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(C) Let the coordinates of the corners be $A(0,0)$,$B(\ell, 0)$,$C(\ell, \ell)$,and $D(0, \ell)$.The diagonal $BD$ connects $(\ell, 0)$ and $(0, \ell)

Vedclass · Accepted Answer

$3\,m\ell^2$

Vedclass · Answer

(C) Let the coordinates of the corners be $A(0,0)$,$B(\ell, 0)$,$C(\ell, \ell)$,and $D(0, \ell)$.The diagonal $BD$ connects $(\ell, 0)$ and $(0, \ell)$. The slope of $BD$ is $m_{BD} = \frac{\ell - 0}{0 - \ell} = -1$.The axis passes through $A(0,0)$ and is parallel to $BD$,so its equation is $y = -x$,or $x + y = 0$.The perpendicular distance $r$ of a point $(x, y)$ from the line $Ax + By + C = 0$ is $r = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$.For our axis $x + y = 0$,the distance $r = \frac{|x + y|}{\sqrt{1^2 + 1^2}} = \frac{|x + y|}{\sqrt{2}}$.Calculating distances for each mass:$r_A = \frac{|0 + 0|}{\sqrt{2}} = 0$$r_B = \frac{|\ell + 0|}{\sqrt{2}} = \frac{\ell}{\sqrt{2}}$$r_D = \frac{|0 + \ell|}{\sqrt{2}} = \frac{\ell}{\sqrt{2}}$$r_C = \frac{|\ell + \ell|}{\sqrt{2}} = \frac{2\ell}{\sqrt{2}} = \sqrt{2}\ell$The moment of inertia $I = \sum m_i r_i^2 = m(r_A^2 + r_B^2 + r_C^2 + r_D^2)$.$I = m(0^2 + (\frac{\ell}{\sqrt{2}})^2 + (\sqrt{2}\ell)^2 + (\frac{\ell}{\sqrt{2}})^2) = m(0 + \frac{\ell^2}{2} + 2\ell^2 + \frac{\ell^2}{2}) = m(3\ell^2) = 3\,m\ell^2$.

Four point masses each of mass $m$ are placed at the corners of a square $ABCD$ of side length $\ell$. What is the moment of inertia about an axis passing through $A$ and parallel to $BD$?

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