CM_2

Consider cube I:
Let there be n-particles.(all are identical for simplicity)
Let the position (r) of i_th particle be r_i
So, the COM = (Σmr_i)/nm
= Σr_i/n
Now, Σr_i is maximum when each r_i = max = a (side of cube) ... (I am just calculating x-coordinate)
So, Σr_{i max} = na
So, R_{COM max} = na/n = a

So, the max value of x-coordinate of COM = a i.e. on the cube

its obvious the ans is yes how are u saying No pls provide ur argument

proof

frm CM Σmr =0

now if its outside cube , all the r vectors are torwards the cube and hence their summation cant be 0 . (as the net vector is always pointing towards cube and cant be null )

Yes, for symmetrical and uniform bodies, the COM is the geometrical centre of the body which for a cube lies inside it. Even if the mass distribution is not uniform, the com would lie somewhere inside the cube only because the mass is inside the cube only.

how does the CM go outside the cube here? It is inside isnt it?

will it be inside or outside ? wont it go a bit outside the cube ??

Copied from net..
A center of mass (CM) of a boomerang, depending upon its particular shape and design, is not located on the device, itself. If you orient the boomerang so that it appears to form the letter V, its CM will be located directly over the vertex. The distance above the vertex will vary depending upon the boomerang's design. When you toss the boomerang, it will spin about its CM.

You may be able to determine a boomerang's CM by doing the following: Suspend the device from a string. The CM will align with the string. Since the CM is also located directly above the vertex, the intersection of the line bisecting the vertex and the line extending downward from the string should locate the CM.

So clearly in ur question,,,CM will lie inside the cube

ok then it will b inside.. r u sure ??i thought it would b in the same pos but above the boomerang ??