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A square plate $ABCD$ of mass $m$ and side $a$ is suspended from point $A$ in vertical plane. A disc of the same mass and $\sqrt2a$ diameter is attached at point $C$ as shown in the figure. The disc can freely rotate about point $C$.

Now my doubt is when they mention that a body is free to rotate about an axis, why do we consider the moment of inertia of that body about that axis as $0$? I thought probably this would make the body non-rigid about the axis, hence it's moment of inertia is $0$.

But this seems counter intuitive to me that a free to rotate about actually "doesn't rotate" and vice versa.

Is this true? How do I convince myself this?

In this particular example, the moment of inertia of the system about a horizontal axis passing through $A$ is

$(ma^2/6 + m(a/\sqrt{2})^2) + (0 + m(\sqrt{2}a)^2) = \frac{8ma^2}{3}$,

by parallel axis theorem. (My doubt is regarding the $0$ in the second term, which is the moment of inertia of the disc).

enter image description here

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  • $\begingroup$ Not clear what you are asking. Please provide a better description of your difficulty. The disk can rotate about C while C rotates about A. $\endgroup$ – sammy gerbil Mar 14 '20 at 18:14
  • $\begingroup$ If the disk can rotate freely about point C, then from the point of view of an axle perpendicular to the sketch at point A, the disk appears to be a point mass at C. $\endgroup$ – R.W. Bird Mar 14 '20 at 19:11
  • $\begingroup$ @sammy gerbil Could you please explain which part is unclear? As mentioned, I have trouble understanding why the moment of inertia of the disc, about an axis passing through it's centre of mass and parallel to the horizontal axis passing through A, through which the system is suspended, is equal to 0. $\endgroup$ – user600016 Mar 15 '20 at 1:42
  • $\begingroup$ @R.W. Bird sorry but I can't quite understand what you mentioned. Why exactly does the disk appear to be a point mass at C? All points on the disc aren't equidistant from the point A/axis passing through A isn't it? $\endgroup$ – user600016 Mar 15 '20 at 1:46
  • $\begingroup$ The force from the frame acts at the center of the disk but exerts no torque. The center of mass of the disk moves in response to that force, but it does not rotate. $\endgroup$ – R.W. Bird Mar 15 '20 at 16:44
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There seems to be some information in the question which you have not provided. We are not told whether the disk is spinning when the square frame is set swinging. If the disk is not spinning about its centre C initially, and if there are no torques which start it spinning when the square frame swings (because the axle is frictionless) then the disk doesn't start spinning later.

The disk maintains the same orientation in space throughout the swinging motion. Its rotational KE starts at zero and remains zero. For this reason its moment of inertia does not have any effect on the motion and can be ignored. We can regard the disk as a point mass which only has translational KE as C swings about A.

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