Question: A car has fourwheels. When the car is moving, what fraction of its total kinetic energy is due to rotation of the wheels about their axles? Assume that the wheels have the same rotational inertia as uniform disks of the same mass and size. Why do you not need to know the radius of the wheels?

$$\begin{gathered} \text{massofcar}=1000\text{kg} \\ \text{massofWheel}=10\text{kg} \end{gathered}$$

Using the formula for Kinetic energy of rotation and total kinetic energy, we can write the equations for kinetic energy of the car and wheels and find the fraction. Simplifying these equations would give us the reason why we do not need the radius for these calculations.

The kinetic energy of rotation is-

$\mathbf{K}\mathbf{.}\mathbf{E}{\mathbf{.}}_{\mathbf{rotation}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}$

The kinetic energy of translation is-

$\mathbf{K}\mathbf{.}\mathbf{E}\mathbf{.}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Here, l is the moment of inertia, w is the angular velocity, v is the linear velocity and m is the mass.

Rotational kinetic energy of the wheel can be written as

$K.E{.}_{rotation}=\frac{1}{2}I{\omega }^2$

We have 4 wheels, so

$K.E{.}_{rotation}=4\times \frac{1}{2}I{\omega }^2$

Wheels can be assumed to be a disc, moment of inertia can be written as

$$\begin{gathered} I=\frac{1}{2}m{R}^2 \\ K.E{.}_{rotation}=2\times \frac{1}{2}m{R}^2{\omega }^2 \end{gathered}$$

We know that$v=R\omega$. Car and wheels would have same linear velocity. So we have

$K.E{.}_{rotation}=m{v}^2$

Total Kinetic energy of the car is given as

$K.E{.}_{tot}=\frac{1}{2}M{v}^2$

The fraction of the total energy is-

$$\begin{gathered} f=\frac{K.E{.}_{rotation}}{K.E{.}_{tot}} \\ =\frac{m{v}^2}{\frac{1}{2}M{v}^2} \\ f=\frac{m}{0.5M} \end{gathered}$$

Substituting the values of masses, we have

$$\begin{gathered} f=\frac{10\text{kg}}{0.5\times 1000\text{kg}} \\ =0.02 \end{gathered}$$

As seen from the above calculations, final equation for rotational kinetic energy would not have the term radius. This is because the motion is assumed to be a perfect rolling motion where there is no skidding or slipping. This would give us the linear velocity of the points on the wheel same as linear velocity of the car.

Hence, we do not need the radius of the wheels to calculate the fraction.