Question

A square plate of mass $M$ and edge $L$ is shown in figure. The moment of inertia of the plate about an axis in the plane of plate, passing through one of its vertex, making an angle ${15}^{\circ }$ from horizontal is

• A. $\frac{M{L}^2}{12}$
• B. $\frac{11M{L}^2}{24}$
• C. $\frac{7M{L}^2}{12}$
• D. None

Answer 1

The correct option is B $\frac{11M{L}^2}{24}$



Let $x-$ axis be is parallel to given axis and passing through centre of square $O$, Moment of inertia about this axis be $I_x$.
$AO$ is the half of its diagonal.
$AO=\frac{L}{\sqrt{2}}$
$AE$ is the perpendicular distance of $x-$ axis from given axis.
So, $△AOE$ is a right angled triangle.
$AE=r=AO\cos {30}^{\circ }=\frac{L\sqrt{3}}{2\sqrt{2}}$

$x-$ axis and $y-$ axis are perpendicular but are in same plane, $z-$ axis is perpendicular to the plane of square.
So, according to perpendicular axis theorem,
$I_x+I_y=I_z$
For a square plate $I_x=I_y=\frac{I_z}{2}$
we know that, $I_z=\frac{M{L}^2}{6}$
$\Rightarrow I_x=I_y=\frac{I_z}{2}=\frac{M{L}^2}{12}$
Applying parallel axis theorem
$I_1=I_x+M(AE{)}^2$
$I_1=\frac{M{L}^2}{12}+M{\left(\frac{L\sqrt{3}}{2\sqrt{2}}\right)}^2$
$I_1=\frac{11M{L}^2}{24}$