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The doubt arose because of this question-

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). Which of the following statement(s) is(are) true? enter image description here
(A) The magnitude of angular momentum of the assembly about its center of mass is $17 \mathrm{ma}^{2} \omega / 2$
(B)The magnitude of the $z$-component of $\vec{L}$ is $55 \mathrm{ma}^{2} \omega$
(C) The magnitude of angular momentum of center of mass of the assembly about the point $\mathrm{O}$ is $81 \mathrm{ma}^{2} \omega$
(D) The center of mass of the assembly rotates about the $z$-axis with an angular speed of $\omega / 5$

It is a popular question from JEE $2016$, an exam annually held for engineering aspirants in India. My confusion is regarding point (D), one of the correct answers.

There are a couple of ways of arriving at $\frac{\omega}{5}$. One way, would be to find the angular velocity in the direction perpendicular to its spin angular velocity, (by finding the velocity of the centre of the first disc, and dividing by $l$) and taking the component of this angular velocity along the $z$ axis. (See $9$:$00$ of this)

Another way, would be to consider the circle the bottom of the first disc traces, and realising that the distance covered on the disc's circumference equals the distance covered on the ground. Dividing the angle covered in time $t$ by $t$ gives us the required answer. (See $4$:$50$ of this)

However, we can also recognise, that the centre of the first disc, moves in a circle about the $z$ axis, with a uniform speed, just like the centre of mass.

enter image description here

If we divide this uniform speed by the distance of the point from the $z$ axis, we arrive at a completely different answer for the angular velocity.

enter image description here

In the shown diagram, noting that the point of contact of the first disc with the ground is at rest, we find that the linear velocity of the centre of the first disc equals $a\omega$ in magnitude (where $a$ is the radius of the first disc and $\omega$ is the spin angular velocity mentioned in the question) and points outside the plane. The component of net angular velocity perpendicular to the rod does not contribute to this speed.

Then, recognising that this centre moves in a circle about the $z$ axis, with a uniform speed $a\omega$, we find the angular velocity of this point , by dividing this velocity with the radius (in the diagram, it equals the perpendicular distance of the centre in the $x-z$ plane, or $l\cos(\theta)$, where $\theta=\arctan\frac{a}{l}$. This distance stays constant.)

This equals the orbital angular velocity of the whole set-up, and the method is what we generally use for finding the angular velocity of any point about an axis. This is the (incorrect) method used in the first 3 minutes of this video

Why is there a discrepancy in answers? Why is this method incorrect in this scenario? I did not find the answers to this question satisfactory.

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    $\begingroup$ Angular velocity is usually expressed in rad/s, and all points on the line running down the axis of the two disks are experiencing the same angular velocity. Due to this, I would divide 2 pi by the number of seconds required to make one complete revolution, to obtain an angular velocity. Have you considered trying this method? $\endgroup$ May 22 at 20:33
  • $\begingroup$ How many times does this same question get asked here. However, this question in particular was well written and I liked reading it. $\endgroup$
    – Buraian
    May 22 at 23:14
  • $\begingroup$ "If we divide this uniform speed by the distance of the point from the z axis, we arrive at a completely different answer for the angular velocity. This method is what we generally use for finding the angular velocity of any point about an axis." Pl add how you did this calculation. Since the disc is slightly tilted, finding the distance from axis to it's center will not be easy $\endgroup$
    – Buraian
    May 22 at 23:21
  • $\begingroup$ @Buraian Added diagram and working. Did not do so initially, for fear of making the question veer towards a "check my working" type question, which the site tends to close down. $\endgroup$
    – Aspirant
    May 23 at 6:10
  • $\begingroup$ @DavidWhite This is similar to the second method I highlighted. I'd also prefer an answer which sheds some light on the concept of angular velocity, and explains why the third method doesn't work in this case, as opposed to a different solution. $\endgroup$
    – Aspirant
    May 23 at 6:13
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After a lot of thought, I found the problem. The issue is that you haven't taken the net angular velocity when doing the cross product. After calculations we find net angular velocity is given as:

$$ \vec{\Omega}= \omega \cos \theta \hat{i}$$

Now, the tangential velocity of the com of the inner disc as it moves in a circle is given as the cross product of the above vector with vector joining the point of contact of disc with ground to it's center:

$$| v_{t}| = |\vec{\Omega} \times \vec{r}| = (\omega \cos \theta)(a)\sin(\frac{\pi}{2} - \theta)= a\omega \cos^2 \theta \tag{1}$$

We can write the same velocity by writing the cross product of the angular velocity about z-axis with the vector joining perpendicular from z-axis to the COM we have, this is expressed as:

$$ |v_t| = | \vec{\omega}' \times \vec{r}_z|= \omega' \cdot l \cos \theta \tag{2}$$

Now, I equate the two different definitions:

$$ a \omega \cos^2 \theta = \omega' l \cos \theta \tag{3}$$

Recall that by geometry $ \frac{a}{l} = \tan \theta$, this leads to the following equation after cancellations:

$$ \tan \theta \omega \cos \theta= \omega' $$

Simplfying more and recalling that $\sin \theta = \frac{1}{5}$ (geometry result):

$$ \frac{\omega}{5} = \omega'$$

Which was required to be shown :)

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    $\begingroup$ Thank you so much! You've resolved a massive misconception, and I have no doubt many future students will be grateful for this answer. $\endgroup$
    – Aspirant
    Jun 13 at 15:39

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