Question

A $\mathbf{32}\mathbf{.0}\mathbf{\text{\hspace{0.17em}kg}}$wheel, essentially a thin hoop with radius$\mathbf{1.20}\mathbf{\text{\hspace{0.17em}m}}$, is rotating at$\mathbf{280}\mathbf{\text{\hspace{0.17em}rev}}\mathbf{/}\mathbf{\text{min}}$. It must be brought to a stop in$\mathbf{15.0}\mathbf{\text{\hspace{0.17em}s}}$. (a) How much work must be done to stop it? (b) What is the required average power?

Short answer

1. Work done to stop the wheel is$-1.98\times {10}^4\text{\hspace{0.17em}J}$
2. An average power to stop the wheel is $1.32\times {10}^3\text{\hspace{0.17em}W}$.

Step-by-step solution

Given

The mass of the wheel is,$\text{m}=32\text{\hspace{0.17em}kg}$.

The radius of wheel is,$\mathrm{R}=1.20\text{\hspace{0.17em}m}$.

The rotating speed is,$\omega =280\text{\hspace{0.17em}rev}/\text{min}$$\mathrm{\omega }=280\text{\hspace{0.17em}rev}/\text{min}$.

The time is,$\mathrm{t}=15.0\text{\hspace{0.17em}s}$.

Understanding the concept

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:

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

(a) Calculate how much work must be done to stop the hoop

M.I of the wheel is

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

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{I}=\left(32\text{\hspace{0.17em}kg}\right){\left(1.20\text{\hspace{0.17em}m}\right)}^2\\ \mathrm{I}=46.1{\text{\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

Angular speed of the wheel is

$\begin{array}{c}\mathrm{\omega }=280\frac{\text{rev}}{\text{min}}\times \frac{2\mathrm{\pi }\text{\hspace{0.17em}rad}/\text{rev}}{60\text{\hspace{0.17em}s}/\text{min}}\\ =29.3\text{\hspace{0.17em}rad}/\text{s}\end{array}$

According to the work-energy theorem,

$\mathrm{W}=\mathrm{\Delta K}.\mathrm{E}$

In this case,

$\begin{array}{c}\mathrm{W}=\mathrm{\Delta K}.\mathrm{E}\\ =0-\frac{1}{2}{\mathrm{I\omega }}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{W}=-\frac{1}{2}(46.1{\text{\hspace{0.17em}kg.m}}^{\text{2}}){(29.3\text{\hspace{0.17em}rad}/\text{s})}^2\\ \mathrm{W}=-19788\text{\hspace{0.17em}J}\\ \simeq -1.98\times {10}^4\text{\hspace{0.17em}J}\end{array}$

Therefore, work done to stop the wheel is $-1.98\times {10}^4\text{\hspace{0.17em}J}$.

(b) Calculate the required average power

Average power is

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

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{P}=\frac{1.98\times {10}^4\text{\hspace{0.17em}J}}{15\text{\hspace{0.17em}s}}\\ \mathrm{P}=1.32\times {10}^3\text{\hspace{0.17em}W}\end{array}$

Therefore, an average power to stop the wheel is $1.32\times {10}^3\text{\hspace{0.17em}W}$.