the conservation of energy formula for a dipole(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\Delta[/tex]K=-[tex]\Delta[/tex]U

(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=[tex]\sqrt{(2qED/I) * (1-cos(theta0)}[/tex]

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

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# Homework Help: Harmonic motion of the dipole?

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