Consider first, the simple case of a disc rolling horizontally on the x-axis, having angular velocity $\mathbf{\omega}$ enter image description here.

Let p be any point on the edge of the disc. Let $\mathbf{r_{c}}$ be the position of the center, and $\mathbf{X}$ be the vector joining the centre to point p. Then, if the position of point p is given by: $$\mathbf{r_{p}}= \mathbf{r_{c}}+ \mathbf{X}$$ Which further gives: $$\dot{\mathbf r}_{\mathbf p}= \dot{\mathbf r}_{\mathbf c}+ \mathbf{\omega}\times \mathbf{X}$$

Now, I define "rolling without slipping" to mean:

The point of contact is instantaneously at rest with respect to the surface.

I believe this is equivalent to:

$\dot{\mathbf r}_{\mathbf p}=0$ at the instant when $\mathbf p$ becomes the point of contact.

For this particular case, at the instant $\mathbf p$ is the point of contact, $\mathbf{X}$ becomes $-R \hat{z}$, and we end up with the scalar equation $v_{\mathbf c}=\omega R$.

More often than not, situations that involve pure rolling are very very similar to this, and so $v=\omega R$ becomes almost synonymous with rolling without slipping.

However, this answer by Selene Routley is a clear example of a case where $v=\omega R$ does not hold true.

She writes: "The instantaneous speed of a point on the cone's axis of symmetry a distance $h$ from the base is $|\Omega|h\sin\alpha = \omega_0 h\cot\alpha\sin\alpha = \omega_0 h\cos\alpha$"

This is why I believe my definition in bold, is more suited to analyse rolling without slipping in general. This brings me to my question:

enter image description here

This is a fairly popular question and has been asked on this site numerous times, however in this post of mine I wish to analyse only a certain aspect of this problem, pertaining to my definition of rolling without slipping.

The question: For this setup, find the angular velocity $\Omega$, about the z-axis. Let $\theta$ be the angle made by the line joining the discs, with the horizontal.

Now, there are many valid approaches that give the correct answer, i.e $\Omega= \omega/5$. However, my approach gives a slightly different answer.

Consider the smaller disc only. Consider the vectors $\mathbf{r_{c}}$ and $\mathbf{r_{p}}$ and $\mathbf{X}$ similarly as before (again, p is some random point on the edge).

Let $\hat{n}$ be the unit vector normal to the circle that the disc traces on the ground. (basically, $\hat{n}$ is the "horizontal" direction). Then, $\mathbf{r_{c}}$ is $l\cos(\theta)\hat{n} + l\sin(\theta) \hat{z}$. Now, its rather obvious that $d\hat{n}/dt= \Omega \times \hat{n}$. With this, the equation:

$$\dot{\mathbf r}_{\mathbf p}= \dot{\mathbf r}_{\mathbf c}+ \mathbf{\omega}\times \mathbf{X}$$


$$\dot{\mathbf r}_{\mathbf p}= l\cos(\theta) \Omega \times \hat{n} +\mathbf{\omega}\times \mathbf{X}$$

Setting this to zero at the instant P becomes the point of contact, we get the scalar equation $\Omega l\cos(\theta)= \omega a$. Which results in $\Omega= \omega / 4.8$, not $5$.

I just don't see why this approach fails. It seems to use a very simple definition of rolling, and at the end of the day it's just geometry followed by derivatives. I don't make any complicated/controversial assumptions about the dynamics of the system.

Also, we get $\dot{\mathbf r}_{\mathbf c}= \Omega l\cos(\theta)$, which is the same as Selene got , for a situation which is extremely similar. This made me believe that the approach is correct, but somehow in the end it gives a wrong answer ... Just what am I missing here? Is my definition not correct?

  • 2
    $\begingroup$ physics.stackexchange.com/questions/638765/… This was pretty much the same as this. Funny enough, while everyone in YT comments is praising the difficulty of the JEE exam, no one could point out this conceptual flaw. GG. However, I appreciate your search for the truth for this even after clearing JEE. That's how everyone should aspire to be. $\endgroup$ Commented Aug 10, 2021 at 14:55

2 Answers 2


You have missed a subtle but important fact, but it's alright since this is a very common mistake. Your equation is not wrong. The key point is that you forgot that the total angular velocity of the disk is not $\boldsymbol{\omega}$, but $\boldsymbol{\omega}+\boldsymbol{\Omega}$. In other words, relative to the lab frame, the disk is doing three things:

  1. The center of the disk revolves around the $z$-axis with angular frequency $\Omega$.
  2. The disk rotates around the $\boldsymbol{\omega}$-axis with angular frequency $\omega$.
  3. The disk rotates around the $z$-axis with angular frequency $\Omega$.

You have missed out the third one. Therefore, your equation should be $$\mathbf{0}= (l\cos\theta) \boldsymbol{\Omega} \times \hat{\mathbf{n}} + (\boldsymbol{\omega}+\boldsymbol{\Omega}) \times\mathbf{X}$$ which you can check that it gives the correct answer.

  • $\begingroup$ Ah yes, indeed. Thanks. Just to confirm, the fact that Vc=aw is false for this case, still holds ,right? $\endgroup$
    – satan 29
    Commented Aug 2, 2021 at 13:36
  • $\begingroup$ @satan29 What do you mean? $\endgroup$ Commented Aug 2, 2021 at 13:40
  • $\begingroup$ $|\mathbf{\dot{r_{c}}}| \neq a\Omega$ is true right? Just as a confirmation. I'm pretty sure its true. $\endgroup$
    – satan 29
    Commented Aug 2, 2021 at 13:42
  • $\begingroup$ @satan29 Yes, they are not equal. You already said it is equal to $\Omega l \cos\theta$ $\endgroup$ Commented Aug 2, 2021 at 13:44
  • $\begingroup$ Right. I just wanted to stress on this because many so called "solutions" were using $v=a\omega$ casually. The main reason I took this approach was to try and debunk that. Thanks a lot again. $\endgroup$
    – satan 29
    Commented Aug 2, 2021 at 13:45

Note that in the sketch, the distance $L$ is perpendicular to the smaller disk, and $L = d\cos(θ)$ where $d$ is the horizontal distance from the z-axis to the point of contact, and $θ$ is measured from the horizontal to $L$. In one cycle of rotation, $T = 2π/ω$, the disk lays down its circumference $2πa$ on the horizontal surface. Then the angular velocity about the z-axis $$Ω = \frac{2πa}{Td} = \frac{ωa\cos(θ)}L.$$

  • 1
    $\begingroup$ Your analysis is certainly correct, however the purpose of my question was to investigate my particular approach. $\endgroup$
    – satan 29
    Commented Aug 2, 2021 at 16:01
  • $\begingroup$ Your approach included an incorrect equation. $\endgroup$
    – R.W. Bird
    Commented Aug 2, 2021 at 18:48

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