A small disc of radius 2, cm is cut from a di | vedclass

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(A) Let the radius of the original disc be $R = 6\, cm$ and the radius of the removed small disc be $r = 2\, cm$. The distance between their centres i

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$0.4$

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(A) Let the radius of the original disc be $R = 6\, cm$ and the radius of the removed small disc be $r = 2\, cm$. The distance between their centres is $d = 3.2\, cm$.Let the surface mass density be $\sigma$. The mass of the original disc is $M = \sigma \pi R^2$ and the mass of the removed disc is $m = \sigma \pi r^2$.The shift in the centre of mass $(x)$ is given by the formula:$x = \frac{m \cdot d}{M - m}$Substituting the values:$x = \frac{(\sigma \pi r^2) \cdot d}{\sigma \pi R^2 - \sigma \pi r^2} = \frac{r^2 \cdot d}{R^2 - r^2}$$x = \frac{2^2 \cdot 3.2}{6^2 - 2^2} = \frac{4 \cdot 3.2}{36 - 4} = \frac{12.8}{32} = 0.4\, cm$.Thus,the shift in the centre of mass is $0.4\, cm$.

$A$ small disc of radius $2\, cm$ is cut from a disc of radius $6\, cm$. If the distance between their centres is $3.2\, cm$,what is the shift in the centre of mass of the disc in $cm$?

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