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As we know, rotational inertia is the mass-equivalent in rotation.

For a discrete body, it is measured as $$I = \sum m_i{r_i}^2 $$ . But when a continuous body comes, $$I = \int r^2 .dm$$ which implies $$I = \phi + C$$ where $\phi$ is the function that gives the change and C is the initial point.

This is what my book writes. Ok, that's the pure definition of indefinite integral. But at what I am confused is that what is $\phi$ dependent on and what is $C$ here. $C$ , in general, gives us the initial condition so as to give us the original function. But what is $C$ representing here?? What is $\phi$ dependent on?? If we use proper limits , then it becomes $$ I = \int_{a}^{b} r^2.dm \implies I = [\phi(b) + C] - [\phi(a) + C]$$ . It is different that $C$ is cancelled here. But it does exist. So what does this $C$ represent in case of calculation of moment of inertia?? Please explain.

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    $\begingroup$ That is a very terse book you have there. $\endgroup$ – BMS Nov 23 '14 at 4:15
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Unless I have the rest of the book, it is difficult to understand what he meant. But many books have equations that could make not much sense or whose explanations are very unclear (no author is perfect). The only thing you really need to know is what you already seem to know. In an actual calculation you use a definite integral, and the constant $C$ doesn't matter, it always goes away. My guess is that the author meant that $C = \phi(a)$, because you can write $$I = \int_{a}^{x} r^2.dm=\phi(x) - \phi(a)$$

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  • $\begingroup$ +1 . Thanks for the response. Yes, $C$ does represent the initial value of the function ie. $$f(x_0) = C$$ where $x_0$ is the initial independent variable. You said $x_0$ is $a$ here. That's where all my confusion started. What does $a$ or $x_0$ mean in this case?? Can you please explain,sir?? $\endgroup$ – user36790 Nov 23 '14 at 11:48
  • $\begingroup$ It is just the lower limit of integration. It depends on the particular example. In the case of a ring, if you origin of coordinates is at the center, then $a$ will be the internal radius of the ring, and b the external. Instead of be you can leave the expresion using $x$, but this will give a function, that is, the $I$ of a ring with internal radius $a$ and arbitrary external radius $x$. $\endgroup$ – Wolphram jonny Nov 23 '14 at 11:54
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I think the book authors might have the Huygens-Steiner theorem in mind when saying that.

The theorem says that if the body has moment of inertia $I_{cm}$ with respect to axis crossing its center of mass, then moving the reference axis to another parallel axis gives a new moment of inertia $I'$, related to original one by

$$I'=I_{cm}+md^2,$$

where $d$ is distance between axes. So I suppose the $C$ in your quotation is $md^2$.

I wouldn't think that it really is the constant of integration, because normally moment of inertia is not an indefinite integral, and usually not even a definite one: it's instead a multiple integral.

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Where are you getting these books? Here it makes absolutely no sense to mention indefinite integrals and I have never seen such confusing statements.

You need definite integral, usually in multiple dimensions (ideally 3D, as we live in 3D space, but you can simplify in some cases). The integration goes over the entire volume of the object, you have $dm=\rho\,dV$ and if the density is constant, it's just a regular integral over volume. No additional constants, nothing. Indefinite integral in multiple dimensions is a weird concept anyway.

However, result depends on selection of your rotational axis (usually you make it go through coordinate origin so $r$ can be the $\sqrt{x^2+y^2}$ if you are looking for the moment of inertia around axis in the direction $z$). If you calculate the momentum of inertia around the center of mass, then you get the inertia around another axis by using Steiner's law:

$$J=J^* + mr^{*2}$$ if $r^\ast$ is the distance between the axis and the object's center and $J^*$ is the moment of inertia around the mass center. However, this relation has nothing to do with indefinite integrals. It follows immediately from $$\int(r+r^*)^2dm=\underbrace{\int r^2 dm}_{J^*}+2r^* \underbrace{\int r dm}_{\text{mass center}=0}+r^{*2}\underbrace{\int dm}_m$$ You see that you only get Steiner law if your $r$ coordinate frame is put in the object's center (otherwise the middle term would not be 0).

So... find another book.

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  • $\begingroup$ +1 . Really indebted for the response. Regarding $C$ , I think the book is telling about the initial condition ie. moment of inertia at distance $x_0$ from the axis. It will be generalized if $x_0 = 0$ , right?? Really a horrendous book! Definitely will keep it away!!! $\endgroup$ – user36790 Nov 23 '14 at 17:03
  • $\begingroup$ Moreover, in $\int F(x).dx$ , $F$ is a function of $x$ . But in $\int r^2 .dm$ , how can I say $r^2$ is a function of $m$ ?? Is it possible to say this??? If so, why?? $\endgroup$ – user36790 Nov 23 '14 at 17:52
  • $\begingroup$ r is a function of spatial coordinates and $dm=\rho\, dV=\rho\,dx\,dy\,dz$. Notice how $dm$ is actually abuse of notation because its actually differential in three dimensions (sometimes it's written $d^3 m$ to make you aware that you will need a triple integral, but even this is just a notational shortcut). Of course you may want to use other coordinates instead of cartesian to make it easier to write the integration limits (spherical, cylindrical or whatever fits best). $\endgroup$ – orion Nov 23 '14 at 18:55

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