from the base of a hemisphere, a right circular c

Question

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.

Arun · Accepted Answer

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)

Mechanics> from the base of a hemisphere, a right ci...

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.

Siddharth Kumar , 7 Years ago

Arun

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Mechanics> from the base of a hemisphere, a right ci...

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.

Siddharth Kumar , 7 Years ago

Arun

LIVE ONLINE CLASSES

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