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From a uniform disc of radius `R`, a circular section of radius `R//2` is cut out. The centre of the hole is at `R//2` from the centre of the original disc. Locate the centre of mass of the resulting flat body. |
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Answer» Let from a bigger uniform disc of radius R with centre O a smaller circular hole of radius `R//2` with its centre at `O_(1)` (where R `O O_(1)=R//2)` is cut out. Let centre of gravity or the centre of mass of remaining flat body be at `O_(2)`, where ` O O_(2)=x`. If `sigma` be mass per unit area, then mass of whole disc `M_(1)=T T R^(2) sigma` and mass of cut out part `M_(2)=pi((R )/(2))^(2) sigma=(1)/(4)piR^(2) sigma=(M_(1))/(4)` `x=(M_(1)xx(0)-M_(2)(O O_(1)))/(M_(1)-M_(2))=(0-(M_(1))/(4)xx(R )/(2))/(M_(1)-(M_(1))/(4))=-(R )/(6)` i.e., `O_(2)` is at a distnce `R//6` from centre of disc on diametrically opposite side to centre of hole. |
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