centre of gravity of the sector of a circle

Find the centre of gravity of the sector of a circle subtending an angle 2α at the centre of the circle?

Ans: To find centre of gravity of the sector of a circle first draw the figure

centre of gravity of the sector of a circle
centre of gravity of the sector of a circle

AOB is the sector of the circle subtending an angle 2α at the centre. Ox is the bisector of angle AOB.

Equation of the curve is r=a, referred to O as pole and Ox as initial line, a being the radius of the circle.

Ox is the line of symmetry. Hence, the C.G. of the sector lies on Ox. Hence ȳ=0.

And    x=2/3[( ʃαr3cosθdθ)/( ʃαdθ)]

=2/3.a[( ʃαcosθdθ)/( ʃαdθ)]

=2a/3[( sinθ)α]/[(θ) α]

= 2a/3[( sinα)/α]

So, r= a

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