Two balls carrying the same charge

Doc Al

I would use r for both, where r is the distance between the centers of the balls.
What's the velocity at the initial position? And at the highest position? (You're solving for that highest position.)

You'll take advantage of the fact that the total mechanical energy remains constant. Evaluate it at the initial position.

jgridlock

Okay so the r is going to be equal to .06m.

The velocity at the initial position is zero, so I set v = 0?

1/2(.012)(0) + (.012)(9.8)(.06) + (8.99e9(.45uC*.45uC)/(.06)) = .03739725

Doc Al

Yes, at the initial position.
Right.
I didn't check your arithmetic, but that's the right idea. Now set up a general equation and solve for the value of r at the final position.

jgridlock

So the number I just calculated is for the TME at r initial?

jgridlock

Thanks for all the help, sorry I need so much support.

jgridlock

Should the new equation be:

1/2(.012)(0) + (.012)(9.8)(.06) + (8.99e9(.45e9*.45e9)/(r)) = .03739725

Where I solve for r?

Doc Al

Yes, but since TME is conserved it is the value for any r.

Doc Al

Almost. But don't plug in any values for r. (Like your second term.)

(And make sure you use the right charges. .45μC = .45e-6C, not .45e9C.)

jgridlock

Alright, my final question is about the non mathematical part. Letters b-f (Correct me if im wrong). When the balls are released the top one should reach a max height and then float above the second one because they repel each other. But then what happens when the friction comes into play?

Doc Al

I'd say that the top ball will reach a max height then fall back down to the lowest point again. It will keep oscillating, as energy is conserved.
Eventually friction will dissipate the energy and the ball will end up at some equilibrium position.

SammyS

This depends upon the height of the cylinder versus the equilibrium position of the top ball.

It may continue to oscillate above the cylinder.

Doc Al

Good point.