# If a car moves at a certain speed, how fast is its wheel going?

Say in a given instance a car moves with speed $v$ and consider any wheel of the car. How fast is it going? Is it the case that the center of the wheel moves at the same speed as the car i.e. $v$? Why is that?

The wheel is connected to the car through an axle which goes through the center of the circle that is our wheel. The point which is in contact with the ground is the instantaneous centre of rotation and that very point on the wheel has a speed of zero. Clearly the center of the wheel has non-zero velocity.

So in any instant the movement is a rotation about a fixed axis, where the fixed axis in the considered instant lies in the contact point with the ground and the speed is zero on the fixed axis. If the angular velocity of the wheel is $\omega$ and the wheel has radius $R$, then the speed at the center of the wheel is $v_C = \omega R$.

Why is this speed $v_C = \omega R$ the same as the speed $v$ of the car? Maybe it is self-evident but I really do not see why.

• You want to know why the point that is rigidly fixed to the entire body of the car is moving at the same velocity as what it is rigidly fixed to? – Jim Feb 3 '15 at 19:24
• Glad to be of help – Jim Feb 3 '15 at 19:26

In the reference frame of the car, the axle is stationary, but the ground is moving below at speed $v_C$. If the car doesn't skid, then the surface of the tyre must move at the same speed, but with a velocity that is directed backwards in the bottom and forwards in the top.
At half the distance between the tyre surface and the hub, the speed is $v_C/2$.
Thus, in the reference frame of the ground, the point in the bottom has a speed zero, while in the top it has a speed $2v_C$.
• Yes. Interesting side note - because the wheel rotates, it adds to the inertia of the car: both for accelerating and braking, it is "a little bit heavier". If the wheel is a uniform disk with mass $m$, its moment of inertia is $\frac12 m r^2$ and the rotational kinetic energy $\frac12 I \omega^2 = \frac14 m v^2$. In other words, it increases the kinetic energy by 50%. – Floris Feb 3 '15 at 20:28