91Ó°ÊÓ

The mass of the wheel is,$\text{m}=32\text{ k²µ}$.

The radius of wheel is,$\mathrm{R}=1.20\text{″¾}$.

The rotating speed is,$\mathrm{Ó\neg }=280\text{ r±ð±¹}/\text{min}$$\mathrm{Ó\neg }=280\text{ r±ð±¹}/\text{min}$.

The time is,$\mathrm{t}=15.0\text{ s}$.

Find the M.I using the formula for it. Then using the work-energy theorem, find the work done from rotational K.E. Using the relation between power and work done, findtherequired average power to stop the wheel.

Formula:

$\begin{array}{c}\mathbf{I}\mathbf{=}{\mathbf{mR}}^{\mathbf{2}}\\ \mathbf{W}\mathbf{=}\mathbf{Δ°­}\mathbf{.}\mathbf{E}\\ \mathbf{P}\mathbf{=}\left|\frac{W}{t}\right|\end{array}$

M.I of the wheel is

$\mathrm{I}={\mathrm{mR}}^2$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{I}=\left(32\text{ k²µ}\right){\left(1.20\text{″¾}\right)}^2\\ \mathrm{I}=46.1{\text{ k²µ.m}}^{\text{2}}\end{array}$

Angular speed of the wheel is

$\begin{array}{c}\mathrm{Ó\neg }=280\frac{\text{rev}}{\text{min}}×\frac{2\mathrm{Ï€}\text{ r²¹»å}/\text{rev}}{60\text{ s}/\text{min}}\\ =29.3\text{ r²¹»å}/\text{s}\end{array}$

According to the work-energy theorem,

$\mathrm{W}=\mathrm{Δ°­}.\mathrm{E}$

In this case,

$\begin{array}{c}\mathrm{W}=\mathrm{Δ°­}.\mathrm{E}\\ =0−\frac{1}{2}{\mathrm{IÓ\neg }}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{W}=−\frac{1}{2}(46.1{\text{ k²µ.m}}^{\text{2}}){(29.3\text{ r²¹»å}/\text{s})}^2\\ \mathrm{W}=−19788\text{ J}\\ ≃−1.98×{10}^4\text{ J}\end{array}$

Therefore, work done to stop the wheel is $−1.98×{10}^4\text{ J}$.

Average power is

$\mathrm{P}=\left|\frac{\mathrm{W}}{\mathrm{t}}\right|$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{P}=\frac{1.98×{10}^4\text{ J}}{15\text{ s}}\\ \mathrm{P}=1.32×{10}^3\text{ W}\end{array}$

Therefore, an average power to stop the wheel is $1.32×{10}^3\text{ W}$.