Question

A circle with radius R, has a smaller circle removed from the left side that has a diameter of R (goes from left edge to the center along the x-axis). If the center is the origin, where is the center of mass?

Answer 1

Center of mass for two masses m1 and m2

Rcm= (m1r1+m2r2) / (m1+m2)

where,r1& r2 are COM (Center of mass) of m1 & m2.

Note here r1& r2 are vectors.

When a part is removed from a mass, the formula changes to

Rcm= (m1r1-m2r2) / (m1-m2)

And if areas of the original and removed part are known,we can use the alternative formula,

Rcm = (A1r1-A2r2) / (A1-A2)

A1 =Original area

r1 = A1's COM

A2= removed area

r2 = A2's COM

COMING BACK TO THE QUESTION,

A1= $\pi$R²

r1 =(0,0) [Given that center is origin)

A2 = $\pi$(R/2)² = $\pi$R²/4

r2= (-R/2, 0) [ smaller circle goes from left edge i.e. -R,0 to center i.e. 0,0 ]

Now we cam independentally find x and y coordinates of Rcm.

Rcm(x) = A1r1(x) - A2r2(x) / A1-A2

= ($\pi$R²)*(0) - ($\pi$R²/4)*(-R/2) / $\pi$R² - $\pi$R²/4

= ( -$\pi$R³/8) / (3$\pi$R²/4) i.e. -R/6

Similarly,

Rcm(y)= A1r1(y) - A2r2(y) / A1-A2

= 0 ( as both r1(y) & r2(y) are 0)

So, the required COM is ( -R/6 ,0).