Centre of mass of a solid hemisphere. What am I doing wrong?

In summary, Marv was trying to calculate the centre of mass of a drawn hemisphere, but he ran into difficulty. He did not provide any explicit information about what he was doing, so it is difficult to understand where he is making mistakes. He appears to have used the wrong integration limits, and may need to do the integration himself to check.
  • #1
Marvin94
41
0
2sayzr6.jpg


I have to calculate the centre of mass of the drawn hemisphere. I don't understand where I make mistakes. The process I followed on the above image is the following:

(I) Here I simply translated position vector components from cartesian to polar coordinates.
(II) Formula of x-component of position vector of centre of gravity (this position vector clearly lies on the x-axis, so I'm interested only in this component)
(III) This is the volume element for polar coordinates system.

After that, I just put (III) into (II).

The problem is that, the result of the last (triple) integral (computable easily with wolfram alpha) seems to result zero! I thought that the mistake could be in the integration limits, but they look to be correct. Where am I doing wrong?

Thank you very much in advance.
 
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  • #2
Hi Marv,

Check your integration limits anyway. Looks to me that you aren't integrating over the righthand half of the sphere...
 
  • #3
I tried to understand why what I wrote should be wrong, but I don't see the mistake. Which limits do you think are correct?
 
  • #4
Well, you haven't shown what ##\theta## ande ##\phi## are, so I can't tell, but you can !
[edit] but it's implicit in your coordinate transformation. Check if Rx > 0 always during the integration...
 
  • #5
θ = angle between position vector and x-axis
ϕ = angle between position vector and z-axis
 
  • #6
I used an ambiguous notation: I should precise, that the initial vector r (see (I) ) is a general position vector. The vector rx I use after that refers to the position vector of the centre of mass (not the general one!).
 
  • #7
Check if your rx is always positive.
Check if your ry really runs from -1 to +1
Check if your rz really runs from -1 to +1

As an exercise, do the integration yourself, instead of with the Wolfram crutches.
 
  • #8
Marvin94 said:
θ = angle between position vector and x-axis
ϕ = angle between position vector and z-axis
Nope. Not if ##z = r\cos \theta##
 
  • #9
Check terms in $$ r_{cm} = \frac{1}{V}\int\int\int \,r\,dV $$
 
  • #10
Thanks you all a lot :)
 

FAQ: Centre of mass of a solid hemisphere. What am I doing wrong?

1. What is the formula for calculating the center of mass of a solid hemisphere?

The formula for calculating the center of mass of a solid hemisphere is (3/8)R, where R is the radius of the hemisphere.

2. How do I find the radius of a solid hemisphere?

The radius of a solid hemisphere can be found by measuring the distance from the center to the edge of the hemisphere. Alternatively, if you know the volume of the hemisphere and the density of the material, you can use the formula R = (3V/(4πρ))^(1/3), where V is the volume and ρ is the density.

3. Can the center of mass of a solid hemisphere be outside of the object?

No, the center of mass of a solid hemisphere will always be located within the object. This is because the center of mass is the point at which the mass of the object is evenly distributed, and in a solid hemisphere, the mass is evenly distributed around the center point.

4. What are some common mistakes when calculating the center of mass of a solid hemisphere?

Some common mistakes when calculating the center of mass of a solid hemisphere include using the wrong formula, not accounting for the density of the material, and not measuring the radius correctly. It is also important to make sure that the hemisphere is symmetrical, otherwise the center of mass may not be accurate.

5. Is the center of mass of a solid hemisphere the same as the centroid?

No, the center of mass and the centroid are not the same for a solid hemisphere. The centroid is the geometric center of the object, while the center of mass is the point at which the mass is evenly distributed. In a symmetrical object like a solid hemisphere, the two points may be the same, but this is not always the case for more complex shapes.

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