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Q.
A small disc of radius $2\, cm$ is cut from a disc of radius $6$ cin. If the distance between their centres is $3.2 \,cm$, what is the shift in the centre of mass of the disc?
JIPMERJIPMER 2011System of Particles and Rotational Motion
Solution:
The situation can be shown as :
Let radius of complete disc is $a$ and that of small disc is $b$.
Also let centre of mass now shifts to $O _{2}$ at a distance $x_{2}$ from original centre.
The position of new centre of mass is given by
$X_{ CM }=\frac{-\sigma \cdot \pi b^{2} \cdot x_{1}}{\sigma \cdot \pi a^{2}-\sigma \cdot \pi b^{2}}$
Here, $a=6 \,cm , b=2\, cm , x_{1}=3.2 \,cm$
Hence, $X_{ CM }=\frac{-\sigma \times \pi(2)^{2} \times 3.2}{\sigma \times \pi \times(6)^{2}-\sigma \times \pi \times(2)^{2}}$
$=\frac{12.8 \pi}{32 \pi}$
$=-0.4 \,cm$
