Understanding velocity of points of wheel

Question

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 ?

Answer 1

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?

ADD: Answer to your little experiment:

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.

Answer 2

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	$-r\omega\ cos \frac 3 2 \pi + r\omega= 0 + r\omega = r\omega$	$r\omega\ sin \frac 3 2 \pi = -r\omega$