Question

from the base of a hemisphere a right cone of height R/2 and same base has been scooped out. find the CM of the remaining part

Answer 1

 Center of mass for hemisphere: 
dm = ρπr^2 dz 
dm = ρπ(R^2 - z^2) dz 

int[z dm] / M = (ρπ/M)int[z(R^2 - z^2) dz] (0 < z < R) = 

u = R^2 - z^2 
du = -2z dz 

(ρπ/M)int[z(R^2 - z^2) dz] (0 < z < R)
= (-ρπ/(2M))int[u du]
= -ρπu^2/(4M)
= -ρπ(R^2 - z^2)^2/(4M)
= ρπR^4/(4M) 

M = ρ(2/3)πR^3 

ρπR^4/(4M) = (3/8)R 

Center of mass of cone: 
dm = ρπr^2 dz 
dm = ρπ(R - 2z)^2 dz 

int[z dm] / M = (ρπ/M)int[z(R - 2z)^2 dz] (0 < z < R/2) 

u = R - 2z 
du = -2 dz 

(ρπ/M)int[z(R - 2z)^2 dz] (0 < z < R/2)
= (-ρπ/(4M))int[(R - u)u^2] du
= (-ρπ/(4M))[Ru^3/3 - u^4/4]
= (-ρπ/(4M))[R(R - 2z)^3/3 - (R - 2z)^4/4]
= (ρπ/(4M))[R^4/3 - R^4/4] = ρπR^4/(48M) 

M = ρ(1/3)πR^2(R/2) 
M = ρ(1/6)πR^3 

ρπR^4/(48M) = (1/8)R 

Total center of mass: 
{[ρ(2/3)πR^3][(3/8)R] - [ρ(1/6)πR^3][(1/8)R]} / [ρ(2/3)πR^3 - ρ(1/6)πR^3] = {(1/9 - 1/48)R} / [1/2]
= 11R/24