Distance from Instantaneous Centre of Rotation in Bicycle Kinematic Model

Question

I'm not sure whether this is the correct community to post this, so pardon me.

I was studying the Bicycle Kinematic Model and came across 3 possible reference points for analysis - the rear tire, front tire and the centre of gravity. For all these reference points, Instantaneous Centre of Rotation was applied and velocities were derived.

I have depicted the velocities of the wheels, the steering angle and the angular velocity of the system.

Under no-slip condition, velocity of the rear wheel can be expressed as

$$v = \omega R.$$

The front wheel velocity can also be expressed similarly,

$$v = \omega R'.$$

For both these equations to hold, $R$ must be equal to $R'$, which is not true. Where am I going wrong? Either both wheels must have different velocities or the no-slip condition must be false, both of which the Bicycle Kinematic Model refutes. I believe I'm missing something right under my nose. Any help will be appreciated. Thanks!

Answer 1

If you were riding the bike on a snow covered surface, you would note that the tracks of the front and rear wheels are two concentric circles, with the front wheel following the larger circular track. When you have completed a full circle, the front wheel will have travelled a further distance than the back wheel in the same time, so the average velocity of the front wheel must be greater than the rear wheel.

Either both wheels must have different velocities or the no-slip condition must be false,

\your first conclusion that they must have different velocities is correct. What is the same at the front and rear is the angular velocity ($\omega$) because the bike is a rigid rotating body. If $v = \omega R$ at the rear wheel and $v' = \omega R'$ at the front wheel, we can equate the two expressions as:

$$ \frac{v}{R} = \omega = \frac{v'}{R'}$$

$$ v = v' \frac{R}{R'}$$

Answer 2

The assumption that the wheels have to have the same velocity is wrong (and in general is not true for two points on a rotating object).

It helps here to think about the extreme case where the front wheel $\delta$ angle is 90 degrees, so that the car pivots about its rear wheel. The front wheel is rolling and the rear wheel is not, so clearly they are rolling at different speeds.

Answer 3

The diagram is incorrect. A bicycle is a rigid body that has a shared rotational velocity of $\omega$.

Indeed, the rear tire axle has translational velocity found with $v = R \omega$ as it prescribes an arc of radius $R = L / \tan \delta$.

But the front tire has a higher velocity of $v' = R' \omega$ since it prescribes the larger arc of $R' = L / \sin \delta $.

Answer 4

If the radius of the wheels are different, the angular velocities are also different. So both can have the same linear velocity and fulfill the condition of no-slip.