If a bike moves through a muddy area, mud gets on its tires. Then the mud flies off from the tires.
Which forces are acting on it? In which direction does it fly off?
On my physics test, I wrote that it flies off along tangential velocity at that point, but it was marked incorrect.
You were probably expected to note that the path of any point of the tire is a cycloid. At the point of contact with the ground, the tire is not moving. As it rises from the ground it is moving faster, with a speed that is $v(1+\sin \theta)$ where $v$ is the bike speed and $\theta$ is measured counterclockwise from horizontal. The centrifugal force is rising as $\theta$ decreases from $270^\circ$ and at some point it exceeds the strength of the attachment to the wheel. When it lets go it should get the tangential velocity of that point of the wheel, which is not the same as the speed of the bike.
It is because there aren't any forces acting on the mud keeping turning with the tire that it flies off. At whatever point the mud comes off, it will travel tangent to the tire at first and the follow a parabola due to earths gravity.
It is most likely the more loose mud will come off first and at that point the tangential direction of the tire points straight at your eyes! If there was an equal probability the mud would come loose around the tire then it would be flung forwards, and either upwards or downwads with equal probability.
The tangential velocity of the tire as a function of the azimouth angle (position angle) $\theta$ is $$[v_x,v_y] = [ v (1-\cos \theta), -v \sin\theta]$$ where $v$ is the bike speed and $\theta=0$ is at the contact point moving CCW for positive angle. Interestingly the acceleration is always pointing towards the center of the tire with magnitude $v^2/r$. So the force of adhesion needed to keep the mud on the tire is constant all along the tire unless the bike is accelerating also.
If the bike is accelerating with $\dot{v}$, then define the dimensionless acceleration as $\alpha = \frac{\dot{v}}{ v^2/r}$ and you can find the peak acceleration of the tire surface occurs at $$\theta = \frac{3\pi}{2}-\arctan(\alpha)$$ with peak acceleration magnitude $$\dot{v} = \frac{v^2}{r} \sqrt{\left(1+2 \alpha^2 + 2 \alpha \sqrt{1+\alpha^2}\right)}$$ This corresponds to an area near where the tire goes downwards before it contacts. But if the bike is deccelerating the peak acceleration is at $\theta = \frac{\pi}{2}-\arctan(\alpha)$ which is when the tire just leaves the ground. So you are most likely to get sprayed with mud when on the brakes hard.
When we increase the speed of bicycle then adhesive force between the mud and the tyre is not sufficient to provide the necessary centripetal force so the mud leaves the tyre and moves along tangent to the tyre.
centrifugal force, is the force from the tire that causes the mud to fly off, and it will travel in the opposite direction that the wheel is moving. So if you wheel is propelling you North the centrifugal force will be sending the mud South.
