Harmonic motion of the dipole?

[StephenDoty]

the conservation of energy formula for a dipole
\DeltaK=-\DeltaU
(1/2)Iw^2=-(-qEDcos(theta))
(1/2)Iw^2=qEDcos(theta)

E-field going parallel to the y-axis. if the positive end of the dipole was in the first quad with angle theta from the y-axis and the negative end of the dipole was in the third quad, what would be the angular velocity of the dipole at the y-axis? using theta0= theta or the angle the dipole is released.U(pi/2)=0. And what is the period of the harmonic motion of the dipole?

The answer is w=\sqrt{(2qED/I) * (1-cos(theta0)}
1/2Iw^=qEDcos(theta). then to find the answer the change in potential energy has to go to theta0 to pi/2?? Do this work?? How do you prove the U(pi/2)=0 since at pi/2 the potential energy from theta0 has turned to kinetic energy? qED(cos(pi/2)-cos(theta0))= 0?

For the period, The harmonic formula d^2x/dt^2=-w^2x replacing x with theta d^2theta/dt^2 = w*theta
and since I*angular acceleration= torque and torque=qEdsin(theta) but my teacher changed torque to qED*theta. Why??
Then I just used I*d^2theta/dt^2 = torque or I*-w^2*theta=torque to find w. And w=2pi/T to find T.

Why was the torque changed from qEdsin(theta) to qEd*theta? And is the harmonic formula d^2x/dt^2=-w^2x the same no matter what?Thanks for the help.
Stephen

 

[StephenDoty]

I have a test on this tomorrow, so could someone please look at the equations and the questions about how to prove U(pi/2)= 0 and why qEDsin(theta) turned into qED*theta and if the second derivative is always equal to -w^2*x for harmonic motion?


Thank you.
Stephen

 

[StephenDoty]

bump.. Can anyone answer my questions?

Please! I do not understand why qEDsin(theta) turned into qED*theta

 

[alphysicist]

Hi StephenDoty,

The approximation \sin\theta\to \theta is used when the angle \theta is assumed to be small.

I don't understand what you are asking about U(\pi/2). The formula for U shows that it is zero at the angle \pi/2. However, the dipole never gets to that angle, so I'm not sure what you are asking.