1
$\begingroup$

I saw the derivation of moment of inertia for a rigid body through an axis of rotation. The derivation did integration of $r^2dm$, which I understood. But my doubt is why can I do the derivation in this way?

First we will find the centre of mass of the object given. And then we can represent the whole body by the centre of mass. What I am exactly saying is say I have a rod of length $L$ and its axis of rotation is at its end. Then I would find out the COM of the rod of mass $M$ which is at a distance $L/2$ from one of its end and then replace the whole rod with just a single dot ( dot represents the centre of mass) placed $L/2$ from its end. As axis of rotation is at the end, I can say the dot is placed $L/2$ away from its axis of rotation. Now we know for a particle of mass $m$ moving in a circle of radius $R$, the moment of inertia is given by $mR^2$, similarly in the above case I can say the dot is moving around the axis of rotation with radius = $L/2$. So the moment of inertia of that dot revolving around the axis of rotation is $M(L/2)^2$ .

But my teacher said this is wrong because the actual formula says $M(L^2)/3$ . So why can’t I do in the method which I did. We know that COM( centre of mass) represents the whole body. So without doing the whole body, I substituted the whole body with its centre of mass which is actually a dot or a point mass of mass same as the mass of the rod and then without saying the rod is rotating around the axis, I substituted the word rod with its centre of mass and I said that the centre of mass which is a dot/point mass is revolving around the axis of rotation. And then I used the formula for the moment of inertia of a revolving particle around the axis.

$\endgroup$
0
$\begingroup$

This is because for rigid bodies moment of inertia means how the mass is distributed for a rigid bodies and that's the meaning of moment of Inertia is. For instance just take MI of a hollow and solid sphere or cylinder they both differ in their properties that's because of the distribution of their mass. COM is the assumed point where all mass of the object is assumed to be concentrated but in actuality you have to also consider shape of that object.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.