Finding the Centre of Mass of a Hemisphere

[akrill]

Homework Statement

Given a hemisphere of radius r and uniform density, find the centre of mass of the hemisphere

Relevant Equations

None given.

• Place hemisphere in xyz coordinates so that the centre of the corresponding sphere is at the origin.
• Then notice that the centre of mass must be at some point on the z axis ( because the 4 sphere segments when cutting along the the xz and xy planes are of equal volume)
• y² + x² = r²
• We want two volumes, V₁ and V₂ which are equal, only cutting parallel to the flat side of the hemisphere at some distance h.
• Recall volume of revolution formula, V = π∫y²dx
• V₁ = ∫₀^h r² - x² dx
• Similarly, V₂ = ∫_h^r r² - x² dx
• Then, by equating the two integrals and doing some rearrangement, I got to: h³-3r²h +r³ = 0
• Also, 0 < h < r obviously.

Not really sure how to solve for h here, I dont think I made a mistake while rearranging. Is my method valid and/or is there a simpler way to do this?

 

[.Scott]

How do you calculate a moment. And how is moment related to the center of gravity?
If you simplify by taking r=1, then you will be calculating h/r.

 

[PeroK]

Your cubic equation looks correct. How to solve a cubic?

 

[vela]

Is my method valid and/or is there a simpler way to do this?

Your method is incorrect. Two equal volumes doesn't work because the center of mass of each piece is a different distance from the plane $z=h$.

 

[akrill]

Your cubic equation looks correct. How to solve a cubic?

To be honest I don't think it is right. I've read some more based on what @.Scott said, and I think the correct strategy is to integrate the moments of the "slices" of the hemisphere (thickness dz) and then divide by the mass.

 

[akrill]

Your method is incorrect. Two equal volumes doesn't work because the center of mass of each piece is a different distance from the plane $z=h$.

Alright, thanks for clarifying :)

 

[PeroK]

To be honest I don't think it is right. I've read some more based on what @.Scott said, and I think the correct strategy is to integrate the moments of the "slices" of the hemisphere (thickness dz) and then divide by the mass.

Yes, my mistake. I just checked your integration.

In my defence I was cooking and doing physics at the same time.

 

[PeroK]

Your cubic equation looks correct. How to solve a cubic?

BTW, the solution to the cubic equation
$$\alpha^3 -3\alpha +1 = 0$$Where $0<\alpha <1$ is:$$\alpha =2\cos(\frac 4 9 \pi) \approx 0.3473$$

 

[.Scott]

BTW, the solution to the cubic equation
$$\alpha^3 -3\alpha +1 = 0$$Where $0<\alpha <1$ is:$$\alpha =2\cos(\frac 4 9 \pi) \approx 0.3473$$

Just to be clear, that's the solution to the cubic, but not the center of gravity.

 

[PeroK]

Just to be clear, that's the solution to the cubic, but not the center of gravity.

If you wanted to divide the hemisphere horizontally into two parts of equal mass, that's the ratio of $\frac h r$ that you would use.