# Can moment of inertia be defined as function of mass?

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.

• Related question here. – knzhou May 8 at 4:03

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.
• Sir, why are some moment of inertia like here $I_i(m)$ excluded in measuring the moment of inertia of the body? – user36790 Dec 12 '14 at 8:08