Question

A rod of mass $M$ and length $L$ is hinged about a end to swing freely in a vertical plane. Its density increases linearly from hinge to the free end, doubling is value at free end. The moment of inertia of the rod about the hinge is

• A. $\frac{1}{3}M{L}^2$
• B. $\frac{5}{24}M{L}^2$
• C. $\frac{7}{18}M{L}^2$
• D. $\frac{9}{35}M{L}^2$

Answer 1

The correct option is C $\frac{7}{18}M{L}^2$

Let the density at the hinged bed and at extreme end is $2d$ density at the distance $x$ from hinge $=d+\frac{dx}{L}$ take $dx$ width element at distance $x$ from hindge. The mass of element is $dx$.

$\int dx={\int }_0^L(d+\frac{dx}{L})dx$

$M=dL+\frac{{dL}^2}{2L}=\frac{3dL}{2}$

$d=\frac{2M}{3L}$

for moment of inertia the $MI$ of the $dx$ with element

$dI=x^{2}dx$

$\int dI={\int }_0^L{x}^2(d+\frac{dx}{L})dx$

$I=\frac{d}{L}[\frac{L^4}{3}+\frac{L^4}{4}]=\frac{{7dL}^3}{12}=\frac{{7L}^3}{12}\left[\frac{2M}{3L}\right]=\frac{{7ML}^2}{18}$