Question

# A 145-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 145-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? (State the magnitude of the force.)
N

Given:

Mass, m=145 kg

Angular speed=0.6 rev/s

t=2 s

Calculating the angular speed in rad/s:

The angular speed in rad/s is given by,

$⇒\mathrm{\omega }=0.6×2\mathrm{\pi }$

On solving,

Using the expression,

$⇒\mathrm{\omega }={\mathrm{\omega }}_{0}+\mathrm{\alpha t}$

On substituting the values,

$\left({\mathrm{\omega }}_{0}=0\right)$

On solving,

The torque is given by,

$⇒\mathrm{\tau }=\mathrm{I\alpha }$

$⇒\mathrm{F}×\mathrm{r}=\frac{1}{2}{\mathrm{mr}}^{2}\mathrm{\alpha }$

On simplifying,

$⇒\mathrm{F}=\frac{1}{2}\mathrm{mr\alpha }$

On substituting the values,

On solving,

Hence, the required force is 204.885 N.

Similar Questions

Recent Questions