Question

Find the moment of inertia of a plate cut in shape of a right angled triangle of mass M, side $AC=BC=a$ about an axis perpendicular to the plane of the plate and passing through the midpoint of side AB

• A. $\frac{{Ma}^2}{18}$
• B. $\frac{{2Ma}^2}{3}$
• C. $\frac{{Ma}^2}{3}$
• D. $\frac{{Ma}^2}{6}$

Answer 1

The correct option is A $\frac{{Ma}^2}{18}$

Given: a plate cut in shape of a right angled triangle of mass M, side AC=BC=a about an axis perpendicular to the plane of the plate and passing through the midpoint of side AB

To find the moment of inertia of the plate

Solution:

The center of mass of the triangle lies at the centroid.

The length of CO is $\sqrt{a^2-\frac{2a^2}{4}}=\frac{a}{\sqrt{2}}$

The centroid of the triangle lies on this line at a distance $\frac{\sqrt{2}}{3}=\frac{a}{3\sqrt{3}}$ from the point O.

The moment of inertia of the triangle about the pointO is

$I=M{r}^2$, where r is the distance of the center of mass from O.

The moment of inertia is

$I=M{\left(\frac{a}{3\sqrt{2}}\right)}^2=\frac{M{a}^2}{18}$