(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:
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.