Answer
(a) $M = \frac{1}{2}~\gamma ~L^2$
(b) $I = \frac{1}{2}ML^2$
(c) $I = \frac{1}{6}ML^2$
Work Step by Step
(a) $\frac{dm}{dx} = \gamma ~x$
$M = \int_{0}^{L}\gamma ~x~dx$
$M = \frac{1}{2}~\gamma ~x^2\vert_{0}^{L}$
$M = \frac{1}{2}~\gamma ~L^2$
(b) $I = \int ~r^2~dm$
$I = \int_{0}^{L}r^2~(\gamma~r~dr)$
$I = \frac{1}{4}\gamma~r^4\vert_{0}^{L}$
$I = \frac{1}{4}\gamma~L^4$
$I = \frac{1}{4}(\frac{2M}{L^2})~L^4$
$I = \frac{1}{2}ML^2$
The moment of inertia for a uniform rod is $I = \frac{1}{3}ML^2$. The moment of inertia for the rod with a varied mass is greater because more of the mass is located farther from the axis of rotation.
(c) $I = \int ~r^2~dm$
$I = \int_{0}^{L}r^2~(\gamma)~(L-r)~dr$
$I = (\frac{1}{3}~\gamma ~L~r^3-\frac{1}{4}~\gamma ~r^4)\vert_{0}^{L}$
$I = (\frac{1}{3}~\gamma ~L^4-\frac{1}{4}~\gamma ~L^4)$
$I = \frac{1}{12}~\gamma ~L^4$
$I = \frac{1}{12}~(\frac{2M}{L^2})~L^4$
$I = \frac{1}{6}ML^2$
The moment of inertia for part (c) is less than the moment of inertia for part (b) and a uniform rod. Because in part (c), more of the mass is closer to the axis of rotation and there is less mass farther from the axis of rotation.
