5
$\begingroup$

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.

$\endgroup$

4 Answers 4

2
$\begingroup$

Landau & Lifshitz explain an equivalent problem. Some observations:

  1. The Instataneous Axis of Rotation (IAR) is a straight line passing for the two poins of the disks, in contact with the flat floor. Thus, the angular velocity $\boldsymbol{\omega} = \omega \boldsymbol{\hat{\imath}}$ will be a vector along this line.
  2. At any of the contact points $C_i$ velocity is zero. Using the fundamental decompositiof of velocities for a rigid body $\boldsymbol{v}_P = \boldsymbol{V}_{C_i} + \boldsymbol{\omega}\times \boldsymbol{r}_{P/C_i}$, where $\boldsymbol{r}_{P/C_i}$ is a vector from $C_i$ to $P$.
  3. The center $P$ of any of the two disks follow a circular movement, so $\boldsymbol{V}_P = -d_{ax}\dot{\alpha}\boldsymbol{\hat{\jmath}}$.
  4. By equalling the velocity of P in both cases we have the condition of no slipping: $d_{ax}\boldsymbol{\hat{\jmath}} = \boldsymbol{\omega}\times \boldsymbol{r}_{P/C_i}$, so we have:

$$-d_{ax}\dot{\alpha}\boldsymbol{\hat{\jmath}} = \omega\boldsymbol{\hat{\imath}}\times \boldsymbol{r}_{P/C_i} \Rightarrow \boxed{d_{ax}\dot{\alpha} = \omega\ x_{P/C_i}} $$

$\endgroup$
1
$\begingroup$

The general no-slip condition for a body is expressed as a dot product between the translational velocity of the body at the point of contact $\boldsymbol{v}_A$, and the contact normal vector $\boldsymbol{n}$.

There are two forms of this equation

$$ \boldsymbol{v}_{A}-\boldsymbol{n}\left(\boldsymbol{v}_{A}\cdot\boldsymbol{n}\right)=\boldsymbol{0} \tag{1}$$

and

$$ \boldsymbol{n} \times \boldsymbol{v}_{A}=\boldsymbol{0} \tag{2}$$

both of which yield the same result, saying that the velocity vector shall have no tangential components at the contact point.

To find the translational velocity at the contact point, one can use a known reference point like the center of mass, located at $\boldsymbol{r}_{C}$ relative to the contact, and the motion of the body at the center of mass

$$ \boldsymbol{v}_{A}=\boldsymbol{v}_{C}+\boldsymbol{r}_{C}\times\boldsymbol{\omega} \tag{3}$$ where $\boldsymbol{v}_C$ is the translational velocity of the center of mass, and $\boldsymbol{\omega}$ the rotational velocity of the body.

or use the axis of rotation if it goes through a known point $\boldsymbol{r}_R$ and has a known pitch value $h$ (the ratio of translational velocity parallel to the rotation over the rotation rate) to get

$$ \boldsymbol{v}_{A}=\boldsymbol{r}_{R}\times\boldsymbol{\omega}+h\,\boldsymbol{\omega} \tag{4} $$

For the problems mentioned the pitch value is assumed it be zero as rolling on a wheel produces no helical motion, but that is not the general case for all rolling.

All the methods you will encounter are some form of projection of the above. This is why you see equations in component form because the author does not want to overstimulate the audience with the full 3D treatment.

But if you are asking about general cases, you need to go to 3D.

In the case of a rolling disk, the above produces the following

$$\begin{array}{r|l} \text{Quantity} & \text{Value}\\ \hline \text{Center of Mass Location} & \boldsymbol{r}_{C}=\begin{pmatrix}-a\sin\theta\\ a\cos\theta\\ 0 \end{pmatrix}\\ \text{Center of Mass Velocity} & \boldsymbol{v}_{C}=\begin{pmatrix}0\\ 0\\ v \end{pmatrix}\\ \text{Body Rotation} & \boldsymbol{\omega}=\begin{pmatrix}\omega\cos\theta\\ \omega\sin\theta\\ 0 \end{pmatrix}\\ \text{Contact Normal} & \boldsymbol{n}=\begin{pmatrix}0\\ 1\\ 0 \end{pmatrix} \end{array}$$

Combine (1) and (4)

$$\begin{aligned}0 & =\boldsymbol{n}\times\left(\boldsymbol{v}_{C}+\boldsymbol{r}_{C}\times\boldsymbol{\omega}\right)\\ \begin{pmatrix}0\\ 0\\ 0 \end{pmatrix} & =\begin{pmatrix}0\\ 1\\ 0 \end{pmatrix}\times\left(\begin{pmatrix}0\\ 0\\ v \end{pmatrix}+\begin{pmatrix}-a\sin\theta\\ a\cos\theta\\ 0 \end{pmatrix}\times\begin{pmatrix}\omega\cos\theta\\ \omega\sin\theta\\ 0 \end{pmatrix}\right)=\begin{pmatrix}v-a\,\omega\\ 0\\ 0 \end{pmatrix} \end{aligned}$$

So the velocity of the center of mass is simply $v=a\,\omega$ which is a bit obvious.

The tricky part is differentiating (3) to get to the acceleration of the body (which is required for the equations of motion).

If you do a material derivative with $\boldsymbol{a}_{A}=\boldsymbol{a}_{C}+\boldsymbol{r}_{C}\times\boldsymbol{\alpha}+\boldsymbol{v}_{C}\times\boldsymbol{\omega}$ and use it in (2) as $$ \boldsymbol{n} \times \boldsymbol{a}_A =0 $$ it will yield incorrect results. This is because the contact point isn't subject to cetrifugal accelerations since it is not a material point (not fixed to the body).

The correct way to bring the no-slip conditions to the acceleration level is to consider the contact point as not moving and use the spatial acceleration $\boldsymbol{\dot{v}_A} = \boldsymbol{a}_{C}+\boldsymbol{r}_{C}\times\boldsymbol{\alpha}$ in (1) or(2)

more specifically:

$$\boldsymbol{0}=\boldsymbol{n}\times\underbrace{\left(\boldsymbol{a}_{C}+\boldsymbol{r}_{C}\times\boldsymbol{\alpha}\right)}_{\text{spatial accel.}} \tag{5}$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
15
  • $\begingroup$ 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 $\endgroup$ Commented May 24, 2021 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 $\endgroup$ Commented May 24, 2021 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 $\endgroup$ Commented May 24, 2021 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) $\endgroup$ Commented May 24, 2021 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? $\endgroup$ Commented May 24, 2021 at 23:08
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 $\endgroup$ Commented May 25, 2021 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
    Commented May 25, 2021 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 $\endgroup$ Commented May 25, 2021 at 8:14
  • $\begingroup$ I have edited the question $\endgroup$ Commented May 25, 2021 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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