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
Disc radius, r=1.5 m
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,
$\Rightarrow \mathrm{\omega}=0.6\times 2\mathrm{\pi}$
On solving,
$\Rightarrow \mathrm{\omega}=3.768\mathrm{rad}/\mathrm{s}$
Using the expression,
$\Rightarrow \mathrm{\omega}={\mathrm{\omega}}_{0}+\mathrm{\alpha t}$
On substituting the values,
$\Rightarrow 3.768\mathrm{rad}/\mathrm{s}=0+\mathrm{\alpha}(2\mathrm{s})$ $\left({\mathrm{\omega}}_{0}=0\right)$
On solving,
$\Rightarrow \mathrm{\alpha}=1.884\mathrm{rad}/{\mathrm{s}}^{2}$
The torque is given by,
$\Rightarrow \mathrm{\tau}=\mathrm{I\alpha}$
$\Rightarrow \mathrm{F}\times \mathrm{r}=\frac{1}{2}{\mathrm{mr}}^{2}\mathrm{\alpha}$
On simplifying,
$\Rightarrow \mathrm{F}=\frac{1}{2}\mathrm{mr\alpha}$
On substituting the values,
$\Rightarrow \mathrm{F}=\frac{1}{2}\times \left(145\mathrm{kg}\right)\times \left(1.5\mathrm{m}\right)\times \left(1.884\mathrm{rad}/{\mathrm{s}}^{2}\right)$
On solving,
$\Rightarrow \mathrm{F}=204.885\mathrm{N}$
Hence, the required force is 204.885 N.