Which way does a stone move when hit by a wheel/tire?

Question

I don't think this exact question has been asked before in the physics section. I am mostly interested in understanding it conceptually rather than calculation. I have asked the same question in the mathematics section before, but I've already voted to close it since I believe it's more suitable for physics: Link. In this post I received a wonderful mathematics answer, but I am still quite confused about the situation conceptually.

My question is: Which way does a stone move when hit by a wheel/tire? Does the general direction (forward, backward, sideways) depend furthermore on the shape of the stone? Whether it is for example perfectly spherical or asymmetric? I am kind of struggling how I would answer such a basic question, since there seems to be many factors at play here. Can you even answer this question without fully knowing all the nuances and details?

The mathematical answer that was given (by TurlocTheRed) states the following: Suppose a rock is lodged in the tread of the tire.

$x=v_xt + R\cos(\omega t)$

$dx/dt =v_x-\omega R\cos(\omega t)$

Typically, $v_x=\omega R$

So it generally wouldn't go backward, but can be overtaken by a car behind it effectively having the same result.

So my conceptual interpretation of this, is that generally, the stone won't move backward since the translational velocity dominates over the rotational velocity in $dx/dt$. Is that the correct interpretation? I am also wondering what you can say about the general direction in which the stone will then move. What can you say about the angles? Are there any additions to this answer, in particular form a physics and conceptual perspective? I am mostly interested in the conceptual way on how you would answer such a question, from a physics perspective.

Answer 1

As the link in the question says, it typically goes straight up, and any car following too closely drives into it. This is because on asphalt the wheel does not slip. The tread or dual wheels grab a stone and move it in the direction the tread is moving. At the bottom of the wheel, the tread is motionless. It is accelerating straight up.

Acceleration can be huge. For a $16$" = $0.4 \space m$ wheel at $60 \space MPH = 27$ m/s, $$a = v^2/r > 1800 \space m/s^2 \approx 200 \space g.$$ A stone can acquire a surprising upward velocity in a short distance. This means it can have a surprising hang time.

On a dirt road, the wheel can slip if you step on the gas. In this case, the bottom of the wheel moves straight backward. It can throw a stone straight back.

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Exercise for the reader: A stone is flung with an upward velocity component of $30$ MPH. How far had the wheel rotated when the stone leaves? How many milliseconds did it take to reach this velocity? How far has the car moved by the time the stone hits the ground?

Answer 2

The equations for the wheel are correct. As regards angles, I observe:

• This is a fairly low energy interaction in which a stationary rock encounters a mostly-stationary wheel (since the speed of the bottom of the wheel is zero relative to the road surface) and is then picked up and moved with the surface of the wheel for a short distance before it is released. So the launch angle will always be approximately tangent to the motion of the surface of the wheel.

• The rock is constrained to move only where the car and the ground are not.

• To make finding the allowed launch trajectories easy, I would assign the axle's velocity to 0 (taking the frame of the car) so that the launch velocity of the rock is the tangent velocity of the wheel relative to the axle, which we can break down into sine and cosine parts for uniform circular motion. Then you have a projectile motion problem with a range of $0<y<y_{car}$ over the domain $0<x<x_{car}$

• if you want to know the trajectories relative to the ground: add the axle velocity to vector to everything to boost back into the ground's frame. Draw the new x velocity and the y velocity vector. Calculate the angle between them.