Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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.
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 RegardsArun (askIITians forum expert)
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -