A bicycle on a rotating disk

Relevant wiki: uniform circular motion

Consider the motion of the cyclist during one revolution of the disk.

If the cyclist didn't pedal at all he would be carried right around the perimeter of the big disk.

So in order to stay stationary, he must actually cycle exactly once (in the opposite direction to the disk's rotation) around the disk.

Since the circumference of his wheels is less than the circumference of the disk, the wheels must rotate more than once on this journey, while the disk has rotated exactly once.

And so the wheels have a greater angular velocity than the disk.

Relevant wiki: uniform circular motion

Since the tires don't slip, the velocity of the wheel where it makes contact with the disk and the velocity of the disk at the same point will be the same.

Let's call this velocity $v$ .

$\implies v = \omega_{tire} r = \omega_{disk} R$

$\implies \omega_{tire} = \dfrac{R}{r}\omega_{disk}$

So, since $R > r$ the $\boxed{\text{tire of the bicycle}}$ will have the greater rotating (angular) speed.

First of all velocity(v) = (radius (r) )(angular velocity ( $\omega$ ))

Bicycle tire speed = $\textit{r}\omega _{2}$

Disk speed at distance $\textit{R} = \textit{R}\omega _{1}$

Both the tire and the disk need to be moving at the same speed so,

$\textit{r}\omega _{2}$ = $\textit{R}\omega _{1}$

solving for r

$\frac{\textit{r}\omega _{2}}{\omega _{2}} = \frac{\textit{R}\omega _{1}}{\omega _{2}}$

$\textit{r} = \frac{\textit{R}\omega _{1}}{\omega _{2}}$

Using substitution and $\textit{R} > \textit{r}$

$\textit{R} > \frac{\textit{R}\omega _{1}}{\omega _{2}}$

Dividing by $\textit{R}$

$\frac{\textit{R}}{\textit{R}} > \frac{\textit{R}\omega _{1}}{\omega _{2}\textit{R}}$

1 > $\frac{\omega _{1}}{\omega _{2}}$

Multiplying by $\omega _{2}$

1 $\omega_{2} > \frac{\omega _{1}}{\omega _{2}} \omega _{2}$

Yields

$\omega _{2} > \omega _{1}$

So the tire rotation speed is greater than the disc.

Small gears spin faster than big gears.

Tangential speed is the same at disc and bicycle's tire $\Rightarrow \omega_{1}R=\omega_{2}r$ and we can see that $R>r \Rightarrow \omega_{2}>\omega_{1}$

The cyclist travels tangentially along the perimeter of the disc. The radius of the disc is larger than the bike's wheel radius. Therefore, the perimeter of the disc is larger than the perimeter of either of the bike's wheels. So, proportionally, it would take more "wheel perimeter" distance to catch up to the perimeter of the disc.

In a way, consider if you took off the tire of one of the bike's wheels, chopped it in one place, and unrolled it along the perimeter of the disc -- you'd need more than 2 tires to cover the entirety of the disc's perimeter. Well, speaking proportionally, if Distance = Rate * Time, and the bike's wheel's distance [unrolled tire length] is decreased compared to shorter than] the disc's, then the wheel's rate would need to increase to match the disc's rate.