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.
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?