what is the centre of gravity of semicircle of diameter 12cm
Answers
Answer:
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
y
=
0
∫
6
A
2xy
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
y
=
0
∫
6
18π
2
36−y
2
y
dy
Let assume 36 - y² = z
=> -2ydy = dz
=> 2ydy = -dz
on substituting
\int{\frac{-\sqrt{z} }{18 \pi}} \, dz∫
18π
−
z
dz
on integrating
\begin{gathered}{\dfrac{-z^{\frac{3}{2}} }{\frac{3}{2}.18 \pi}}+c \\{\dfrac{-(36-y^2)^{\frac{3}{2}} }{27 \pi}}+c \\\end{gathered}
2
3
.18π
−z
2
3
+c
27π
−(36−y
2
)
2
3
+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