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

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

Grade:11

1 Answers

Arun
25750 Points
5 years ago
Dear Siddharth
 
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
u = R^2 - z^2 
du = -2z dz 
(ρπ/M)int[z(R^2 - z^2) dz] (0
= (-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
u = R - 2z 
du = -2 dz 
(ρπ/M)int[z(R - 2z)^2 dz] (0
= (-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
 
 
Regards
Arun (askIITians forum expert)

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