where is the c.g of the half circle?

Question

iwant the math formula in another word the distance from the centere

Answer 1

Its center if you mean by "cg" the concept center of gravity. The center of gravity is the center even for non-uniform materials as long as the density varies as the radius, density =f(radius). In other words, as long as an impurities are radially distributed.

Whoops, sorry you said "half circle" whooooooops....... It is definitely on the line x=0 up the y-axis a bit. You would integrate 2xydy from 0 to r to get 2/3r^3 then average out over the whole mass integral 2xdy = 1/2*Pi*r^2 and divide these to get 4/(3Pi)*r. That is about 4/9ths r, so just under 1/2 of r but I would've guessed it to be farther down (closer to 0). Hmmmmm

Answer 2

Here x=Rcost and y=Rsint is equation of a circle; if 0 Let us put our semicircle on a knife blade parallel to x-axis so that semicircle is in equilibrium position. Assume b is the distance of the knife from center of the circle, i.e. b is y-coordinate of mass center to be found, while x-coordinate of mass center =0;
The elementary torque with respect to knife blade is dT = L*ds, where L(t)=y-b is lever of elementary “mass” ds = sqrt((dx)^2 +(dy)^2);
Also dx = -Rsint*dt, dy = Rcost*dt, ds= R*dt, L(t) = R(sint -(b/R));
Assume R=1, then dT=(sint –b)*dt; for mass center the integral{for t=0 until pi} of dT must be equal to zero;
Thence T = -cost –bt = {for t=0 until pi} = (1 –b*pi) –(-1 –b*0) =0; hence 2 –b*pi =0, so b = 2/pi; and finally b=2R/pi;

Answer 3

The centre of mass of a uniform semicircular lamina of radius r lies on the axis of symmetry at a distance of 4r/3p from the straight edge