Problem finding the center of mass of remaining portion ,when a part of it is removed

  • #1
rahul.mishra
7
0
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????
 

Answers and Replies

  • #2
36,101
13,024


The center of mass of the cone is at R/8.
R*(3/8 - 1/4*1/8)/(1-1/4)=11/24 R
 
  • #3
rahul.mishra
7
0


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..!!
 

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