Question

A car of mass 1000 kg has four wheels (disc) of mass 20 kg each. What fraction of kinetic energy is contributed by rotation of wheels?
A. $\dfrac{1}{{25}}$
B. $\dfrac{1}{{50}}$
C. $\dfrac{1}{{10}}$
D. $\dfrac{1}{{100}}$

Answer 1

Calculate the kinetic energy of the car. Then determine the rotational kinetic energy of the wheels and take the ratio of rotational kinetic energy of the wheel to the kinetic energy of the car to determine the fraction of kinetic energy contributed by the rotation of each wheel.

Formula used:
$K.E = \dfrac{1}{2}m{v^2}$
Here, m is the mass and v is the velocity.

$I = \dfrac{1}{2}m{r^2}$
Here, I is the moment of inertia of the disc, m is the mass and r is the radius of the disc.

Complete step by step answer: We have to calculate the kinetic energy of the moving car. Express the kinetic energy of the car as follows,
${\left( {K.E} \right)_{car}} = \dfrac{1}{2}{m_{car}}v_{car}^2$

Here, ${m_{car}}$ is the mass of the car and ${v_{car}}$ is the velocity of the car.

Substitute 1000 kg for ${m_{car}}$ in the above equation.
${\left( {K.E} \right)_{car}} = \dfrac{1}{2}\left( {100} \right)v_{car}^2$
$\Rightarrow {\left( {K.E} \right)_{car}} = 500{v^2}$ …… (1)

Expression for the rotational kinetic energy is,
${\left( {K.E} \right)_{rot}} = \dfrac{1}{2}I{\omega ^2}$

Here, I is the moment of inertia of the disc and $\omega$ is the angular velocity of the disc.

The moment of inertia of the disc of mass m and radius r is,
$I = \dfrac{1}{2}{m_{disc}}{r^2}$

Therefore, the rotational kinetic energy of the disc becomes,
${\left( {K.E} \right)_{rot}} = \dfrac{1}{2}\left( {\dfrac{1}{2}{m_{disc}}{r^2}} \right){\omega ^2}$
$\Rightarrow {\left( {K.E} \right)_{rot}} = \dfrac{1}{4}{m_{disc}}{r^2}{\omega ^2}$

Substitute 20 kg for ${m_{disc}}$ in the above equation.
${\left( {K.E} \right)_{rot}} = \dfrac{1}{4}\left( {20} \right){r^2}{\omega ^2}$
$\Rightarrow {\left( {K.E} \right)_{rot}} = 5{r^2}{\omega ^2}$ …… (2)

The angular velocity is related to the linear velocity of the body as,
$\omega = \dfrac{v}{r}$

Therefore, equation (2) becomes,
${\left( {K.E} \right)_{rot}} = 5{r^2}{\left( {\dfrac{v}{r}} \right)^2}$
$\Rightarrow {\left( {K.E} \right)_{rot}} = 5{v^2}$ …… (3)

This is the rotational kinetic energy of the single wheel. The rotational kinetic energy of the 4 wheels will be,
${\left( {K.E} \right)_{rot}} = 4 \times 5{v^2}$
$\Rightarrow {\left( {K.E} \right)_{rot}} = 20{v^2}$ …… (4)

Divide equation (4) by equation (1).

$\dfrac{{{{\left( {K.E} \right)}_{rot}}}}{{{{\left( {K.E} \right)}_{car}}}} = \dfrac{{20{v^2}}}{{500{v^2}}}$
$\Rightarrow \dfrac{{{{\left( {K.E} \right)}_{rot}}}}{{{{\left( {K.E} \right)}_{car}}}} = \dfrac{1}{{25}}$

This is the fraction of energy contributed by each wheel.

So, the correct answer is option (A).

Note: The moment of inertia of the disc and the moment of inertia of the ring are different and the wheel of the car is sort of disc. After you calculate the rotational kinetic energy of the single wheel, do not forget to multiply it by 4 to get the rotational kinetic energy of 4 wheels of the car.