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Centroid of Semicircular Arc

arccent.html

Exploration

Show that the centroid of the area bounded by a semicircular arc of radius r and its chord is on the axis of symmetry at a distance

from the chord of the arc.

Approach

Construct representative geometry for the semicircular arc. Using symbolic computations, compute the width of a horizontal rectangle spanning the arc having infinitesimal height. Use integration to find moments of inertia on each side of the centroid. Equate the moments of inertia on each side of the centroid and solve for the y-coordinate of the centroid.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

Construct a semicircular arc of radius r (the portion of the circle above the x-axis).

Clear[r];

C1=Circle2D[{0,0},r];

Construct a horizontal line at height y.

Clear[y];

L1=Line2D[Point2D[0,y],0];

Compute the intersection points of the horizontal line with the arc.

pts=Points2D[L1,C1]

The width of the arc is the difference between the abscissas of the intersection points.

L=XCoordinate2D[First[pts]]-XCoordinate2D[Last[pts]]

By integrating L d find the moment of inertia of an area, where d is the distance from the centroid line (). is the expression for the moment of inertia of the upper area with respect to .

Clear[yB];

int1=Integrate[L*(yB-y),y] //Simplify

The next few steps simplify the solution.

I1=(int1 /. y->yB)-(int1 /. y->0) /. (r^2)^(3/2)->r^3

is the expression for the moment of inertia of the upper area with respect to .

int2=Integrate[L*(y-yB),y] //Simplify

I2=-Limit[int2, y->r]-(int2 /. y->yB) //FullSimplify

The moments of inertia must be the same on each side of the centroid line.

eq1=I2-I1==0//FullSimplify

Solve for .

Solve[eq1,yB]

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