Question

The moment of inertia of a triangular lamina of mass M and height h about its base is

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

Answer 1

Answer: (B)

$\frac{1}{3}M{H}^2$

Mass per unit area $=M1/2aH$

base $=a$

Moment of inertia$=I=1/2\times M/1/2aH\times 1/ax\times xdx$

$x$is the small strip along $h$from base.

So, moment of inertia at base $=1/2\times M/hlimit{x}^{2}dx=M/2h\left(x^3/3\right)=M/2h.x^3/3=Mh2/6=1/3M{h}^2$