B) Center of Mass of semi circular wire:
Total length of semi circular wire = R
Elemental length = Rdq
C) Center of Mass of a uniform semi circular plate:
Here the element chosen is a thin wire (semi circular) of radius r.
As derived earlier, the for this is at .
1) Why is here?
Ans: The mass dm is a semi circular thin wire whose position is variable (y is not unique), so we concentrate dm mass on the COM of this wire that is at .
D) Center of Mass of hemispherical shell:
1) Why is y = Rsinq here?
Ans: The Center of Mass of elemental ring is at its center (by symmetry) which is at a height of Rsinq from origin; hence y = Rsinq.
E) Center of Mass of a hemisphere:
1) Why is volume of elemental disc = Rdq (cosq) (pR2cos2q) and not
Ans: This is a solid hemisphere. If we consider the curvature at the ends of the disc as negligible (as we do in a hollow shell as in the previous case) then the integration starts yielding wrong results. So as a tip we can take this that whenever the integration is being done over solid objects, the curvature effects cannot be neglected!