Question

The moment of inertia of a square lamina about the perpendicular axis through its centre of mass is $20kg-m^2$. Then, its moment of inertia about an axis touching its side and in the plane of the lamina will be:

• A. $20kg-m^2$
• B. $30kg-m^2$
• C. $40kg-m^2$
• D. $25kg-m^2$

Answer 1

The correct option is C $40kg-m^2$

Given,

Inertia at center is $20kg-m^2$

Let,

Mass of square plate, $=m$

Side of square, $=a$

Inertia about perpendicular axis at center of square, $I_z=\frac{m{a}^2}{6}=20kg-m^2$

Apply perpendicular axis theorem

$I_z=I_x+I_y=2I_x$(square has same side)

$I_x=\frac{I_z}{2}=\frac{m{a}^2}{12}$

Edge of square is at distance, $\frac{a}{2}$ from center.

Apply parallel axis theorem $I_{edge}=I_x+m{\left(\frac{a}{2}\right)}^2$

$I_{edge}=\frac{m{a}^2}{12}+m{\left(\frac{a}{2}\right)}^2$

$I_{edge}=\frac{m{a}^2}{3}=2\times \frac{m{a}^2}{6}=2I_Z$

$I_{edge}=2\times 20=40kg-m^2$

$I_{edge}=40kg-m^2$