Let there be hollow hemisphere (there is no solid bottom) with radius $r$. Find the center of mass of the hollow hemisphere, assuming the mass distribution is uniform.
My attempt:
Let there be a hollow hemisphere centered at $(0,0,0)$ with the bottom plane of the hemisphere resting on the $x-z$ plane. Let it's total mass be M. By symmetry the center of mass of the hemisphere must lie on the $y$ axis.
Now let us divide the hemisphere into thin rings, starting from the bottom. Let $dA$ be the area of a thin ring, where is the center of mass is the ring's center. Also, let $dM$ be the mass of the ring.
From the diagram it is clear that:
$$x^2+r^2=R^2$$
$$dA=2\pi r\cdot dx \implies dA = 2\pi\cdot \sqrt{R^2-x^2}\cdot dx$$
Hence, $dM=\frac{M}{2\pi R^2} \cdot dA=\frac{M}{R^2} \cdot \sqrt{R^2-x^2} \cdot dx$
$$y_{com}=\frac{1}{M} \cdot \int dm \cdot x=\frac{1}{R^2}\int_0^Rx \cdot \sqrt{R^2-x^2} \cdot dx $$
Evaluating this gives $y_{com} = R/3$, which is proven wrong by a quick google search.
The method shown here uses an angle subtended by a small surface area on a ring, which after computation does give the correct answer of $R/2$. My question is, which step am I doing wrong? Why is the answer coming one way but not the other?
I did the same thing with a cone, and the same thing happened, if I approximated a hollow cone as a series of thin hollow cylinders on top of each other, I get the incorrect answer, but with the angle approach, it gives the correct answer. When is an approximation considered "good enough" mathematically? I would really appreciate an explanation.
Best Answer
I think you want to say $y^2+r^2=R^2$ as you have assumed the base in xz-plane.
Now you write $dA = 2\pi r \ dy \ $ but that is surface area of a ring of width $dy$ on a cylinder of radius $r$, not on a spherical surface. This is the step where you have a mistake. The correct surface area of a ring of infinitesimal width $dy$ on a spherical surface is rather $dS = 2 \pi r\sqrt{(dr)^2+(dy)^2} = 2\pi r \left(\sqrt{\left(\cfrac{dr}{dy}\right)^2+1}\right) dy$
$ = 2 \pi r \left(\sqrt{\left(\cfrac{y}{r}\right)^2+1}\right) dy = 2 \pi R \ dy$
So the center of mass is $(0, \overline y, 0)$ where,
$ \overline y = \displaystyle \cfrac{1}{2 \pi R^2} \int_0^R 2 \pi R \ y \ dy$