Question

Consider a semicircular ring of radius R. Its linear mass density varies as $\lambda ={\lambda }_{0}sin\theta$. Locate its centre of mass.

Answer 1

$\lambda ={\lambda }_0\sin \theta dm=\left(Rd\theta \right)\lambda ={\lambda }_{0}R\sin \theta d\theta X_{com}=\frac{\int xdm}{\int dm}=\frac{\int (R\cos \theta )\left({\lambda }_{0}R\sin \theta d\theta \right)}{\int {\lambda }_{0}R\sin \theta d\theta }{X}_{com}=\frac{\frac{{\lambda }_0{R}^2}{2}{\int }_0^{\pi }\sin 2\theta d\theta }{{\lambda }_{0}R{\int }_0^{\pi }\sin \theta d\theta }=\frac{R}{2}\times \frac{{[-\frac{\cos 2\theta }{2}]}_0^{\pi }}{{-}_0{\left[\cos \theta \right]}_0^{\pi }}=0Y_{com}=\frac{\int ydm}{\int dm}=\frac{\int \left(R\sin \theta \right)\left({\lambda }_{0}R\sin \theta d\theta \right)}{{\lambda }_{0}R{\int }_0^{\pi }\sin \theta d\theta }=\frac{R}{2}=\left[\frac{{\int }_0^{\pi }{\sin }^2\theta d\theta }{{\int }_0^{\pi }\sin \theta d\theta }\right]=\frac{R}{2}\left[\frac{{\int }_0^{\pi }(1+\cos 2\theta ]d\theta }{{[-\cos \theta ]}_0^{\pi }}\right]=\frac{R}{2}\frac{{[\theta -\frac{\sin 2\theta }{2}]}_0^{\pi }}{2}=\frac{\pi R}{4}$