Determine the x-coordinate of the mass center of the homogeneous hemisphere

In summary, the problem is asking for the x-coordinate of the mass center of a homogeneous hemisphere with the smaller hemispherical portion removed. The center of mass is defined as the average of the positions of the particles weighted by their masses. The attempt at a solution involves using an integral to determine the x-coordinate, but it is unclear if the smaller hemisphere was properly removed and if the correct integration was used.
  • #1
Godblessyou
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Homework Statement


Determine the x-coordinate of the mass center of the homogeneous hemisphere with the smaller hemispherical portion removed?

I know what the answer should be it's Xcm = 45/112 R

Homework Equations



The center of mass R of a system of particles is defined as the average of their positions, ri, weighted by their masses, mi


The Attempt at a Solution


dm = pi*p*(R^2 - x^2)
and then doing the (integral of x*(dm/dx)*dx)/integral from 0 to R (dm/dx)*dx
which finally gives me 3/2R but this is wrong. Please someone help thanks
 

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  • #2
Godblessyou said:

Homework Statement


Determine the x-coordinate of the mass center of the homogeneous hemisphere with the smaller hemispherical portion removed?

I know what the answer should be it's Xcm = 45/112 R

Homework Equations



The center of mass R of a system of particles is defined as the average of their positions, ri, weighted by their masses, mi


The Attempt at a Solution


dm = pi*p*(R^2 - x^2)
and then doing the (integral of x*(dm/dx)*dx)/integral from 0 to R (dm/dx)*dx
which finally gives me 3/2R but this is wrong. Please someone help thanks

It's a little hard to tell without the integral in Latex notation, but did you remember to remove the smaller hemisphere? Why is your integration from 0 to R? And why isn't it a triple integral?
 
  • #3
Yes I did remove smaller hemisphere from bigger which gave me 3/2R. I don't understand what you mean and what you are asking?
 
  • #4
Godblessyou said:
Yes I did remove smaller hemisphere from bigger which gave me 3/2R. I don't understand what you mean and what you are asking?

I mentioned Latex to help you post your equations in a format that is more readable. Look at the Latex Tutorial thread in the Learning Materials area of the PF:

https://www.physicsforums.com/forumdisplay.php?f=151

I mentioned the triple integral, because what you wrote looks like a 1-D integration, and it would seem that you need to integrate over the volume of the object, no?
 

Related to Determine the x-coordinate of the mass center of the homogeneous hemisphere

1. How do you determine the x-coordinate of the mass center of a homogeneous hemisphere?

The x-coordinate of the mass center of a homogeneous hemisphere can be determined by using the formula x = (3/8)r, where r is the radius of the hemisphere. This formula assumes that the origin is at the center of the flat base of the hemisphere.

2. What is the definition of a homogeneous hemisphere?

A homogeneous hemisphere is a solid object with a curved surface and a flat base, where the material is distributed evenly throughout the hemisphere. This means that the density of the material is constant at any point within the hemisphere.

3. Can the x-coordinate of the mass center of a homogeneous hemisphere be negative?

Yes, the x-coordinate of the mass center can be negative if the origin is not at the center of the flat base of the hemisphere. In this case, the formula x = (3/8)r will give a negative value for the x-coordinate.

4. How does the density of the hemisphere affect the x-coordinate of the mass center?

The density of the hemisphere does not affect the x-coordinate of the mass center, as long as it is constant throughout the hemisphere. This is because the formula x = (3/8)r only takes into account the radius of the hemisphere and not the density.

5. What is the significance of determining the mass center of a homogeneous hemisphere?

Determining the mass center of a homogeneous hemisphere is important in many engineering and physics applications, as it helps in understanding the stability and balance of the object. It also allows for accurate calculations of forces and moments acting on the hemisphere.

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