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I'm not asking you to solve this question, but here it is anyways for clarification purposes:

A rod of mass $m$ and length $2R$ can rotate about an axis passing through $O$ in vertical plane. A disc of mass $m$ and radius $R/2$ is hinged to the other end $P$ of the rod and can freely rotate about $P$. When disc is at lowest point both rod and disc has angular velocity $\omega$. If rod rotates by maximum angle $\theta = 60°$ with downward vertical, then find $\omega$ in terms of $R$ and $g$ (all hinges are smooth).

enter image description here

(Answer: $\omega = \sqrt{\frac{9g}{16R}}$ )

Now, what I concluded is that the disc spin isn't going to complicate the problem as no force is opposing it's rotation and it's going to be rotating with same angular velocity throughout.

So, my flow of solution was to calculate torque about upper most $O$ point at an arbitrary angle $\theta$ for the entire system (for the entire system because that way I don't need to deal with hinge forces on either bodies), find angular velocity as a function of $g$ and $R$.

BUT, for calculating torque, I need their moment of inertia about $O$. And that's where the problem lies.

If I calculate $Io$ as

$Io = I(rod\,\,about\,\,O) + [ I(disc\,\,about\,\,its\,\,COM) + m(2R)^2]$, where $m$ is the mass of disc and $2R$ is its distance from $O$, I don't get the answer.

IF, however, I ignore the I(disc about its COM) and treat disc as a point mass at $P$, the solution works out. Why is that the case? Is it just a coincidence or can we actually do this? And why can we do this?

My solution is posted below.

And I've verified the answer via solving it by energy conservation, so the answer is correct.

Sorry for the fact that in one step, I've randomly used $L$ in place of $R$ and also forgot a minus sign. Thankfully the mistake doesn't propagate forward in the solution.

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  • $\begingroup$ Could you format this in latex? , It's much harder to read if you send a picture of your work on paper. $\endgroup$ – JustJohan Nov 10 '20 at 18:35
  • $\begingroup$ I'm sorry, I'm just a high school student and have no experience writing latex. $\endgroup$ – Vivek Yadav Nov 10 '20 at 18:36
  • $\begingroup$ Why do you think there is a printing mistake? $\endgroup$ – G. Smith Nov 10 '20 at 18:51
  • $\begingroup$ @G. Smith It does look like the radius of the disc is R/2 from the image associated with the question, but this isn't neccesarily true. (I'm currently editing the question into mathjax and will leave it as original for now) $\endgroup$ – Jonas Nov 10 '20 at 19:14
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    $\begingroup$ IF the disk is freely pivoting about P then it won't share $\omega$ and it can be treated as point mass, because only forces pass through the pivot. $\endgroup$ – JAlex Nov 10 '20 at 20:10
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I take this question as follows:

Why can I ignore the MMOI of the disk when calculating the kinetic energy of the system when $\theta = 0$?

The quick answer is because the KE of the disk is the same regardless of the swing angle, since it is freely pivoted and there is zero torque applied to it.

So you can include it, but you have to maintain $\omega$ even when the swing angle is 60° causing identical terms on left hand side as the right hand side of the energy balance.

Or you can ignore it altogether and make simplify the problem. But since the center of mass of the disk is moving, the $m(2R)^2$ has to be included.

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    $\begingroup$ Thing is, this gets apparent when you solve it via energy conservation. But how would I have the insight to know this if I calculate angular velocity by calculating torque as a function of g and R? What must I observe in the problem to realise that MOI of disc will be ignorable? $\endgroup$ – Vivek Yadav Nov 11 '20 at 4:39
  • $\begingroup$ Recognize that the disk is not rotating. $\endgroup$ – Bill Watts Nov 12 '20 at 4:15
  • $\begingroup$ @BillWatts I quote from op: "When disc is at lowest point both rod and disc has angular velocity ω" $\endgroup$ – JAlex Nov 12 '20 at 18:11
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    $\begingroup$ @JAlex my comment was not in refute of your answer, but the OP seems to not recognize that the disk is not rotating. The COM of the disk is revolving about the top pivot $O$ which of course gives it angular velocity, but its orientation stays constant so that it is not rotating about $P$. If the disk were tightly clamped so that it was forced to rotate with the rod, the way he originally wanted to work the problem would be correct. $\endgroup$ – Bill Watts Nov 12 '20 at 18:31
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    $\begingroup$ Not gonna lie, the comments were as much as if not more helpful than the answers. I'm still going to upvote this answer. $\endgroup$ – Vivek Yadav Nov 12 '20 at 20:02

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