# Dynamics rigid body question -- Velocity of points on a car's wheel

Pipsqueakalchemist
Homework Statement:
An automobile travels to the right at a constant speed of 48 mi/hr. If the diameter of a wheel is 22 in., determine the velocities of points B,C,D, and E on the rim of the wheel.
Relevant Equations:
V_b = V_a + (w)X(R_ab)
The solutions used the spend of the wheel and its radius to find the angular velocity. I’m confused because I thought to find angular velocity you use the speed at the points of the radius not the translation speed of the wheel itself. Can someone explain this to me plz

## Answers and Replies

Gold Member
Think about what happens if the wheel is moving faster or slower than the translational velocity where the wheel touches the road. If it is moving slower, it is being dragged. If it is moving faster, it is slipping (or spinning). Therefore under smooth rotation, it is moving at the translational velocity.

Think about the extreme cases:
1) wheel not turning, car moving
2) wheel turning, car not moving

Pipsqueakalchemist
So in a scenario when we have say a disk spinning, say we wanted to find the speed of a point that’s some radius away from the centre we would use V=wr and w is angular velocity and r is radius. But in this case since the wheel is moving at constant speed so is that why we can use the speed of the wheel and not the speed of the points on the edge of the wheels to find the angular velocity? Is that correct?

Pipsqueakalchemist
Think about what happens if the wheel is moving faster or slower than the translational velocity where the wheel touches the road. If it is moving slower, it is being dragged. If it is moving faster, it is slipping (or spinning). Therefore under smooth rotation, it is moving at the translational velocity.

Think about the extreme cases:
1) wheel not turning, car moving
2) wheel turning, car not moving
When you say translation velocity, you mean velocity at the points of B,C and D right?

Gold Member
Since B,C,D,E are at the same radius they have the same angular speed wr (with respect to point A) which has the magnitude of the translational velocity. Velocity also has direction associated with it so they are moving in different direction with respect to point A. Point A also has a directional velocity which has to be added to these velocities.

Pipsqueakalchemist
Since B,C,D,E are at the same radius they have the same angular speed wr (with respect to point A) which has the magnitude of the translational velocity. Velocity also has direction associated with it so they are moving in different direction with respect to point A. Point A also has a directional velocity which has to be added to these velocities.
But why can we use the speed of wheel to find angular velocity of the points? Is it because the wheel is moving at constant speed?

Gold Member
Do you know what the magnitude of wr is in the frame where point A is not moving?

Pipsqueakalchemist
Ummm not really sure what you’re talking about can you explain please

Gold Member
Imagine that you are moving with the translational velocity of the car. In this frame, point A does not move and it looks like the wheel is just rotating in a circle without any translation.

Gold Member
I am going to try and start again. For a point on the edge of the wheel, there are 2 velocities, the translational velocity of the car (and point A) and rotational velocity about point A. This problem is essentially asking you to add these two directional quantities. Do you understand this?

Pipsqueakalchemist
I understand there are 2 velocities for the points in the rotation and tangent direction, but why is the tangential velocity of the points the same as the velocity of the wheel that’s heading to the right

• italicus
Pipsqueakalchemist
Sorry I’m kinda dumb

Gold Member
No you are inexperienced.

Let’s say the car travels a distance equal to the circumference of the wheel. Do you believe that the wheel has undergone 1 rotation?

Pipsqueakalchemist
Yes

Gold Member
Ok. Since the car travels 1 circumference we know
vt=2PiR
Since the wheel rotates once during the same time we also know
w=2Pi/t

Substituting to cancel t gives w=v/r

Pipsqueakalchemist
Yea appreciate it, thx

Pipsqueakalchemist
So it’s not always the case that in order to find angular velocity that you have to use the velocity of the points that are some radius away from the centre

Gold Member
That’s not how I would put it. The wheel is rotating at a constant rate, so w is still determined by the velocity of the points that are some radius away from the centre. In this case, the wheel is also translating, so one has to figure out how this modifies the velocities of the points on the wheel.

Gold Member
The velocity of a point on the wheel is v_rotational plus v_translational. The non-slipping wheel condition gives a relationship between v_rotational and v_translational.

Pipsqueakalchemist
Non slipping mean no acceleration right?

Gold Member
Non-slipping is when the number of wheel circumferences traveled by the car is equal to the number of wheel rotations. There can be acceleration. Slip conditions are everything else (for example: the extreme cases I proposed in post #2).

Last edited:
Pipsqueakalchemist
Ok so V_wheel=wr is when the wheel is rolling without slipping

Pipsqueakalchemist
Also how would Ik that this problem is rolling without slipping since it’s not directly stated and there’s no mention of the arc length travelled being equal to the displacement of the wheel

Gold Member
The no-slip condition describes how we think a wheel normally works. The problem assumes that you know this. I could see a hw problem where you were expected to develop the condition yourself, but from the solition this wasn’t it.

One more thing for you to think about. Imagine you had a toy car and you spun one of its wheels with a velocity v. Now imagine that you did this in a car moving at the same velocity v. To you, the motion of the wheel would be identical in both cases. Now imagine, a pedestrian watching you spin the toy car wheel while you were in the moving car. He would see the same motion described in this problem.