Question

Find the moment of inertia of the triangular lamina of mass M about the axis of rotation AA' shown in the figure:

• A. $\frac{M{l}^2}{3}$
• B. $\frac{M{l}^2}{6}$
• C. $\frac{2}{3}M{l}^2$
• D. $\frac{1}{3}M{l}^2$

Answer 1

The correct option is B $\frac{M{l}^2}{6}$

mass per unit area = $M\times \frac{al}{2}$ where a= base which is AA' =l(given)

M.O.I $I=\frac{1}{2}\times \frac{M}{2}al\times \frac{1}{2}axdx$

where $x$ is te small strip along 'l' from the base a(AA')

so $M.O.I=\frac{1}{2}\times \frac{M}{l}\int x^{2}dx$ =$\frac{M}{2l}\left[\frac{x^3}{3}\right]=\frac{M}{2l}\left[\frac{l^3}{3}\right]=\frac{M{l}^2}{6}$