Question

A non-uniform rod $AB$ has a mass $M$ and length $2l$.The mass per unit length of the rod is $mx$ at a point of the rod distant $x$ from $A$. Find the moment of inertia of this rod about an axis perpendicular to the rod through $A$.

• A. $2M{l}^2$
• B. $4M{l}^2$
• C. $6M{l}^2$
• D. $8M{l}^2$

Answer 1

The correct option is A $2M{l}^2$


Calculating the moment of inertia of the rod, we know that $M.I$ for the small element $dx$ at the distance $x$ from the axis is $dI=mx{x}^{2}dx$,

Now for calculating the $M.I$ at $A$ we integrate it from $0\text{to}2L$

$I=\int dI=\underset{0}{\overset{2l}{\int }}mx{x}^2=m\underset{0}{\overset{2l}{\int }}{x}^{3}dx$
$=\frac{1}{4}m(2l{)}^4=4m{l}^4$

Now for calculating the total mass of the rod $\left(M\right)$
$M=\underset{0}{\overset{2l}{\int }}mxdx=\frac{1}{2}m(2l{)}^2=2m{l}^2$
$\Rightarrow m=\frac{M}{2l^2}$

$\therefore I=4.\frac{M}{2l^2}.l^4=2M{l}^2$

Hence option A is the correct answer