0
$\begingroup$

The original scenario describes an object of mass M rotating about a parallel axis d distance away from the center of mass. I wonder how this scenario differs from the rotation of a mass point of the exact same mass M in radius d about an axis, as illustrated in the following image: enter image description here

The two situations produce totally different moments of inertia. But I simply cannot see how the scenarios differ! Thanks in advance for any clarification and help!

$\endgroup$
2
  • 1
    $\begingroup$ Just to clarify, you cannot see how an extended object differs from a single point mass? Or are you asking why the moment of inertia depends on the object's shape? $\endgroup$
    – J. Murray
    Jul 2, 2021 at 10:23
  • $\begingroup$ Sorry about the unclarity. I can't see how an extended object differs from a single mass point. $\endgroup$
    – K-V
    Jul 2, 2021 at 10:55

3 Answers 3

1
$\begingroup$

These two cases are essentially of the same spirit. Your right hand side is a particular case of the right hane side. In the left hand side, the inertial moment at the cernter of mass $I_{CM}=0$, thus $$ I_d = I_{CM} + M d^2 = 0 + M d^2 = M d^2. $$

$\endgroup$
0
$\begingroup$

The parallel axis theorem considers two factors contributing to resistance to rotation about the new axis. You can think of all of the mass being concentrated at the center of mass. This gives the M$d^2$. But remember, as a ridig body goes about the new axis, it must also still rotate about its center of mass.

$\endgroup$
0
$\begingroup$

In the left case, one can think that the rotating objects are infinitesimally small points called the mass density. Actually, the mass density is rotating about the axis, not only the center of mass itself. One should normally consider all those small points (and their distance to the rotation axis) and integrate them over the volume of the object in order to calculate the moment of inertia. For the continuous mass distributions the formula is the following.

$I = \int r^2 dm$

However, the parallel axis theorem makes it easier to calculate the moment of inertia of a continuous mass distributions if the moment of inertia of the center of mass is known. Thus, instead of calculating the above integral over an odd positioned rotation axis for an odd shaped object, one can calculate the moment of inertia for the center of mass (which is easier), then calculate the moment of inertia for the rotation axis using parallel axis theorem.

Therefore, for your question, you can think that the left case consists of infitintely many point particles, but the right one has only one point particle.

$\endgroup$

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.