urgent help with this M3 question please!!!

Need help with this centre of mass question. its from the Jan 2005 paper.
it should be pretty easy as its only question 2, but i can't seem to get it! :frown:

A child’s toy consists of a uniform solid hemisphere, of mass M and base radius r, joined to a uniform solid right circular cone of mass m, where 2m < M.
The cone has vertex O, base radius r and height 3r.
Its plane face, with diameter AB, coincides with the plane face of the hemisphere.

(a) Show that the distance of the centre of mass of the toy from AB is

3r(M-2m)
8(M+m)

your assistance would be much appreciated! :smile:

Reply 1

Christophicus

Mass of cone
= m

Mass of hemisphere
= M

Centre of mass of cone from AB = -3r/4
Centre of mass of hemisphere from AB = 3r/8

Total mass of shape
= m+M

Centre of mass of shape
= y-bar

Emy = y-bar x Em
-mr/4 + 3rM/8 = (m+M)y-bar
-6mr + 3rM = 8(m+M)y-bar
y-bar = (3rM - 6rm)/8(m+M) = 3(M-2m)r/8(m+M)

I know this is a weird way of doing it.
If you think about it, the question tells you that 2m<M
Therefore the mass of the hemisphere is more than twice as heavy as the cone, so the centre of mass lies in the hemisphere section. So you calculate the centre of mass with the base being the vertex of the cone.

Reply 2

Christophicus

Method 2

Mass of cone
= m

Mass of hemisphere
= M

Centre of mass of cone from base of hemisphere = r + 3r/4 = 7r/4
Centre of mass of hemisphere from AB = 5r/8

Total mass of shape
= m+M

Centre of mass of shape
= y-bar

Emy = y-bar x Em
7mr/4 + 5rM/8 = (m+M)y-bar
14mr + 5rM = 8(m+M)y-bar
y-bar = (14mr + 5rM)/8(m+M)

So distance of C.O.M from AB
= r - (14mr + 5rM)/8(m+M)
= [8(m+M)r - (14mr + 5rM)]/8(m+M)
= [8mr + 8Mr - 14mr - 5Mr]/8(m+M)
= (3Mr - 6mr)/8(m+M)
= 3(M-2m)r/8(m+M)

Reply 3

r1sh1_165

thanks, this ques. took the piss!