# What is KE of the system using F = ma ?

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1. Dec 9, 2014

### Karina

1. The problem statement, all variables and given/known data
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.

2. Relevant 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)

3. 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?

2. Dec 9, 2014

### haruspex

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.
This isn't earth surface gravity.
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?

3. Dec 9, 2014

### Karina

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..

4. Dec 9, 2014

### haruspex

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.
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?
This is not in Earth's gravitational field. g does not enter into it, only G.

5. Dec 9, 2014

### Karina

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.

6. Dec 9, 2014

### haruspex

My apologies - I misunderstood your notation. I now agree with what you have written.
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?

7. Dec 9, 2014

### Karina

It's acceleration is f/m...? Idk. I'm lost now.

8. Dec 9, 2014

### haruspex

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?

9. Dec 9, 2014

### Karina

Acceleration = 0

10. Dec 9, 2014

### haruspex

No.
A bob swung around on the end of a string at constant speed is nonetheless accelerating, yes?