Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of$\mathbf{200}\mathbf{\pi rad}\mathbf{/}\mathbf{s}$. Suppose that one such flywheel is a solid, uniform cylinder with a mass of 500kg and a radius of 1.0m. (a) What is the kinetic energy of the flywheel after charging? (b) If the truck uses an average power of 8.0kW, for how many minutes can it operate between chargings?

1. The top speed of the flywheel is, $\omega =200\pi rad/s$.
2. The mass of the flywheel is, m = 500kg .
3. The radius of the flywheel is, r = 1m .
4. The average power of the truck is, P = 8.0kW .

By using the concept of rotational kinetic energy and power we can find the required values and the formula for rotational kinetic energy is given as.

$$\begin{gathered} \mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{l}{\mathit{\omega }}^{\mathbf{2}} \\ \mathit{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{r}}^{\mathbf{2}} \\ \mathit{P}\mathbf{=}\frac{\mathbf{d}\mathbf{E}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{f}}\mathbf{-}{\mathbf{K}}_{\mathbf{i}}}{\mathbf{d}\mathbf{t}} \end{gathered}$$

As you know the angular velocity, the radius, and mass of the cylindrical flywheel,you can use the formula for rotational kinetic energy.

$K=\frac{1}{2}l{\omega }^2$ …(1)

Now, calculatethemoment of inertia of flywheel which is cylindrical.

$\begin{array}{rcl}l& =& \frac{1}{2}m{r}^2\\ & =& \frac{1}{2}\times 500kg\times {\left(1m\right)}^2\\ & =& 250kg.m^2\end{array}$

Using in equation 1, we get

$\begin{array}{rcl}K& =& \frac{1}{2}\times 250kg.m^2\times {\left(200\mathrm{\pi rad}/\mathrm{s}\right)}^2\\ & =& 49.25\times {10}^{6}J\\ & =& 49.25MJ\\ & \approx & 49MJ\end{array}$

Hence the kinetic energy of flywheel is 49MJ .

You know, Power is the rate of transfer of energy.

$\begin{array}{rcl}P& =& \frac{dE}{dt}\\ & =& \frac{K_f-K_i}{dt}\end{array}$

At t = 0 , Initial kinetic energy K_i = 0

$P=\frac{K_f}{dt}$

Substitute all the value in the above equation.

$\begin{array}{rcl}8\times {10}^{3}W& =& \frac{49.25\times {10}^{6}J}{dt}\\ dt& =& \frac{49.25\times {10}^{6}W}{8\times {10}^{3}J}\\ & =& 6125s\end{array}$

But you are asked to find the time in minutes. Therefore,

$\begin{array}{rcl}dt& =& 6125s\times \frac{1min}{60s}\\ & =& 102.08min\\ & =& 1\times {10}^{2}min\end{array}$

For about $1\times {10}^{2}min$, the flywheel will operate between the charging.