At 2:12 of this video by MIT OCW, rolling without slipping condition for a disc rotating along an axis is given as:

$$ v_{cm} = \omega r \tag{1}$$

Where $r$ is the radius, $\omega$ is the angular velocity.

How would I extend this for motions of rigid bodies? After reading this post it seems to me that I set the condition that at point of contact, the velocity as zero. It also seems that this is the way considered by Selene Routley in this answer. It may be noted that this is the following expression arrived at by Selene Routley for the question to be mentioned: $$v_{cm}=\omega_0\,l\,\cos\alpha$$

However, In a video by famous youtuber Tibees on a question of JEE where this concept comes up, she at 9:16 uses (1) to define the no slip condition. Here is the question's statement:

Two thin circular discs of mass $m$ and $4 m$, having radii of $a$ and $2 a$, respectively, are rigidly fixed by a massless, rigid rod of length $l=\sqrt{24} a$ through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $\omega$. The angular momentum of the entire assembly about the point ' $O$ ' is $\vec{L}$ (see the figure). enter image description here

My question is why was it justified for her to use eqtn (1) in this case? I think she may be correct because she ended up with the 'correct' final answer. However, I tried working back and having $v=\omega *r$ seems to violate the no slip condition (?) in this case as given by Selene Routley.

Further it also seems strange that there can are two velocities of centre of mass in each approach but ultimately both give the same answer. Both by Tibees and Selene Routleys method , one can reach the correct relation between angular velocity(one of the statements required to be shown)

I had written an answer on reaching the answer following Routley's calculations here

Related questions: (1) , (2) , (3)

All videos are time stamped and I will highly appreciate someone who goes through all the links carefully before answering.


On the MIT 'blackboard video' he derives the formula $v_{cm}=r\omega$ but that is for a wheel that's rolling with it's plane perpendicular to the ground.

If we imagine the same wheel tilted away from us (as we view it on the video) and towards the person who's drawing - the distance he calls $\Delta x_{cm}$ would be lower, i.e. the centre of mass doesn't move as far.

You can imagine this if there is a big tilt, the point of contact of the wheel with the ground traces out a big circle on the ground, but because of the big tilt, the cm point traces out a much smaller circle.

So for a tilted wheel $v_{cm}=r\omega$ is no longer valid. What is still valid is the velocity of the point of contact of the wheel with the ground is still $v=r\omega$.

On the other video at about 10.20, Tibees adjusts her formula with a $ cos\theta$ term, after applying $v_{cm}=r\omega$ for an imaginary path along an imaginary ground with the wheel perpendicular to it (thus making the formula valid).

You might find it easier to use the circle traced out by the point of contact of the small wheel and the floor. The radius of that is $R= \sqrt{L^2+a^2}$ - it's valid to apply $v=r\omega$ for the point of contact and then use $\Omega = \frac{v}{R}$, it also gives the right answer.

  • $\begingroup$ John, I understand what your argument for getting the right answer but I'm more interested in how the tibee's approach is consistent with the mentioned things in post rather than the actual answer itself $\endgroup$
    – Buraian
    May 24 '21 at 22:14
  • 1
    $\begingroup$ The answer has been edited, if it still doesn't clarify things please explain a bit more about what you think confusing with Tibees approach. To be honest it seems strange to use the circle that she used and then adjust the angular velocity vector later, when we know the system remains on the floor and does a different circle $\endgroup$ May 24 '21 at 22:32
  • $\begingroup$ My problem is when she writes the velocity , of what I think is centre of inner disc, the radius times omega at the time stamped point. This equation is required for rolling about one axis but I don't think it should be applied here. Particularly speaking I noticed that there are two angular velocities; one about the rods axis and one about z. One would need their vector sum and do omega cross produced with distance from point of contact from axis to get the right answer. However even without this tjbees got the right answer.. which is making me pill my hair $\endgroup$
    – Buraian
    May 24 '21 at 22:39
  • $\begingroup$ At first an alternative way was tried and gave the wrong answer, now it's been edited again and is right, hopefully. It seems that the $v=r\omega$ formula for the centre is only valid when it's to give the velocity around a circle with radius measured perpendicular to the the disc . It doesn't work when we try using it around the circle that it actually moves in (as in this question) $\endgroup$ May 24 '21 at 23:04
  • $\begingroup$ See, that is fine, I am asking right at the moment she writes v=a omega. I think this is problematic because there the angular velocity can be decomposed to have a component along z-axis and along the line segment connecting rod, ultimately pointing along x-axis. It points along x-axis due to it being instantenous axis of rotation, so shouldn't we take the cross product of this net ang. velocity vector with seperation vector between the bottom point and center rather than directly write aw? $\endgroup$
    – Buraian
    May 24 '21 at 23:08

Using the equation $v_{cm}=R\omega$ is equivalent to requiring the condition that "at point of contact, the velocity is zero."

  • Firstly, notice that moving the wheel's centre-of-mass forwards with $v_{cm}$ as seen from the ground frame is equivalent to the ground moving backwards with $v_{cm}$ as seen from the centre-of-mass frame.

  • Secondly, notice that in order to avoid slipping, as seen from the centre-of-mass frame the wheel periphery must be moving just as fast backwards as the ground when coming in contact with the ground.

  • Thirdly and finally, apply the geometric bond between rotation and speed of particles on the wheel: $$v=r\omega,$$where $r$ is the distance from the centre. Since the ground is moving backwards with $v_{cm}$, $v$ in this general geometric bond is in the no-slipping scenario equal to $v_{cm}$, and we now have a relationship for the angular speed required to make a periphery particle at a distance of the radius $R$ from the centre move just as fast as the ground while in contact, corresponding to no slipping: $$v_{cm}=R\omega.$$

This only requires us to beforehand having proved the geometric bond that is used.

  • $\begingroup$ If this applies for all kind of bodies, could you explain the approach described by selene routley in this answer? It seems to me that she gets something else as the center of discs velocity for a very similar scenario $\endgroup$
    – Buraian
    May 25 '21 at 8:07
  • $\begingroup$ @Buraian That seems to be a slightly different scenario. I have tried to answer your question "My question is why was it justified for her to use eqtn (1) in this case?" from your post. Would you mind elaborating on your question if have misunderstood what you are looking for? $\endgroup$
    – Steeven
    May 25 '21 at 8:11
  • $\begingroup$ The scenario is different but I am fairly certain the principle applied there is applicable here as well. I'll work a bit on editing the question $\endgroup$
    – Buraian
    May 25 '21 at 8:14
  • $\begingroup$ I have edited the question $\endgroup$
    – Buraian
    May 25 '21 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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