Question

The moment of inertia of a thin uniform rod of mass $M$ and length $L$, about an axis passing through a point, midway between the centre and one end, perpendicular to its length is ___________.

• A. $\frac{48}{7}M{L}^2$
• B. $\frac{7}{48}M{L}^2$
• C. $\frac{1}{48}M{L}^2$
• D. $\frac{1}{16}M{L}^2$

Answer 1

The correct option is B $\frac{7}{48}M{L}^2$

The moment of inertia of the given rod is

$\Sigma M_i{r}_i^2$

$=\int \left(dm\right)x^2$

$={\int }_{-L/4}^{3L/4}\left(\lambda dx\right)x^2$

$={\int }_{-L/4}^{3L/4}\lambda x^{2}dx$

$=\lambda \frac{7L^3}{48}$

$=\frac{M}{L}.\frac{7L^3}{48}$

$=\frac{7}{48}M{L}^2$