Q60P - University Physics with Modern Physics

Question

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{\text{ cm}}$. Starting from rest at $\mathit{t}\mathbf{=}\mathbf{0}$, the flywheel rotates with constant angular acceleration  $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{ }\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$ about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of , what thickness must it have to store $\mathbf{\text{800 J}}$ of kinetic energy at $\mathit{t}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\text{ s}}$?

Step-by-Step Solution

Answer

The thickness of flying wheel is $0.05267\text{ m}$.

Step 1: Angular velocity and acceleration:

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

Step 2: Given data:

Consider the given data as below.

The angular acceleration,  $\alpha =3 \raisebox{1ex}{$\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, $\rho =8600 \raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$  

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

Step 3: Determine angular speed:

The expression which relates angular speed and angular acceleration is,

$\omega =\alpha t$

Here, $\alpha$ is the angular acceleration, $\omega$ is the angular speed, and $t$ is time.

 

Substitute known numerical values in the above equation.

$\begin{array}{rcl}\omega & =& \left(3 \raisebox{1ex}{$\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,

$K=\frac{1}{2}I{\omega }^2$

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

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

Step 4: Solve further:

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

 $m=\rho V$

Here, $m$ is the mass, $\rho$ is the density, and $V$ is the volume.

 

So, the rotational kinetic energy will be,

$K=\frac{1}{4}\left(\rho V\right)R^2{\omega }^2$

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

$V=\pi R^{2}t$

Therefore, the kinetic energy will become,

$\begin{array}{rcl}K& =& \frac{1}{4}\left(\rho \pi R^{2}t\right)R^2{\omega }^2\\ & =& \frac{1}{4}\rho \pi t{R}^4{\omega }^2\end{array}$

Rearrange the above equation for time.

$t=\frac{4K}{\rho \pi R^4{\omega }^2}$ 

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

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

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