91Ó°ÊÓ

1. The top speed of the flywheel is, $\mathrm{Ó\neg }=200\mathrm{Ï€}rad/s$.
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.

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

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

$K=\frac{1}{2}l{\mathrm{Ó\neg }}^2$ …(1)

Now, calculatethemoment of inertia of flywheel which is cylindrical.

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

Using in equation 1, we get

$\begin{array}{rcl}K& =& \frac{1}{2}×250kg.m^2×{\left(200\mathrm{Ï€°ù²¹»å}/\mathrm{s}\right)}^2\\ & =& 49.25×{10}^{6}J\\ & =& 49.25MJ\\ & ≈& 49MJ\end{array}$

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.

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

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

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

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