Finding the Center of Mass for a Hemisphere and Right Cone

[rahul.mishra]

Suppose there's a hemisphere of radius R (say) and a right cone of same radius R but ht. R/2 is scooped out of it then i have to find the center of mass of the remaining part.

Here's how i approached...

clearly by symmetry, Xcm = 0

Now, Let M be the mass of the hemisphere so,

Density per unit volume, ρ = M/(2/3.π .r³) x 1/3.π.r².(r/2) = M/4

Now, Ycm of remaining portion = {M(3R/8) - M/4(R/6)}/{M-M/4} = 4R/9

Thus, C.M of the remaining portion = (0,4R/9)

But the result given by the source is 11R/24 from base...!

Now where am i wrong?

 

[mfb]

The center of mass of the cone is at R/8.
R*(3/8 - 1/4*1/8)/(1-1/4)=11/24 R

 

[rahul.mishra]

Thanks a lot...
my fault was actually i assumed the cone to be hollow but it is solid...!
so h/4 not h/3 is the location of its center of mass from base... got it now..!