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.

 

[Walter Black]

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?

hope this clears it up ,got any doubt then ffta,sorry for bad handwriting

 

[berkeman]

hope this clears it up ,got any doubt then ffta,sorry for bad handwriting

Welcome to PF.

Just an FYI, we do not solve students' schoolwork problems here at PF. We offer hints, ask questions, find mistakes, etc., but the student must do the bulk of the work. Having said that, this thread is over a year old, so the OP has presumably finished this class and moved on. So in cases like this, it's okay to post a full solution or an alternate solution to help the thread for future viewers.

I'll send you a DM with some tips on how to use LaTeX to post math at PF. We generally discourage showing math in pictures of hand-written work.

 

[Walter Black]

Welcome to PF.

Just an FYI, we do not solve students' schoolwork problems here at PF. We offer hints, ask questions, find mistakes, etc., but the student must do the bulk of the work. Having said that, this thread is over a year old, so the OP has presumably finished this class and moved on. So in cases like this, it's okay to post a full solution or an alternate solution to help the thread for future viewers.

I'll send you a DM with some tips on how to use LaTeX to post math at PF. We generally discourage showing math in pictures of hand-written work.

My bad. ,i am sorry for that, i registered today and didnt know about this

 

[berkeman]

No worries!

 

[mathwonk]

I hope from the comments here it is ok to post a solution. this can be done by noting that, (just as one gets a solid 3 ball by revolving a half disc around its straight edge), revolving a (solid) hemisphere (i.e. a half 3-ball) around the plane of its equator in 4 space, gives a solid 4 ball. Let the radius be 1. Since that 4ball volume is π^2/2, this gives π^2/2 = 2πh.vol(half 3-ball) = 2πh(2π/3), where h is the height of the center of mass. Hence h = 3/8, so for radius R, h = 3R/8. (But to me, the word "hemisphere" denotes an object of two dimensions, not three.)

 

[Mark44]

But to me, the word "hemisphere" denotes an object of two dimensions, not three.

Albeit, embedded in three-space. If I'm remembering the definitions correctly, a sphere is two-dimensional while a ball (the version in three-space) is three-dimensional.

 

[mathwonk]

Yes. If the question referred to the center of mass of a hemispherical shell, (embedded in 3 -space), then since as I believe Archimedes knew, the surface area of a hemisphere equals that of the circumscribing cylinder, and the same holds for horizontal bands cut from the two figures by pairs of horizontal and parallel planes, then their centers of gravity are at the same height, i.e. half way up, at height R/2.
What I like is that Archimedes could answer both these questions long before the invention of calculus. (He could compute the center of gravity of half a solid ball from those of the cylinder and cone.)