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From-a-circular-disc-of-radius-R-a-circular-hole-of-radius-R-2-is-cut-out-The-centre-of-the-circular-hole-is-at-R-2-from-the-centre-of-the-original-disc-Locate-the-centre-of-gravity-of-the-resulting-f




Question Number 31476 by NECx last updated on 09/Mar/18
From a circular disc of radius R  a circular hole of radius R/2 is  cut out.The centre of the circular  hole is at R/2 from the centre of  the original disc.Locate the centre  of gravity of the resulting flat  body.    please help me as fast as possible.  Thanks!
$${From}\:{a}\:{circular}\:{disc}\:{of}\:{radius}\:{R} \\ $$$${a}\:{circular}\:{hole}\:{of}\:{radius}\:{R}/\mathrm{2}\:{is} \\ $$$${cut}\:{out}.{The}\:{centre}\:{of}\:{the}\:{circular} \\ $$$${hole}\:{is}\:{at}\:{R}/\mathrm{2}\:{from}\:{the}\:{centre}\:{of} \\ $$$${the}\:{original}\:{disc}.{Locate}\:{the}\:{centre} \\ $$$${of}\:{gravity}\:{of}\:{the}\:{resulting}\:{flat} \\ $$$${body}. \\ $$$$ \\ $$$${please}\:{help}\:{me}\:{as}\:{fast}\:{as}\:{possible}. \\ $$$${Thanks}! \\ $$
Answered by mrW2 last updated on 09/Mar/18
(R/2)×((πR^2 )/4)=e×(πR^2 −((πR^2 )/4))  e=(R/6)
$$\frac{{R}}{\mathrm{2}}×\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}={e}×\left(\pi{R}^{\mathrm{2}} −\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$${e}=\frac{{R}}{\mathrm{6}} \\ $$
Commented by NECx last updated on 09/Mar/18
please can you explain
$${please}\:{can}\:{you}\:{explain} \\ $$$$ \\ $$
Commented by mrW2 last updated on 09/Mar/18

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