Centre of mass(com)

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A square shaped hole of side $l = \frac{a}{2}$ is carved out at a distance $d = \frac{a}{2}$ from the centre ‘O’ of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is $- \frac{a}{X}$ , value of X (to the nearest integer) is ______

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Solution

CM of remaining portion shifts towards -x axis
$\begin{array}{l} M_{r e m a i n i n g} x = M_{r e m o v e d} d \\ \Rightarrow (M_{t o t a l} - M_{r e m o v e d}) x = M_{r e m o v e d} d \\ \Rightarrow x = \frac{M_{r e m o v e d} d}{M_{t o t a l} - M_{r e m o v e d}} = \frac{σ l^{2} d}{σ π a^{2} - σ l^{2}} \\ [M_{r e m o v e d} = σ l^{2}, M_{t o t a l} = σ (π a^{2})] \\ \Rightarrow x = \frac{l^{2} d}{π a^{2} - l^{2}} = \frac{a}{8 π - 2} = \frac{a}{23.12} \end{array}$

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Centre of mass(com)

Remember concepts with our Masterclasses.

A square shaped hole of side l=a2 is carved out at a distance d=a2 from the centre ‘O’ of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is −aX, value of X (to the nearest integer) is ______

CM of remaining portion shifts towards -x axisMremainingx=Mremovedd​⇒Mtotal−Mremovedx=Mremovedd​⇒x=Mremoved dMtotal−Mremoved=σl2dσπa2−σl2 Mremoved=σl2,Mtotal=σπa2​⇒x=l2dπa2−l2=a8π−2=a23.12

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A square shaped hole of side $l = \frac{a}{2}$ is carved out at a distance $d = \frac{a}{2}$ from the centre ‘O’ of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is $- \frac{a}{X}$ , value of X (to the nearest integer) is ______