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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.

hft
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Mustaghees
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4 Answers4

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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.

Ross Millikan
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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.

John Alexiou
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  • Thanks for the explanation ( I understood a little of it as I have not studied that high physics ). I just needed the short answer, which you gave at beginning. Thanks! – Mustaghees May 19 '15 at 06:35
  • I disagree with this analysis. The linear velocity of a point on the wheel varies from zero at the ground to twice the bike's velocity at the top. The mud is accelerated during the bottom-to-top travel, and is ejected when the adhesive force is less than the local linear acceleration. Take a look at any bicylist's jersey on a rainy day and you'll see a line of mud thrown off roughly 3*pi/4 on the way "up." Any mud which remains stuck at the top of the wheel will stay there, as the linear acceleration decreases from there on. – Carl Witthoft May 19 '15 at 08:45
  • Please if you can give a better explanation ( which is easier to understand for high school students, or a little above ) then give it in answer. Thank you. – Mustaghees May 19 '15 at 09:17
  • You are confusing the fact that the mud that is most likely to come off (loose mud) will come off first, and the mud that is embedded in the treads will stay stuck for far longer. The riders jersey is there catch only the mud from that location. Mud flung from the top will not land on the rider. – John Alexiou May 19 '15 at 12:18
  • @CarlWitthoft the acceleration vector is $$\begin{bmatrix} \dot{v}_x \ \dot{v}_y \end{bmatrix} = \begin{bmatrix} -\frac{v^2}{r}\sin\theta+\dot{v} (1-\cos\theta) \ \frac{v^2}{r}\cos\theta-\dot{v} \sin\theta \end{bmatrix}$$ Take the magnitude and see that is is constant when the bike is not accelerating $\dot{v}=0$, with $| a | = \frac{v^2}{r}$. – John Alexiou May 19 '15 at 12:22
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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.

Elaaf fatima
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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.

George Pearson
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