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...

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.