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For continuous bodies, moment of inertia is found as $$ \int dI = \int_{m_i}^{m_f} r^2(m) .dm$$ . Now, $$\int dx = \int_{u_i}^{u_f} f(u).du \implies X_f(u) = \int_{u_i} ^{u_f} f(u) .du + X_i(u)$$ , where $X_f$ is the final value and $X_i$ is the initial value and both are functions of $u$ .

So, $$\int dI = \int_{m_i}^{m_f} r^2(m) .dm \implies I_f(m) - I_i(m) = \int_{m_i}^{m_f} r^2 .dm \implies I_f(m) = \int_{m_i}^{m_f} r^2(m) .dm + I_i(m)$$ . Now , the $\int_{m_i}^{m_f} r^2(m) .dm$ is the moment of inertia of the bodies. If so, what are these $I_f(m)$ & $I_i(m)$? What do they imply?? They must be function of m ie. mass & $I_f(m)$ is the value of the function at mass $m_f$ while $I_i(m)$ is the value of the function at mass $m_i$. If their difference ie. $$I_f(m) - I_i(m)$$ is the moment of inertia of the body. But what are they individually??? Please help.

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  • $\begingroup$ Related question here. $\endgroup$ – knzhou May 8 at 4:03
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Using my understanding of definite integrals, $I_i$ would be the inertia of the mass that you are not including in your calculation, while $I_f$ is the inertia of all the masses, including your mass of interest. $I_f$ - $I_i$ therefore is the inertia of the mass you are interested in.

It appears as if $I_f(m)$ and $I_i(m)$ are not a functions of mass per se (the numerical value of the mass in kg), but they are a function of the actual masses, which includes all the other information about masses, including their radius.

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  • $\begingroup$ Sir, why are some moment of inertia like here $I_i(m)$ excluded in measuring the moment of inertia of the body? $\endgroup$ – user36790 Dec 12 '14 at 8:08
  • $\begingroup$ When you are taking a definite integral, you choose the area that you are integrating. For instance integrating 1/x between 1 and 2, we don't care about the entire integral of 1/x, which is impossible at certain points. I_i is like all he other mass in the system or in the universe, we simply don't care about it. That is my best guess as to the meaning of what you are reading. $\endgroup$ – Richard Dec 12 '14 at 8:10

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