Question

A thin uniform wire is bent in the form of a semi-circle of radius R Distance of its center of mass from the geometric center is
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Answer 1

answer is 2R/R is the answer

Answer 2

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 :

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

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

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

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

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

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

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

So, final answer is:

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