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Consider a car moving along a straight horizontal road at constant speed, $v$.

Also consider one of the tyres/wheel of the car. On it, there are two particles of dust $A$ and $B$ (as shown in the diagram) at the outer edge of the tire.

How would I find the velocity of each dirt particle, relative to the road?

Just a diagram, showing what I'm

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    $\begingroup$ Hint : What part of the wheel is moving with velocity $v$? If the wheel is not slipping against the road, what is the velocity at point A? $\endgroup$
    – sammy gerbil
    Commented Jan 9, 2017 at 4:07
  • $\begingroup$ Um, I'd guess the whole wheel is moving at velocity, v. But if, it's not slipping against the road at A, that would perhaps mean it's at rest at A. However I don't understand how it is not slipping, I can't appreciate that. $\endgroup$
    – Saad
    Commented Jan 9, 2017 at 4:09
  • $\begingroup$ Get hold of a wheel or disk and watch what happens as it rolls. $\endgroup$
    – sammy gerbil
    Commented Jan 9, 2017 at 4:11
  • $\begingroup$ To imagine how it is not slipping imagine that the wheel is a gear and the floor has holes for the teeth. It is not a great analogy, but it show you that there are ways for a translating but also rotating device to move without slipping. $\endgroup$
    – user126422
    Commented Jan 9, 2017 at 5:03

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(I answered a very similar question about rotational acceleration the other day - you might find it useful)

You find the velocities of the particles on the rim by remembering the expression:

$$v=R\omega$$

Some call this a geometric bond because it ties rotation ($\omega$) to translation ($v$).

$v$ is the speed of a particle on the rim seen from the centre.

  • On a freely spinning wheel, particles on the rim move with speed $v$.
  • But for a wheel touching the ground without slipping, there must be no motion of that rim-point that touches the ground! Instead it is the centre of the wheel that moves with $v$. And the opposite upper rim-point moves with $2v$.

The point is that the rim moves with $v$ at all points seen from the centre. So when the centre itself is moving, we have to add that to those rim-points when we look at it from outside.

I can see there has started a small discussion about the no-slipping of the contact-point in the comments.

The point is that the point of the wheel that touches the ground musn't move of the same reason that your foot touching the ground musn't move when you take a step. If your foot was slipping and sliding, you wouldn't be able to push yourself forward. Same for the wheel.

And when that touching contact-point is slipping (from accelerating or braking to violently e.g.), we call it wheelspin and you would hear a squeaking sound and maybe see smoke because of the kinetic friction that happens. And as you maybe know from movies, during the wheelspin the car doesn't accelerate - it only start really speeding up, when the wheelspin is over, because then the wheels grip the ground again and there is no slipping.

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Here is a possible way to solve the problem:

Consider the road is at rest.

The wheel rotates with an angular velocity $\dot\phi$. For polar coordinates, we know

$$ x = r*cos(\phi) \\ y = r*sin(\phi) $$

From this, you can calculate $\dot{x}, \dot{y}$ as a function of $\phi, \dot{\phi}$. Also, dust particle A is at $\phi = -\pi/2$ and particle B is at $\phi = \pi/2$.

To this, add the velocity $v$ with which the wheel is moving and you should have your solution.

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