Question

Linear mass density (mass per unit length) of a rod depends on the distance from one end (say A) as ${\lambda }_x=\alpha x+\beta .Here\alpha and\beta$ are constants. The moment of inertia of this rod about an axis passing through A and perpendicular to the rod. (Length of the rod is l) is

• A. $\frac{\alpha l^4}{4}+\beta \frac{l^3}{3}$
• B. $\frac{\alpha l^3}{3}+\beta \frac{\beta l^2}{2}$
• C. $\frac{\alpha l^4}{4}+\beta l^3$
• D. $\frac{\alpha l^3}{4}+\frac{\beta l^2}{2}$

Answer 1

The correct option is A $\frac{\alpha l^4}{4}+\beta \frac{l^3}{3}$

At a distance x from A, consider an element of length dx. Linear mass density at the element is = $\alpha x+\beta$
$\therefore$ Mass of the element is dm $=(\alpha x+\beta )dx$
Moment of inertia of this element is
$dl=dm.x^2=x^2(\alpha x+\beta )dxordl=(\alpha x^3+\beta x^2)dx\therefore I_A={\int }_0^1(\alpha x^3+\beta x^2)dx=\frac{\alpha l^4}{4}+\beta \frac{l^3}{3}$