Question

A thin strip of material is bent into the shape of a semicircle of radius R.
Find its center of mass. ​

Answer 1

To find:

Centre of mass of a semi-circular ring of Radius R;

Calculation:

Linear mass density of ring

$\lambda = \dfrac{m}{\pi R }$

$= > \dfrac{dm}{dy} = \dfrac{m}{\pi R }$

$= > dm = \dfrac{m}{\pi R } \times (dy)$

$= > dm = \dfrac{m}{\pi R } \times (R \: d \theta)$

$= > dm = \dfrac{m \: d \theta}{\pi }$

So, centre of mass :

$\displaystyle \: \bar{y} = \dfrac{1}{m} \int \: y \: dm$

$= > \displaystyle \: \bar{y} = \dfrac{1}{m} \int \: R \cos( \theta)\: dm$

$= > \displaystyle \: \bar{y} = \dfrac{1}{m} \int \: R \cos( \theta)\: \frac{m \: d\theta}{\pi}$

$= > \displaystyle \: \bar{y} = \dfrac{1}{\pi} \int \: R \cos( \theta) \: d \theta$

$= > \displaystyle \: \bar{y} = \dfrac{R}{\pi} \int \: \cos( \theta) \: d \theta$

Putting limit ;

$= > \displaystyle \: \bar{y} = \dfrac{R}{\pi} \int_{0}^{\pi} \: \cos( \theta) \: d \theta$

$= > \displaystyle \: \bar{y} = \dfrac{2R}{\pi}$

So, final answer is:

$\boxed{ \red{ \large{ \sf{ \: \bar{y} = \dfrac{2R}{\pi}}}}}$