If we take a straight line of length (L), then its midpoint will be at a distance (L/2) from either end of the line. Let us denote the centroid as the point G(Xg,Yg). Suppose we have a straight line AB of length (L) that makes an angle θ with X axis. Let the coordinate of point A is (Xo,Yo). Let the centroid be at G(Xg,Yg), then
Xg = Xo + (Lcos θ)/2
Yg = Yo + (Lsin θ)/2
Xg = X1 - (Lcos θ)/2
Yg = Y1 - (Lsin θ)/2
For Horizontal lines θ = 0° and for Vertical lines θ = 90°
CENTROID OF A CURVED LINE
The steps to derive the centroid of a quarter circular arc of radius R.
Centroid of a curved line can be derived with the help of calculus.
i) Draw the figure in a X-Y coordinate system. Let the curved line has been represented by a function θ
ii) Take an arbitrary point P(X,Y) on the curve. Join the line OP, where O is the origin of the coordinate system. Let OP has a length L and makes an angle (θ) with X axis. Therefore, we can write
X = Rcosθ ----- (a)
Y = Rsinθ ----- (b)
iii) Let another point Q, such that PQ = dL where dL is very very small. Let the angle subtended by (dL) be (dθ). So we can write
Y = Rsinθ ----- (b)
iii) Let another point Q, such that PQ = dL where dL is very very small. Let the angle subtended by (dL) be (dθ). So we can write
dL = Rdθ ------ (c)
Xg = (1/L) ∫(XdL)
= (1/L) ∫ Rcosθ.Rdθ
= (1/L).R² ∫ sinθ.dθ ------- (d)
Yg = (1/L) ∫ YdL
= (1/L) ∫ Rsinθ.Rdθ
= (1/L).R² ∫ sinθ.dθ -------- (e)
Xg = (1/L) ∫(XdL)
= (1/L) ∫ Rcosθ.Rdθ
= (1/L).R² ∫ sinθ.dθ ------- (d)
Yg = (1/L) ∫ YdL
= (1/L) ∫ Rsinθ.Rdθ
= (1/L).R² ∫ sinθ.dθ -------- (e)
CENTROID OF A QUARTER CIRCULAR ARC OF RADIUS R
Suppose we have a quarter circular arc in a co-ordinate system as shown in the figure. Total length of the arc AB = (πR)/2 . We take an arbitrarily small length of the arc CD and denote it as dL.
So, dL = Rdθ ------ (iii) [ as s=Rθ ]
where R = Radius of the quarter circular arc.
Let the co-ordinate of the point D be D(x,y) where
X = Rcosθ -----(iv) and
Y = Rsinθ -----(v)
Hence Xg = (1/L)∫x.dL ; here L = (πR)/2 ;
X = Rcosθ
dL = Rdθ
Xg = (2/πR) 0∫π/2Rcosθ.Rdθ
= (2/πR) R2 0∫π/2cosθ.dθ
= 2R/π
Yg = (2/πR) 0∫π/2Rsinθ.Rdθ
= (2/πR) R2 0∫π/2sinθ.dθ
= 2R/π
Hence, for a quarter circular arc of radius R will be G(2R/π,2R/π)
CENTROID OF A COMPOSITE LINE
In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure.So, a composite line is consisted of several straight or curved lines.
Let a composite line is made of n number of lines, which may straight or curved lines.
STEP-ONE:
Draw the composite line and break it into n number of parts. Let the lengths of these lines are L1, L2, L3 ........ Ln. Let the centroids of these lines are G1(X1,Y1),G2(X2,Y2), G3(X3,Y3) ........ Gn(Xn,Yn).
Calculate length (Li), and coordinates (Xi,Yi) for each and every parts.
Let a composite line is made of n number of lines, which may straight or curved lines.
STEP-ONE:
Draw the composite line and break it into n number of parts. Let the lengths of these lines are L1, L2, L3 ........ Ln. Let the centroids of these lines are G1(X1,Y1),G2(X2,Y2), G3(X3,Y3) ........ Gn(Xn,Yn).
Calculate length (Li), and coordinates (Xi,Yi) for each and every parts.
Now, if the centroid of the composite line be G(Xg,Yg)
Xg = (∑LiXi)/(∑Li)
=> (L1X1 + L2X2 + L3X3)/(L1 + L2 + L3)
Yg = (∑LiYi)/(∑Li)
=> (L1Y1 + L2Y2 + L3Y3)/(L1 + L2 + L3)
:
and the length of side AB of the triangle. Note that there are two possible answers to this question and we have provided only one of them as an answer.
