Question

What is the centre of gravity of a semi circle of diameter 12cm?

Answer 1

Given : a semi circle of diameter 12 cm

To Find : Center of Gravity

Solution:

Diameter = 12 cm

Radius = 6 cm

Taking center at ( 0 , 0)  and semicircle above x axis

Because of symmetry center of gravity will lie of y axis

$\overline{y} = \int\limits^6_0 {\frac{2xy}{A}} \, dy$

A = (1/2)πr² =   (1/2)π(6)²  = 18π  cm² ( area of semicircle)

x² + y² = r² =  6²

=> x² = (6² - y²)

=> x = √(36 - y²)

$\overline{y} = \int\limits^6_0 {\frac{2\sqrt{36-y^2} y}{18 \pi}} \, dy$

Let assume 36 - y²  = z  

=> -2ydy  = dz

=> 2ydy = -dz  

on substituting  

$\int{\frac{-\sqrt{z} }{18 \pi}} \, dz$

on integrating

${\dfrac{-z^{\frac{3}{2}} }{\frac{3}{2}.18 \pi}}+c \\{\dfrac{-(36-y^2)^{\frac{3}{2}} }{27 \pi}}+c$

Applying limits 0 to 6

216/27π

= 8/π  cm

= 2.5465  cm

Hence centre of gravity of a semi circle of diameter 12 cm

will be at a height of 8/π  cm or 2.5465  cm  from the center of Semicircle

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Answer 2

To find:

Centre of mass of a semi-circular ring of diameter 12 cm ?

Calculation:

Let radius of ring be R = 12/2 = 6 cm:

Linear mass density of ring be $\lambda$

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

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

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

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

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

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 ;

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

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

Putting available values:

$= > \displaystyle \: \bar{y} = \dfrac{2\times 6}{\pi}$

$= > \displaystyle \: \bar{y} = \dfrac{12}{\pi}$

$= > \displaystyle \: \bar{y} = 3.819 cm$

So, final answer is:

$\boxed{ \red{ \large{ \sf{ \: \bar{y} = 3.819 cm}}}}$