# Understanding velocity of points of wheel

I am very grateful to all for helping me out. Thank you. Also , It will be quicker for me to understand if you tell where at which point am I going wrong or right ? This way , I know exactly how much I know & where do I need to change.

This is a wheel rotating + moving on ground. (Pure rolling ). We say total velocity of a wheel = rotational + translational.

This is the trajectory of top point of wheel. s3+s4. AO

Like this :

s1 & s2 are v of wheel when Axis of rotation is centre of wheel. AE , EC.

s01 & s02 are translational distances covered by wheel. AD

Now , original position of the top point now is O.

My Q are :

1. We know v of centre of wheel = velocity of wheel. Axis of rotation is bottom point of wheel. Original distance covered by wheel = s3 + s4. When we talk of v of top point of wheel , we have to think is that s3 + s4 = in terms of translational + rotational distance. If we think of velocity of points in terms of rotational + translational velocity. It has to be the same with distance as well then.

2. When we say v of top point of wheel = 2rw. Do we mean to say

Rotational + translational = 2rw

Translational + w = 2rw

2rw - w = translational ?

• If it is rolling with no slip vs ground, then you have another constraint: instantaneous velocity of contact point = 0 ... that allows us to know omega from v_center. Then you can get instantaneous v at any point you want from v = (v_center) + (omega)(r) ... note the are vectors Commented Dec 17, 2021 at 20:52
• @PeteW But you’re not supposed to take axis of rotation as centre of wheel. Axis of rotation is the bottom point of wheel right ? If you take bottom point , you get v=2r*w which is one way I saw it in my textbook but Also , I have seen another text where axis of rotation is taken to be centre like you’re saying & they don’t mention bottom point as axis of rotation. Then , they say every point on wheel has v = vc + wr. Both the answers are totally different from each other too. So , its actually kinda confusing ; If you can proof your answer , I think that will help a lot. Commented Dec 18, 2021 at 8:30
• re: contact point as center of rotation -- if you are looking at it from ground reference frame, then you could say that. But everything becomes much simpler in the constant-velocity reference frame of the wheel's axle. Commented Dec 18, 2021 at 14:28

The relationship between the rotational speed and the linear speed is expressed as:

$$v = \dfrac{2\pi}{60}*r*N$$

N is the rotational speed - revolution per minute. If we set N = 1, then the equation becomes

$$v = \dfrac{2\pi*r}{60} = \dfrac{\Delta x}{time}$$

Does this help?

You said "Yes. Total circumference of circle = distance travelled by wheel on ground." So, now you know, for 1 revolution, $$s = 2\pi r$$, and lets assume you clocked the time interval is $$5s$$.

From definition of velocity, $$v_L = \frac{ds}{dt} = \frac{2*pi*r}{5} = 0.4 pi*r$$

From rotational speed, $$v = \dfrac{2\pi}{60}*r*N$$, in which, $$N = \dfrac{revolution}{time} = \dfrac{1 rev}{5s} * \dfrac{60s}{min} = 12 rpm$$, then

$$v_R = \dfrac{2\pi r}{60}*12 = 0.4\pi r$$

Conclusion: $$v_L = v_R = 0.4\pi r$$. What this tells us?

Comment: From the above, we noticed the main difference between the rotational speed and linear speed (velocity) is that while the former is associated with rotation and radial distance (curve), the latter is the speed of linear motion and linear distance (line) measured. Hope this helps.

• Great. Do you also know where can I get the proof for it , like how it is derived. I have never seen sth like this every before in this chapter. Commented Dec 18, 2021 at 8:33
• You can proof it by yourself. Take a CD disk and mark 2 points on its perimeter, then align one of the points with one on the desk. Now roll the disk until the second point is in touch with the desk, mark this point on the desk, then measure the distance between the two points on the desk. It is the distance traveled within the time interval - velocity.
– r13
Commented Dec 18, 2021 at 11:31
• Yes. Total circumference of circle = distance travelled by wheel on ground. This is 2pir. Then , by how much time. 2pir/t. I didn’t is the ‘N’ . Commented Dec 18, 2021 at 14:48
• K. Yes , this did help. Thanks a lot. Commented Dec 19, 2021 at 8:09

Try this:

Figure 1. Annotated version of diagram.

• Point $$a$$ is at angle $$A$$ from the point of contact with the ground.
• It has two components to its velocity - that of the wheel's general movement, $$v$$, and that due to rotation $$r\omega$$.
• The rotational velocity can be broken into the horizontal and vertical components:
• Horizontal velocity = $$-r\omega\ cos A$$.
• Vertical velocity = $$r\omega\ sin A$$ (taking up as positive direction).
• As you point out in your diagram, $$v = r\omega$$ so now we can say that:
• The horizontal velocity of any point on the wheel = $$-r\omega\ cos A + r\omega$$.
• The vertical velocity of any point on the wheel = $$r\omega\ sin A$$.

Let's test:

A Horizontal Vertical
0 radians $$-r\omega\ cos 0 + r\omega = -r\omega + r\omega = 0$$ $$r\omega sin 0 = 0$$
π/2 radians $$-r\omega\ cos \frac \pi 2 + r\omega= 0 + r\omega = r\omega$$ $$r\omega\ sin \frac \pi 2 = r\omega$$
π radians $$-r\omega\ cos \pi + r\omega= r\omega + r\omega = 2r\omega$$ $$r\omega\ sin \pi = 0$$
3π/2 radians $$-r\omega\ cos \frac 3 2 \pi + r\omega= 0 + r\omega = r\omega$$ $$r\omega\ sin \frac 3 2 \pi = -r\omega$$
• K. Getting it. Thanks a lot. Commented Dec 18, 2021 at 14:45