The case we have is : enter image description here

A wheel is kept in a box. There are two images given. Given below are conditions of the first image:

  1. Velocity at which the wheel is turning.

$\frac{d s}{d t}=\frac{d \vec{\theta} \times \vec{r}}{d t}, \vec{v}=\vec{\omega} \times \vec{r}$

  1. Velocity of the box in which the wheel is = v0

In the second image , we say w.r.t an outside frame. The velocity of the top point of wheel is $v_0 + \omega r$ & at the lowest point of wheel , velocity backward = $\omega r - v_0$. I have got this part. But if we compare this part with an actual wheel turning on ground.

Then, I’m not able to understand it because for a normal wheel. When the wheel turns $\omega r$ at top point, the velocity with which it has turned $\omega r$ will also be the velocity of wheel by which it has covered a distance on ground since the wheel is not in a box now. There is only one speed i.e $\omega r$. According to my textbook, for a normal wheel also moving. Velocity of top point = $v_0 + \omega r$ . I think it will be just $\omega r$.

Please help me understand this part. If any difficulty in understanding the question, please let me know.


Let us imagine a wheel moving & then we will consider all the things that the wheel has. Let us say the wheel moves a angular distance s, so it will also move the same distance s on ground.

s = θr is the angular distance covered by wheel.

v = ωr is the velocity by which this distance s is covered in a time t.

Now, what I mean to say is that velocity of the centre of the wheel will also be v right. There is no v0</sub?. Now, translational motion = rotational motion. They can’t be different, but it doesn’t mean there are two quantities separately. enter image description here

  • $\begingroup$ regarding your edit: the velocity of the center of the wheel is indeed $\omega R$ (when the wheel is not slipping. However, not all point on the wheel travel have the same velocity. Take the point of contact, if the wheel is rolling (not sliding), then its velocity should be the same as the velocity of the contact surface. $\endgroup$
    – NMech
    Commented Dec 8, 2021 at 8:27
  • $\begingroup$ See if this thread helps. physics.stackexchange.com/questions/308491/… $\endgroup$
    – r13
    Commented Dec 8, 2021 at 15:39
  • $\begingroup$ @r13 This did help. Thanks a lot. $\endgroup$
    – S.M.T
    Commented Dec 9, 2021 at 9:13
  • $\begingroup$ @r13 What I see I am getting confused between is the velocity of wheel vs velocity that certain points of wheel have. $\endgroup$
    – S.M.T
    Commented Dec 9, 2021 at 9:14
  • $\begingroup$ @S.M.T. regarding your last question, think of the merry go round as it rotates. It does not move as a whole, so the translational velocity is zero, but each point (along the radius) in the merry go round has a different magnitude of velocity. $\endgroup$
    – NMech
    Commented Dec 10, 2021 at 5:34

2 Answers 2


First of all, at the bottom of the wheel the velocity of the wheel is not $\omega r- v_0$. Instead its $v_0 - \omega r$ in the outside frame of reference. (if you disagree with that I can expand on it.

In the case of a wheel that its rolling (and not slipping), the velocities are:

  • 0 at the contact point
  • $\omega r$ at the center of the wheel which coincides with the general translational velocity of the wheel $v_0$
  • $2*\omega r$ at the top of the wheel (this can be thought as $v_0+ \omega r$

This can be summarised in the following image (I got it from this question but the original should be in Beer and Johnston Vector Mechanics).

enter image description here

Essentially it says that you can think of any motion on a plane as the sum of translational and rotational movement (it doesn't happen like that but its useful to think of it like that).

so the translational component means that all points on the wheel travel with the same velocity ($v_0$), while each point has a different vector velocity $\vec{\omega}\times \vec{r}$. The addition produces the result.

  • $\begingroup$ Do you mean to say that it is v0 - wr in the right direction for lowest point of wheel ? $\endgroup$
    – S.M.T
    Commented Dec 8, 2021 at 7:18
  • $\begingroup$ if you think of it vector terms then I guess what you wrote ($\omega r - v_0$) is OK. I think its better to use the $v_0 + \omega r$ because when v_0 is high then you don't really know the direction of the velocity (and in my experience unknown/uncertain quantities are better drawn pointing towards the positive direction - especially in dyanmics problems). $\endgroup$
    – NMech
    Commented Dec 8, 2021 at 7:22
  • $\begingroup$ Yes. You’re right. I have still not understand this part of v + wr. 1) There are 2 situations : We have a wheel that is turning in a box which is compared to a real life car wheel which moves on ground. Do you agree to say that the HP(Highest point) of these 2 situations have velocity = v+wr ? Because I feel these 2 situations are different. $\endgroup$
    – S.M.T
    Commented Dec 8, 2021 at 7:24
  • $\begingroup$ What is the difference in general translation velocity of wheel & wr at centre of wheel. Isn’t it that when wheel Travels a distance theta. This theta is the same distance covered on road. Therefore , there is only 1 speed wr & not v0 ? I agree v0 = wr but that doesn’t mean at top point of wheel. Velocity is 2*wr. $\endgroup$
    – S.M.T
    Commented Dec 8, 2021 at 7:35
  • $\begingroup$ regarding your comment "Because I feel these 2 situations are different", the only difference -- in my perspective -- is that the LP is constraint to zero velocity in the normal, rolling wheel. Regarding the last comment about "What is the difference in general translation velocity of wheel & wr at centre of wheel.". At the center of the wheel O the velocity is $v_0$ (i.e. only translational). The rotational component of the velocity ($\vec{\omega}\times \vec{r_O}$) is 0 . $\endgroup$
    – NMech
    Commented Dec 8, 2021 at 7:37

The speed of rotation has two components - rotational speed (rotation/time) and linear (tangential) speed (distance/time). As shown on left in the sketches below, the rotation speed is constant about the center of rotation, it describes how many times the red dots will pass points A & B within a given time frame; the linear speed describes how fast are the dots at the points while passing. The linear speed is directly proportional to the radius.

The sketch on the right depicts the wheel rotation without slip. When the wheel is placed against the ground, friction force will occur that reduces the speed at the contact point and turns spinning into rolling. It (the friction) simulates adding the linear velocity vector to the wheel system and causing the imbalance of forces that is responsible for the wheel to move forward.

enter image description here enter image description here

  • $\begingroup$ Can we say for a ball moving on ground with rotating & moving forward. The distance it will cover will be twice the distance covered by a ball just moving forward & not rotating since the velocity of top point of ball = 2v0. $\endgroup$
    – S.M.T
    Commented Dec 9, 2021 at 3:56
  • $\begingroup$ No. As in the baseball game, the pitcher throws a straight ball with uniform speed and it glances over the batter's bat, at the time of the hit, the velocity at the ball-bat contact point could approach zero, while the top is 2V, the imbalance in moving speed causes the ball to rotate and fly sideways. $\endgroup$
    – r13
    Commented Dec 9, 2021 at 15:07
  • $\begingroup$ K. Apart from saying velocity of particular point of ball , can we say velocity of ball overall will be wr or v ? $\endgroup$
    – S.M.T
    Commented Dec 10, 2021 at 5:16
  • $\begingroup$ At the baseball game, they usually call how fast was the ball. But, at Nascar, they call how fast the car is moving while its wheel is rolling ahead to achieve that speed. $\endgroup$
    – r13
    Commented Dec 10, 2021 at 14:22

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