91Ó°ÊÓ

Angular velocity is measured in angles per unit time or in radians per second. The rate of change of angular velocity is angular acceleration.

Consider the given data as below.

The angular acceleration, $\mathrm{Î\pm }=3\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$

The time, $t=8.00\text{s}$

Kinetic energy, $K=800\text{J}$

The density of the flywheel,$\mathrm{ÒÏ}=8600\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$

The radius of the flywheel disk, $R=25\text{cm}$

The expression which relates angular speed and angular acceleration is,

$\mathrm{Ó\neg }=\mathrm{Î\pm }t$

Here,$\mathrm{Î\pm }$is the angular acceleration,$\mathrm{Ó\neg }$is the angular speed, and$t$is time.

Substitute known numerical values in the above equation.

$\begin{array}{rcl}\mathrm{Ó\neg }& =& \left(3\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.\right)\left(8.00\text{s}\right)\\ & =& 24\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

The equation for the rotational kinetic energy of the flywheel is,

$K=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

Here, $I$is the moment of inertia.

As momentum of inertia of the disk is,

$I=\frac{1}{2}m{R}^2$

Here,$m$ is the mass and$R$ is the radius of the disc.

By putting the momentum of inertia of the disc in the rotational kinetic energy of the fly wheel, you have

$\begin{array}{rcl}K& =& \frac{1}{2}\left(\frac{1}{2}m{R}^2\right){\mathrm{Ó\neg }}^2\\ & =& \frac{1}{4}m{R}^2{\mathrm{Ó\neg }}^2\end{array}$

As mass is the product of its density and volume that is,

$m=\mathrm{ÒÏ}V$

Here, $m$is the mass, $\mathrm{ÒÏ}$is the density, and $V$is the volume.

So, the rotational kinetic energy will be,

$K=\frac{1}{4}\left(\mathrm{ÒÏ}V\right)R^2{\mathrm{Ó\neg }}^2$

As the volume of wheel with radius$R$and thickness$t$is,

$V=\mathrm{Ï€}{R}^{2}t$

Therefore, the kinetic energy will become,

$\begin{array}{rcl}K& =& \frac{1}{4}\left(\mathrm{ÒÏ}\mathrm{Ï€}{R}^{2}t\right)R^2{\mathrm{Ó\neg }}^2\\ & =& \frac{1}{4}\mathrm{ÒÏ}\mathrm{Ï€}t{R}^4{\mathrm{Ó\neg }}^2\end{array}$

Rearrange the above equation for time.

$t=\frac{4K}{\mathrm{ÒÏ}\mathrm{Ï€}{R}^4{\mathrm{Ó\neg }}^2}$

Substitute known numerical values in the above equation, and you obtain

$\begin{array}{rcl}t& =& \frac{4×800\text{J}}{8600{\displaystyle \raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.}×3.14×{\left(25\text{cm}×{\displaystyle \frac{1\text{m}}{100\text{cm}}}\right)}^4{\left(24{\displaystyle \raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\right)}^2}\\ & =& 0.05267\text{m}\end{array}$

Hence, the thickness of flying wheel is$0.05267\text{m}$.