91Ó°ÊÓ

The formula for the moment of inertia is the "sum of the product of the mass" of each particle by "the square of its distance from the axis of rotation". The moment of inertia formula is expressed as

\(I = {\int r ^2}dm\) ….. (1)

Here,\(r\)and\(dm\) is the distance and mass respectively.

Obtain the total mass of the rod in terms of \(\gamma \) and \(L\) as follows:

\(\frac{{dm}}{{dx}} = \gamma x\)

Apply the separation method and integrate on both sides of the expression as follows:

\(\begin{aligned}{}\int\limits_0^M {dm} &= \int_0^L {\gamma xdx} \\\left. m \right|_0^M &= \left. {\frac{1}{2}\gamma {x^2}} \right|_0^L\\M &= \frac{1}{2}\gamma {L^2}\end{aligned}\)

Thus, the total mass of the rod is \(M = \frac{1}{2}\gamma {L^2}\).

The moment of inertia is given by

\(I = {\int r ^2}dm\)

Using the previously computed mass, the moment of inertia must be expressed as a function of \(L\) and \(R\).

Assume that the rotational axis would be at the left end of the rod.

Calculate the moment of inertia of the rod as follows:

\(\begin{aligned}{}I &= \int_0^L {{r^2}dm} \\ & = \int_0^L {{r^2}\gamma xdx} \\ & = \gamma \int_0^L {{r^2}xdx} \,\,\,\,\left( {r = x} \right)\\ & = \gamma \int_0^L {{x^2} \cdot xdx} \end{aligned}\)

\(\begin{aligned}{}l & = \gamma \int_0^L {{x^3}dx} \\ & = \gamma \left. {\frac{{{x^4}}}{4}} \right|_0^L\\ & = \frac{{\gamma {L^4}}}{4}\end{aligned}\)

Utilizing the relation discovered in part (a), the previously acquired expression for the moment of inertia of the rod is expressed as follows:

\(\begin{aligned}{}I & = \frac{{\gamma {L^4}}}{4}\\ & = \frac{1}{2}\frac{{\gamma {L^2}}}{2}{L^2}\,\,\,\,\,\left( {M = \frac{{\gamma {L^2}}}{2}} \right)\\ = \frac{{M{L^2}}}{2}\end{aligned}\)

Thus, the moment of inertia of the rod for an axis at the left end of the rod is \(I = \frac{{M{L^2}}}{2}\).

Follow the same procedure as in part (b), however, assume that the rotational axis is at the right end of the rod.

This implies that write \(L - x\) instead of r. Shift the position of the center along the for the length L as follows:

\(\begin{aligned}{}I& = \int_0^L {{r^2}dm} \\& = \int_0^L {{{\left( {L - x} \right)}^2}\gamma xdx} \\ & = \int_0^L {\left( {{L^2} - 2Lx + {x^2}} \right)\gamma xdx} \\ & = \gamma \int_0^L {{L^2}xdx} - \gamma \int_0^L {2L{x^2}dx} + \gamma \int_0^L {{x^3}dx} \end{aligned}\)

\(\begin{aligned}{}l & = \gamma {L^2}\int_0^L {xdx} - 2\gamma L\int_0^L {{x^2}dx} + \gamma \int_0^L {{x^3}dx} \\ & = \left. {\gamma {L^2}\frac{{{x^2}}}{2}} \right|_0^L - \left. {2\gamma L\frac{{{x^3}}}{3}} \right|_0^L + \left. {\gamma \frac{{{x^4}}}{4}} \right|_0^L\end{aligned}\)

Further, simplify the above expression as follows:

\(\begin{aligned}{}I & = \frac{{\gamma {L^4}}}{2} - \frac{2}{3}\gamma {L^4} + \frac{{\gamma {L^4}}}{4}\\& = \left( {\frac{1}{2} - \frac{2}{3} + \frac{1}{4}} \right)\gamma {L^4}\\ & = \left( {\frac{6}{{12}} - \frac{8}{{12}} + \frac{3}{{12}}} \right)\gamma {L^4}\\ & = \frac{1}{{12}}\gamma {L^4}\end{aligned}\)

\(\begin{aligned}{}l & = \frac{1}{2} \times \frac{{\gamma {L^4}}}{6}\\ & = \frac{{M{L^2}}}{6}\end{aligned}\)

When comparing the obtained result from part (b) with the moment of inertia of the uniform rod, it is observed that they are different since the body with non-uniform distribution of mass was taken into account. The difference in the results from parts (b) and (c) would be because of the displacement of the mass.