Here is a situation:

enter image description here

I want to find out the angular velocity vector of the Center Of Mass (COM) of the system.

My approach is as follows :

Method 1:

I calculated the linear velocity of center of mass and used the property that

$\vec V_{point}$ = $\vec V_{com}$ $\pm$ $\vec r$ $\omega_{com}$,

$\vec r$ being the position vector(positive) of the required point with respect to COM.

Using this property about the COM of the system(located on the rod), I get the magnitude of $\omega_{com}$

For the direction,wouldn't it would be perpendicular to both the position vector of the COM wrt O as well as the velocity of the discs(as otherwise velocities along the rod for both the discs would be different,as they have the same inclination)?

Method 2:

I have studied about something called Instantaneous Center(Axis?) Of Rotation, and used the property that $\omega$ about this point would be constant and equal to velocity of the required point divided by the position vector.

Here, I found that the required point is the origin O itself.

I proved that $\omega_{com}$ = $\omega_{O}$ [Using the Theorem Of Parallel Axes as well as Energy considerations]

[Is this the only point other than the COM(for non COM rotation) where $\omega_{point}$ = $\omega_{com}$?]

And so this situation is equivalent to finding angular velocity about O,which is pretty easy to calculate,with the fact that it is also the Instantaneous Center.

So are both my methods right?If so,is there any other easier route to calculate $\omega_{com}$?


The diagram that is drawn should give you the clue as to how to approach this problem.

As the no slipping condition is satisfied for both discs you immediately know the linear speed of the centres of both discs.

$v_{\rm a} = a \omega$ and $v_{\rm 2a} = 2a \omega$

The direction of the velocities being in the negative y-direction $( - \hat y)$

[As an aside you have the x and y axes in your diagram are the wrong way round for a right-handed coordinate system. I mention this because a lot of rotational dynamics is counter-intuitive and such a slip may cause problems in the future]

The linear velocities being as found indicate that the system is rotating about $O$ with the direction of the angular velocity of the centres of the discs being in the z direction $(\hat z)$.
Note it is not the direction of $\vec r \times \vec v_{\rm a}$ which is inclined to $\hat z$ even if the x and y axes have been switched.
This is your Method 2.

All parts of the rod rotate about $O$ with the same angular velocity so you can find the angular velocity of the centre of mass by finding the angular velocity of either of the centres of the two disc.
You know the linear speed so all you need is a distance in terms of $a$ and $l$.

  • $\begingroup$ How are the directions of the velocities of the centre of the discs in the negative y direction? Shouldn't they have an inclination towards -y(Perpendicular to the length of the rod)? $\endgroup$
    – user73157
    Oct 14 '16 at 11:10
  • $\begingroup$ At the instant shown in the diagram with the discs touching the x-axis the centres of the discs have a velocity in a direction which is perpendicular to the x-axis and as there is no motion in the direction of the z-axis (constant height above the xy plane) it must be in the $-\hat y$ direction. $\endgroup$
    – Farcher
    Oct 14 '16 at 11:31
  • $\begingroup$ It's true that there is no motion about the z axis, but can't this situation be thought of as two discs performing uniform circular motion about the origin,with their velocities tangential to the path traversed? I'm saying this, as if it is in the -y direction,there will be a component of velocity along the rod, but since one component is twice the other(they have the same angle-the while system is inclined equally),rigid body properties of the rod indicate that it isn't possible. $\endgroup$
    – user73157
    Oct 14 '16 at 12:49
  • $\begingroup$ Think about the motion of the centres of the discs. They execute circles in a plane which is parallel to the xy plane. The one travelling nearer $O$ travels half the distance of the one further from $O$ but is travelling half as fast, so both centres take the same time for one rotation. Note that the circles are not centred at $O$ but points on the z-axis above $O$. So the centre of mass of the system of two disc and the rod must also travel around a similar circle with centre on the z-axis. $\endgroup$
    – Farcher
    Oct 14 '16 at 12:58
  • $\begingroup$ The whole system was initially at an angle $\theta$ with the x axis. If $\theta$ = 0 then I understand that the velocity will be in negative y direction. But are you saying that the velocity direction is independent of $\theta$? $\endgroup$
    – user73157
    Oct 14 '16 at 13:36

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