Question Number 71506 by ajfour last updated on 16/Oct/19
Commented by ajfour last updated on 16/Oct/19
$${Find}\:{location}\:{of}\:{center}\:{of}\:{mass} \\ $$$${of}\:{the}\:{arc}\:{of}\:{ring}. \\ $$
Answered by mr W last updated on 16/Oct/19
$${let}\:{z}_{{C}} ={distance}\:{from}\:{center}\:{of}\:{mass} \\ $$$${to}\:{the}\:{top}\:{of}\:{arc} \\ $$$${dA}=\rho{Rd}\theta \\ $$$${M}=\mathrm{2}\int_{\mathrm{0}} ^{\alpha} {dA}=\mathrm{2}\rho{R}\int_{\mathrm{0}} ^{\alpha} {d}\theta=\mathrm{2}\rho{R}\alpha \\ $$$${Mz}_{{C}} =\mathrm{2}\int_{\mathrm{0}} ^{\alpha} {R}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\rho{Rd}\theta \\ $$$$=\mathrm{2}\rho{R}^{\mathrm{2}} \int_{\mathrm{0}} ^{\alpha} \left(\mathrm{1}−\mathrm{cos}\:\theta\right){d}\theta \\ $$$$=\mathrm{2}\rho{R}^{\mathrm{2}} \left(\alpha−\mathrm{sin}\:\alpha\right) \\ $$$$\Rightarrow{z}_{{C}} =\frac{\mathrm{2}\rho{R}^{\mathrm{2}} \left(\alpha−\mathrm{sin}\:\alpha\right)}{\mathrm{2}\rho{R}\alpha}={R}\left(\mathrm{1}−\frac{\mathrm{sin}\:\alpha}{\alpha}\right) \\ $$$${for}\:\alpha=\frac{\pi}{\mathrm{2}}:\:{z}_{{C}} ={R}\left(\mathrm{1}−\frac{\mathrm{2}}{\pi}\right) \\ $$$${for}\:\alpha=\pi:\:{z}_{{C}} ={R} \\ $$
Commented by ajfour last updated on 17/Oct/19
$${Nice}\:{and}\:{correct}.\:{Thank}\:{you}\:{Sir}! \\ $$