# A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of...

## Question:

A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg.

a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy.

b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m/s at the top of the hill?

## Principal of kinetic energy:

The principal of kinetic energy stated that the kinetic energy of the system remains constant. It means that the initial kinetic energy of the system is equivalent to the final kinetic energy of the system.

## Answer and Explanation:

**Given Data**

- The mass of the flywheel is: {eq}m = 1500\;{\rm{kg}}. {/eq}

- The radius of the disk is: {eq}r = 0.6\;{\rm{m}} {/eq}.

- The total mass of the bus is: {eq}M = 10000\;{\rm{kg}} {/eq}.

**(A)**

The expression for the moment of inertia of the flywheel is,

{eq}I = \frac{{m{r^2}}}{2} {/eq}

Substitute the given values.

{eq}\begin{align*} I &= \frac{{\left( {1500\;{\rm{kg}}} \right){{\left( {0.6\;{\rm{m}}} \right)}^2}}}{2}\\ &= \frac{{540}}{2}\\ &= 270\;{\rm{kg}}{\rm{.}}{{\rm{m}}^{\rm{2}}} \end{align*} {/eq}

The relation between the rotational kinetic energy of the bus is,

{eq}\begin{align*} \frac{{90}}{{100}}{K_{{\rm{flywheel}}}} &= {K_{{\rm{bus}}}}\\ 0.9{K_{{\rm{flywheel}}}} &= {K_{{\rm{bus}}}}\\ 0.9 \times \frac{1}{2}I{\omega ^2} &= \frac{1}{2}M{v^2} \end{align*} {/eq}

Substitute the given values.

{eq}\begin{align*} 0.9 \times \frac{1}{2}\left( {270} \right){\omega ^2} &= \frac{1}{2}\left( {10000} \right){\left( {20} \right)^2}\\ 121.5{\omega ^2} &= 2000000\\ {\omega ^2} &= \frac{{2000000}}{{121.5}}\\ \omega &= \sqrt {\frac{{2000000}}{{121.5}}} \\ &= 128.30\;{\rm{rad/s}} \end{align*} {/eq}

Thus, the angular velocity of the flywheel is {eq}128.30\;{\rm{rad/s}} {/eq}.

**(b)**

The expression for the energy conservation is,

{eq}\frac{1}{2}Mv_1^2 = Mgh + \frac{1}{2}Mv_1^2 {/eq}

Substitute the given values.

{eq}\begin{align*} \frac{1}{2}\left( {10000} \right){\left( {20} \right)^2} &= \left( {10000} \right)\left( {9.8} \right)h + \frac{1}{2}\left( {10000} \right){\left( 3 \right)^2}\\ 98000h &= 2000000 - 45000\\ h &= \frac{{1955000}}{{98000}}\\ &= 19.948\;{\rm{m}}\\ &\approx {\rm{19}}{\rm{.95}}\;{\rm{m}} \end{align*} {/eq}

Thus, the height achieved by the bus is {eq}19.95\;{\rm{m}} {/eq}.

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Chapter 4 / Lesson 14