Find Center of Mass: Attached Problem

In summary, when calculating the center of mass of a system with non-point masses, each mass's center of mass should be considered. The distances from the reference point to each center of mass should be measured and used in the calculation. In this problem, the distances for masses 1 and 2 are correct, but the distance for mass 3 should be measured from the left edge of mass 1, making it 10m instead of 8m. The final answer for the center of mass is 6.67m.
  • #1
jacy
76
0
hi,
I am finding the center of mass in this problem. I have it as an attachment. Please take a look, thanks.
 

Attachments

  • center_of_mass.doc
    19.5 KB · Views: 154
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  • #2
Your attachment shows three masses, but I can't tell their positions from what's written. (It's just not clear to me.)
 
  • #3
Doc Al said:
Your attachment shows three masses, but I can't tell their positions from what's written. (It's just not clear to me.)

Here is that file again. Thanks for your help.
 

Attachments

  • center_of_mass.doc
    20.5 KB · Views: 135
  • #4
OK, now it's a bit clearer. The first thing to realize is that the masses are not point masses, but have length. Assuming that they are uniform, each mass has its own center of mass, right at its center. When calculating the center of mass of the system, you need to measure the distance of the center of each mass from your reference point.

Give it another shot.
 
  • #5
Doc Al said:
OK, now it's a bit clearer. The first thing to realize is that the masses are not point masses, but have length. Assuming that they are uniform, each mass has its own center of mass, right at its center. When calculating the center of mass of the system, you need to measure the distance of the center of each mass from your reference point.
Give it another shot.


Thanks again, so the distance for mass 1 will be 1m from my reference point, for mass 2 it will be 6m, for mass 3 it will be 8m. Am i correct.

center of mass = (20(1) + 30(6) + 40(8))/ 90
= 5.78 m
 
  • #6
Recheck the distance to the cm of the third mass.
 
  • #7
jacy said:
so the distance for mass 1 will be 1m from my reference point, for mass 2 it will be 6m, for mass 3 it will be 8m. Am i correct.
Measured from the left edge of mass 1, I'd say that your distances are correct for masses 1 and 2, but not for mass 3.
 
  • #8
Doc Al said:
Measured from the left edge of mass 1, I'd say that your distances are correct for masses 1 and 2, but not for mass 3.


Thanks, the distance for mass 3 will be 10m from the left edge of mass 1, correct.

center of mass = (20(1) + 30(6) + 40(10))/90
= 6.67 m
This will be the answer, thanks for ur help.
 

1. What is the center of mass?

The center of mass is the point at which the mass of an object or system is evenly distributed in all directions. This point is also known as the center of gravity.

2. Why is finding the center of mass important?

Finding the center of mass is important for understanding the stability and equilibrium of objects or systems. It is also used in various fields such as physics, engineering, and astronomy to calculate the motion and dynamics of objects.

3. How do you calculate the center of mass?

The center of mass can be calculated by dividing the total mass of an object or system by the total distance of each individual mass from a reference point. Another method is to use the formula xcm = (m1x1 + m2x2 + ... + mnxn) / (m1 + m2 + ... + mn), where x is the distance and m is the mass.

4. Can the center of mass be outside of an object?

Yes, the center of mass can be outside of an object. This usually occurs when the object has an irregular shape or when there are multiple objects involved. In these cases, the center of mass may be located in empty space.

5. How does the center of mass change with different objects or systems?

The center of mass changes depending on the distribution of mass within an object or system. For example, a symmetrical object will have its center of mass at its geometric center, while an asymmetrical object will have its center of mass shifted towards the heavier side.

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