Finding the center of mass of a solid hemisphere

[Hamiltonian]

Homework Statement

find the center of mass of a solid hemisphere of radius R.

Relevant Equations

$Y_{cm} = \frac {1}{M} \int y dm$

for this derivation, I decided to think of the solid hemisphere to be made up of smaller hemispherical shells each of mass $dm$ at their respective center of mass at a distance r/2 from the center of the base of the solid hemisphere.
also, I have taken the center of the base of the solid hemisphere to be the origin. Due to symmetry, I know that the center of mass of the solid hemisphere is along the Y-axis.

I found the mass $dm$ of all the small hemispherical shells
$$dm = \frac {M}{(4/6)\pi R^3} (\frac{4\pi}{3}r^2 dr)$$

substituting $dm$ into $$Y_{cm} = \frac {1}{M} \int \frac {r}{2} dm $$

gives me $Y_{cm} = R/4$ which is wrong.

also, I know this can be done in a slightly easier way if you consider the hemisphere to be made up of solid discs but I thought I'd give this method a go! can someone help point out what exactly I am doing wrong here:)

 

[PeroK]

I decided to think of the solid hemisphere to be made up of smaller hemispherical shells each of mass $dm$ at their respective center of mass at a distance r/2 from the center of the base of the solid hemisphere.

This doesn't sound right. What do you mean by this?

 

[Hamiltonian]

This doesn't sound right. What do you mean by this?

I have considered the large solid hemisphere to be made up of smaller hemispherical shells whose center of mass we know is at $r/2$ and is of mass $dm$

 

[PeroK]

View attachment 269250I have considered the large solid hemisphere to be made up of smaller hemispherical shells whose center of mass we know is at $r/2$ and is of mass $dm$

Why is the CoM of the shell at $r/2$?

 

[Hamiltonian]

Why is the CoM of the shell at $r/2$?

hmm...because you can derive that by breaking the hemispherical shell into rings of mass $dm$ whose CoM lies at their centers and then you end up with $$Y_{cm} = 3R/2\int_0^{\pi/2}cos\theta sin^2\theta d\theta = R/2$$
the origin is at the center of the base of the hemispherical shell and R is the radius of the hemispherical shell.

 

[PeroK]

hmm...because you can derive that by breaking the hemispherical shell into rings of mass $dm$ whose CoM lies at their centers and then you end up with $$Y_{cm} = 3R/2\int_0^{\pi/2}cos\theta sin^2\theta d\theta = R/2$$
the origin is at the center of the base of the hemispherical shell and R is the radius of the hemispherical shell.

Can you see why that doesn't work?

 

[Hamiltonian]

Can you see why that doesn't work?

I thought breaking a solid hemisphere into hemispherical shells was analogous to breaking like a solid disk into smaller rings?

 

[PeroK]

I thought breaking a solid hemisphere into hemispherical shells was analogous to breaking like a solid disk into smaller rings?

That's not the problem. This is the problem:

hmm...because you can derive that by breaking the hemispherical shell into rings of mass $dm$ whose CoM lies at their centers and then you end up with $$Y_{cm} = 3R/2\int_0^{\pi/2}cos\theta sin^2\theta d\theta = R/2$$
the origin is at the center of the base of the hemispherical shell and R is the radius of the hemispherical shell.

 

[Hamiltonian]

That's not the problem. This is the problem:

$$Y_{cm} = \frac {1}{M} \int \frac {r}{2} dm$$ so this is wrong?

 

[etotheipi]

Consider a thin hemisphere of thickness $dr$, and mass $dm = 4\pi r^2 \rho dr$. Since the hemisphere is of infinitesimal thickness, its centroid is at $\frac{r}{2}$ in the $y$ direction. The sum of the first moments of mass of all of these tiny hollow hemispheres, which combine to make a solid hemisphere, in the $y$ direction, $S_y$, will be$$S_y = \int_0^R y dm = 2\pi r^3 \rho dr$$Can you take it from here?

 

[PeroK]

hmm...because you can derive that by breaking the hemispherical shell into rings of mass $dm$ whose CoM lies at their centers and then you end up with $$Y_{cm} = 3R/2\int_0^{\pi/2}cos\theta sin^2\theta d\theta = R/2$$
the origin is at the center of the base of the hemispherical shell and R is the radius of the hemispherical shell.

Actually, my apologies, this calculation is correct.

 

[Hamiltonian]

what exactly is wrong in my method?

 

[PeroK]

I found the mass $dm$ of all the small hemispherical shells
$$dm = \frac {M}{(4/6)\pi R^3} (\frac{4\pi}{3}r^2 dr)$$

I think this is not right. It should be $2\pi r^2 dr$ for a shell of radius $r$ and thickness $dr$.

 

[PeroK]

PS I thought it was simpler to integrate disks of radius $r$ and height $dz$:
$$dm = \rho \pi r^2 dz = \rho \pi (R^2 - z^2)dz$$
And, for the hollow hemisphere I got the integral:
$$R \int_0^{\frac {\pi}{2}}\sin \theta \ \cos \theta \ d\theta$$

 

[Hamiltonian]

I think this is not right. It should be $2\pi r^2 dr$ for a shell of radius $r$ and thickness $dr$.

what exactly should be $2\pi r^2 dr$? $dm$?
also, I am considering the mass per unit volume.

 

[PeroK]

what exactly should be $2\pi r^2 dr$? $dm$?
also, I am considering the mass per unit volume.

In place of the volume formula you used.

 

[Hamiltonian]

In place of the volume formula you used.

how exactly did you get that?
I got the volume of $dm$ as $$=\frac {4\pi}{6}(r + dr)^3 - \frac {4}{6}\pi r^3$$
$$\approx \frac {4\pi}{3}r^2dr$$

 

[PeroK]

how exactly did you get that?
I got the volume of $dm$ as $$=\frac {4\pi}{6}(r + dr)^3 - \frac {4}{6}\pi r^3$$
$$\approx \frac {4\pi}{3}r^2dr$$

That's definitely not right. I took the surface area times $dr$. It should come to the same thing.

What do you get when you expand $(r + dr)^3$?

 

[Hamiltonian]

That's definitely not right. I took the surface area times $dr$. It should come to the same thing.

What do you get when you expand $(r + dr)^3$?

what exactly is wrong here as we are supposed to get the same answer either way? or is the approximation of $(dr)^2 = 0$ and $(dr)^3 = 0$ wrong?

 

[PeroK]

what exactly is wrong here as we are supposed to get the same answer either way? or is the approximation of $(dr)^2 = 0$ and $(dr)^3 = 0$ wrong?

You have to start challenging your own work. And being able to double-check things for yourself.

PS You can be as skeptical as you like, but you're the one who is asking why you got the wrong answer.