Four point masses,each of mass m,are fixed at | vedclass

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(C) Let the corners of the square be at $(0,0)$,$(\ell, 0)$,$(\ell, \ell)$,and $(0, \ell)$. The axis passes through $(0,0)$ and is parallel to the dia

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let the corners of the square be at $(0,0)$,$(\ell, 0)$,$(\ell, \ell)$,and $(0, \ell)$. The axis passes through $(0,0)$ and is parallel to the diagonal connecting $(0, \ell)$ and $(\ell, 0)$. The equation of this diagonal is $x + y = \ell$. The perpendicular distance $r$ of a point $(x, y)$ from the line $x + y - \ell = 0$ is given by $r = \frac{|x + y - \ell|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - \ell|}{\sqrt{2}}$.For the four masses:$1$. At $(0,0)$: $r_1 = \frac{|0 + 0 - \ell|}{\sqrt{2}} = \frac{\ell}{\sqrt{2}}$.$2$. At $(\ell, 0)$: $r_2 = \frac{|\ell + 0 - \ell|}{\sqrt{2}} = 0$.$3$. At $(0, \ell)$: $r_3 = \frac{|0 + \ell - \ell|}{\sqrt{2}} = 0$.$4$. At $(\ell, \ell)$: $r_4 = \frac{|\ell + \ell - \ell|}{\sqrt{2}} = \frac{\ell}{\sqrt{2}}$.The moment of inertia $I$ about the axis is $I = \sum mr_i^2 = m \left( \left(\frac{\ell}{\sqrt{2}}\right)^2 + 0^2 + 0^2 + \left(\frac{\ell}{\sqrt{2}}\right)^2 \right) = m \left( \frac{\ell^2}{2} + \frac{\ell^2}{2} \right) = m\ell^2$.Angular momentum $L = I\omega = m\ell^2\omega$. Therefore,the missing value is $1$.

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