What is KE of the system using F = ma ?

AI Thread Summary
The discussion revolves around calculating the potential energy (PE) and kinetic energy (KE) of a system of three equal mass particles arranged in an equilateral triangle. The potential energy was calculated using the formula U = -GMm/r, with contributions from each pair of masses, resulting in a total PE of approximately -9.0045x10^-10 J. However, the approach to calculating KE using F = ma was questioned, as it incorrectly multiplied force by three to derive energy. The conversation emphasizes the need to consider the gravitational forces acting on each mass and their respective accelerations, rather than relying on Earth's gravitational acceleration. The participants ultimately agree that a clearer understanding of the forces and accelerations involved is necessary for accurate calculations.
Karina
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Homework Statement


Three particles of mass all = 3 kg are located at the vertices of an equilateral triangle and are spinning about their center of mass in an empty space. The sides are length d = 2 m which doesn't change with time.

Homework Equations


What is the PE of the system? What is the KE of the system (using Newton's second law equation f = ma not the shortcut energy equation)

The Attempt at a Solution


Okay so I believe I found the PE of the system. I used U = - GMm/r to find the PE from 2 of the vertices and since they're all the same mass & same distance U1:2 + U1:3 + U2:3 and U1:2 = U1:3 = U2:3 therefore Utotal = 3U1:2. So I found U1:2 by -6.67x10^-11 (3 kg) (3kg)/ 2 m and I got -3.0015x10^-10 and multiplied by 3 since I have 3 equal vertices = -9.0045x10^-10 J
Now for KE using F = MA, do I just do (3kg)(9.81m/s^2) = 29.43 N and multiply that by 3 as well for a system total of 88.29 J?
 
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Karina said:
I used U = - GMm/r to find the PE from 2 of the vertices and since they're all the same mass & same distance U1:2 + U1:3 + U2:3 and U1:2 = U1:3 = U2:3 therefore Utotal = 3U1:2.
It is not apparent that that method is valid. How did you find the PE of one? However you did it, that energy is (minus) the energy required to remove one off to infinity. But having removed that one, the energy to remove the second is less, and no energy is required to move the third. Or do you mean you found the PE of one with respect to one other, ignoring the third, then multiplied by 3? That might give the right result, but it doesn't strike me as evident.
Karina said:
9.81m/s^2
This isn't Earth surface gravity.
Karina said:
29.43 N and multiply that by 3 as well for a system total of 88.29 J
You're multiplying a force by 3 to get energy?!
You need to compute the velocities, based on F = ma. What acceleration is each undergoing?
 
I found the PE of one with respect to the other because they are all the same mass and distance anyway. So the PE of one is equal to the other 2 as well. V = d/t but the time never changes.. I thought the acceleration would just be g m/s^2..
 
Karina said:
I found the PE of one with respect to the other because they are all the same mass and distance anyway. So the PE of one is equal to the other 2 as well.
It's the PE of the total system that you want. Let's start with something simpler, just two equal masses at distance x. If I regard one as fixed and ask what the PE is of the other I get -Gm2/x. But the PE of the system is (minus) the energy required to send everything off to infinity in different directions. Having spent the energy Gm2/x to send one mass to infinity, no further energy is required to send the other mass to infinity, the total PE is also -Gm2/x.
Karina said:
V = d/t but the time never changes.
I have no idea what you mean. The three masses are orbiting a common centre. For that orbit to be maintained, there is a relationship between their gravitational attraction, the radius of the system, and their speed. What forces act on one mass? What is its acceleration?
Karina said:
I thought the acceleration would just be g m/s^2.
This is not in Earth's gravitational field. g does not enter into it, only G.
 
In this case isn't -Gm^2/x the exact same thing as -GMm/r because the masses are the same so we're just multplying 3x3 for the masses or 3^2 which are both equal to 9. Force of gravity acts on one mass, G.
 
Karina said:
I found the PE of the system. I used U = - GMm/r to find the PE from 2 of the vertices and since they're all the same mass & same distance U1:2 + U1:3 + U2:3 and U1:2 = U1:3 = U2:3 therefore Utotal = 3U1:2.
My apologies - I misunderstood your notation. I now agree with what you have written.
Karina said:
Force of gravity acts on one mass, G.
Yes, the force of gravity acts on each mass, but the force isn't G, or Gm, or gm. What will it be?
If the linear speed of each mass is v, what is its acceleration?
 
It's acceleration is f/m...? Idk. I'm lost now.
 
Karina said:
It's acceleration is f/m...? Idk. I'm lost now.
Draw a diagram. Single out one mass and consider the forces on it from the other two. What is the magnitude and direction of each of those forces? What do they add up to vectorially?
 
Acceleration = 0
 
  • #10
Karina said:
Acceleration = 0
No.
A bob swung around on the end of a string at constant speed is nonetheless accelerating, yes?
 
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