Question

A rod of uniform cross-section of mass $M$ and length $L$ is hinged about an end to swing freely in a vertical plane. However, its density is nonl uniform and varies linearly from hinged end to the free end doubling its value. The moment of inertia of the rod, about the rotation axis passing through the hinge point is

• A. $\frac{2M{L}^2}{9}$
• B. $\frac{3M{L}^2}{16}$
• C. $\frac{7M{L}^2}{18}$
• D. None of these

Answer 1

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

$S=$ area of cross- section
${\rho }_x={\rho }_0+\left(\frac{2{\rho }_0-{\rho }_0}{L}\right)x=p_0+\frac{{\rho }_0}{L}x$
Now, $M={\int }_0^{L}dM={\int }_0^L\left(Sdx\right)({\rho }_0+\frac{{\rho }_0}{L}x)$
$={\rho }_{0}S\left(\frac{3L}{2}\right)$
$\therefore {\rho }_{0}S=\frac{2M}{3L}.....\left(i\right)$
Now, $I={\int }_0^{L}dI={\int }_0^L\left(dM\right)X^2$
$={\int }_0^L\left(Sdx\right)({\rho }_0+\frac{{\rho }_0}{L}x)\left(X^2\right)$
After substituting value of ${\rho }_{0}S$ from Eq. (i), then we find
$I=\frac{7}{18}M{L}^2$