I've got a problem from my physics course which I am a little stuck on.
A dumbell consisting of two identical masses m=5.8 kg attached to the ends of a thin (massless) rod of length a=0.4 m that is pivoted at its centre. The masses have charge +q and -q and the system is placed in a uniform electric field E=1.6 N/C in the positive x direction. Initially the dumbell is at rest and makes a 60o angle with the x axis. The dumbell is the released.
a) Calculate the horizontal displacement (in meters) of the mass with positive charge since the dumbell was released until the point where the dipole is temporarily aligned with the electric field (Use 3 significant figures in your answer).
b) Calculate the charge q knowing that when the dipole is momentarily aligned with the electric field, its kinetic energy is K=4.9x10^-3 J.
c)Calculate the angular velocity of the dumbell when its position is temporarily aligned with the electic field.
So I've approached the problem by torques. I've tried to find the perpendicular force due to the electric field, and used that to calculate the torque. I then use angular torque*angular displacement = change in angular kinetic energy. This approach doesn't seem to give me the right answer, and what's even more confusing is part c). If I try to answer part c) using K=4.9*10^-3 J as my angular K to get my angular velocity my answer isn't correct, which suggests to me the K given in b) is just linear kinetic energy.
I suspect this question has something to do with electric potential energy, however we haven't covered this content yet, making me think that I've missed something critical.
For reference here's a picture