Lets take the following example

enter image description here

enter image description here

According to above examples it means that velocity at the above portion is max while the velocity at lower portion is min.

But I think it should be the same at both parts (just opposite in direction).

Why are both different?

  • 4
    $\begingroup$ When the wheel is rotating freely (above ground), the velocities at the same radius are the same. Think about what happens when you drop the rotating wheel :) $\endgroup$ – Luaan Nov 28 '16 at 11:22
  • $\begingroup$ Once you've understood the answer to this, try figuring out what shape the path is that is described by a point on the rim of a rolling (not slipping) wheel, as seen by a stationary observer. :-) $\endgroup$ – nekomatic Nov 29 '16 at 9:51
  • $\begingroup$ I just remembered that there's a scene in the Big Lebowski taken from the perspective a hole in a bowling ball (looking out) as it rolls down a lane. It's a pretty good visual demonstration of this. Another option is to get inside a big tire and roll down a hill but that will fairly likely result in injury or at least some motion sickness. $\endgroup$ – JimmyJames Nov 29 '16 at 15:12

You have to remember that the entire wheel is also moving.

Think of this. Where the wheel meets the ground, the velocity of the contact point must be 0, otherwise the wheel would be skidding. Another way of looking at it is that at the contact point the forward velocity of the wheel is cancelled by the backward velocity of the point. On the other hand, at the top of the wheel these velocities add together: the velocity of the entire wheel with respect to the ground, plus the velocity of that point with respect to the centre of the wheel.

I once tested this, when I drove behind a truck that was trailing a rope on the road. I drove one of my front wheels over the rope and instantly the rope broke. It had to break because one end of the rope was moving at the speed of the truck, while the other was stationary between the road and my tyre.

  • 6
    $\begingroup$ So, really, the frame of reference is key. If the observer is moving alongside the rolling wheel at the same average linear velocity, then the blurring should be uniform... right? $\endgroup$ – hBy2Py Nov 28 '16 at 14:28
  • 3
    $\begingroup$ @hBy2Py Correct. $\endgroup$ – Shufflepants Nov 28 '16 at 16:00
  • 2
    $\begingroup$ @hBy2Py -- the blurring will be more pronounced nearer the rim on each spoke, but the blurring will be the same at the same point along every spoke. $\endgroup$ – Malvolio Nov 28 '16 at 19:55
  • 1
    $\begingroup$ Ah, absolutely right, @Malvolio, I wasn't even thinking about the radial velocity distribution. $\endgroup$ – hBy2Py Nov 28 '16 at 20:38
  • 15
    $\begingroup$ That was probably not the safest experiment ever. $\endgroup$ – Devsman Nov 28 '16 at 21:24

In the no slipping condition the translational speed $v$ of the centre of mass of the wheel and the angular speed of rotation $\omega$ of the wheel are related.
$v= r \omega$ where $r$ is the radius of the wheel.
So one can find the vector sum (blue) of the translational velocity of the wheel at any point (red) and the tangential velocity of the wheel at any point (grey) as in done in the diagram below.

enter image description here

Since at the instant shown on the right hand diagram the wheel is roating about the point of contact the directions of those resultant velocities at each point on the wheel must be at right to the line joining the point to the point of contact between the wheel and the ground (green).

  • 7
    $\begingroup$ The wheel is not necessarily rotating about its centre of mass: I can add a reflector which moves its centre of mass and yet it continues to rotate about the axle. $\endgroup$ – Andrew Morton Nov 28 '16 at 14:12
  • 1
    $\begingroup$ Good point above, but +1 for the picture. When I see this type of question, the first thing that comes to mind is velocities relative to different reference frames. This is a tremendous help IMHO. $\endgroup$ – Charles Nov 29 '16 at 5:44
  • $\begingroup$ Farcher, if you are creating a chatroom, feel free to invite me. Would like to chat with you on the dramatic question... $\endgroup$ – bonCodigo Aug 12 '17 at 4:24

There are two contributions to the speed for which the spokes on the wheel move at. There is translational velocity and rotational velocity.

On the top part of the wheel the vectors corresponding to the translational and rotational velocities add together as they are moving in the same direction (to the right).

Whereas at the bottom section the wheel is rotating in the opposite direction to which the wheel is moving along (as the bike moves); in other words the vector corresponding to the rotational velocity is to the left, but the bike and hence the wheel is moving towards the right. So these two vectors subtract.

This is why the top section is moving faster than the bottom.

  • $\begingroup$ (at point P) Is it inertia that kept the lower part moving (after achieving 0 velocity ) ? $\endgroup$ – Junaid Nov 28 '16 at 4:22
  • 1
    $\begingroup$ @Junaid It's the angular (rotational) velocity that keeps the lower part moving, you have to understand that the velocity of the point of contact is zero with respect to the ground. But the point of contact is constantly changing (the point P will move round the wheel as the bike moves along). $\endgroup$ – BLAZE Nov 28 '16 at 4:34

Rotation about P is equivalent to rotation about O plus translation of O horizontally. This works in reverse to. Any rotation + translation can be described equivalently by a pure rotation about a distant point.

Let's place a coordinate system at P and measure the linear velocity at an arbitrary point $\vec{r}$.

$$ \vec{v} = \vec{\omega} \times \vec{r} $$

  • The velocity at P is $\vec{r}_P = 0$ } $\vec{v}_P = 0$
  • The velocity at O is $\vec{r}_O = R\, \hat{j}$ } $\vec{v}_O = (-\omega) \,\hat{k} \times R \,\hat{j}= (\omega R)\, \hat{i}$

You should be able to imagine this, but if it gives you trouble, use an actual tire, or any flat circular object of reasonable size. Perhaps a lid for a storage container, or a plate (something like a coin is much to small).

Think of it this way:

  • Set the tire (or substitute item) in its normal rolling position.
  • Locate the very top of the tire, which will be directly above the point of contact where the tire is resting.
  • While the tire is still resting on its bottom surface, move the top of the tire about 10 degrees (or if its easier to estimate, about 2-inches or 5-cm, less for a smaller sized object) in a direction that will cause the tire to begin to roll.
  • Move the tire back to the starting point and then move it 10 degrees in the opposite direction.
  • Repeat this (back and forth) and while doing so, notice that while the top of the tire is moving a span of about 20 degrees (about 4-inches or 10-cm), the bottom of the tire (the "middle" contact point) is barely moving at all.

People. Please. All parts of a rotating disc travel at EXACTLY the same speed dependent upon ONLY the distance from its center. The top is NOT moving faster than the bottom--it just looks like it is due to camera blurring--an unintentional photographic effect. If the spokes were traveling at different speeds the wheel would NECESSARILY deform or disintegrate. We have some great attempts to interpret, explain and prove the original thesis--but such proof is itself an illusion based on misinterpretation by some otherwise EXTREMELY intelligent people. If the original conjecture is correct then, again necessarily, THE EARTH ITSELF WOULD DISINTEGRATE. Velocity = 2 Pi * r (from the center) * rotational speed, at EVERY POINT OF THE WHEEL. If you look closely at the photo, the spokes appear bend around rim and at a varying degree everywhere. And I know that on stackexchange someone will call me an idiot. I'm cool with that. :P

  • 1
    $\begingroup$ Great sense of humour :D $\endgroup$ – Junaid Nov 29 '16 at 7:28

Not the answer you're looking for? Browse other questions tagged or ask your own question.