Question

The radius of gyration of a plane lamina of mass $M$, length $L$ and breadth $B$ about an axis passing through its center of gravity and perpendicular to its plane will be

• A. $\sqrt{\frac{(L^2+B^2)}{12}}$
• B. $\sqrt{\frac{(L^2+B^2)}{8}}$
• C. $\sqrt{\frac{(L^2+B^2)}{2}}$
• D. $\sqrt{\frac{(L^3+B^3)}{12}}$

Answer 1

Answer: (A)

$\sqrt{\frac{(L^2+B^2)}{12}}$

The moment of inertia about a rectangular lamina with area density $\rho$ is $\int \int \rho r^{2}dA$. Here r is the distance from each point to the axis of rotation and $dA$ is the differential of area.
Take the center of the rectangle to be at $(0,0)$ so the vertices are $(L/2,B/2)$, etc. Then the distance from (x,y) to the axis of rotation (passing through $(0,0)$ perpendicular to the plate) is $x^2+y^2$ and the moment of inertia is
${\int }_{-L/2}^{L/2}{\int }_{B/2}^{B/2}\rho (x^2+y^2)dxdy$
or
${\int }_{-L/2}^{L/2}\rho (B{x}^2+\frac{1}{12}{B}^3)$
or
$\frac{1}{12}\rho (L^{3}B+L{B}^3)$
Now as $M=\rho LB$, thus we get moment of inertia as $\frac{M}{12}(L^2+B^2)$

so radius of gyration is $K=\sqrt{\frac{I}{M}}$

$⟹K=\sqrt{\frac{1}{12}(L^2+B^2)}$