: $c$