# Electric Dipoles and angular velocity

1. Sep 6, 2008

### StephenDoty

Consider an electric dipole located in a region with an electric field of magnitude E pointing in the positive y direction. The positive and negative ends of the dipole have charges + q and - q, respectively, and the two charges are a distance D apart. The dipole has moment of inertia I about its center of mass. The dipole is released from angle theta=theta_0, and it is allowed to rotate freely. What is the max angular velocity, the magnitude of the dipole's angular velocity when it is pointing along the y axis? ( See picture bellow)

I dont even know where to begin. I know U= - P X E, but I do not know what to do from here.

Any help would be appreciated.
Thanks.
Stephen

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2. Sep 6, 2008

### Gear300

Not sure how much this will help but...dU/d(theta) is torque, t. dt/d(time) is angular momentum, L. L = I*angular velocity. Find the optimal value for L.

3. Sep 6, 2008

### jackiefrost

I replied to your other, similar, post. That should get you started on this one.

When you can arrive at an expression for the potential energy at theta, that is the energy that's available for conversion to rotational kinetic energy when the dipole is set free to rotate. You just need to think a bit on how to calculate angular velocity if you know the rotational kinetic energy.

Also - think about the process of how some particular quantity of potential energy (before being released) will get converted to rotational kinetic energy. At what point in its arc will it "run out" of potential energy? [Think about how it aquired it in the first place].

jf

Last edited: Sep 6, 2008
4. Sep 6, 2008

### StephenDoty

would you find the potential energy by integrating the torque and multiplying by a -1? And then would I set it equal to .5Iw^2?

Thanks for the help.
Stephen

5. Sep 7, 2008

### alphysicist

Hi StephenDoty,

I don't believe the formula for the potential energy in your original post is quite right. It should be

$$U = - \vec p \cdot\vec E$$
and so the potential energy at a particular orientation depends on the dot product of those two vectors. Does that help?

6. Sep 7, 2008

### StephenDoty

I meant the dot product. I mis typed

How would I solve the problem? Should I solve the dot product and set it equal to .5Iw^2?

Thanks for the responses.
Stephen

7. Sep 7, 2008

### alphysicist

The decrease in potential energy in going from the initial to the final point is equal to the increase of kinetic energy. What does that give?

8. Sep 7, 2008

### StephenDoty

max kinetic energy at that final point. So what do I set .5Iw^2 to?

9. Sep 7, 2008

### alphysicist

What is the decrease in potential energy (or change in potential energy, depending on how you write the energy equation) as the dipole goes from the initial point (at angle $\theta_0$) to the final point where it points along the $y$-axis?

10. Sep 7, 2008

### StephenDoty

change in PE = - P dot E

11. Sep 7, 2008

### alphysicist

No, the initial potential energy is found from calculating $-\vec p\cdot \vec E$ when the dipole is in its first orientation, and the final potential energy is found from calculating $-\vec p\cdot \vec E$ when the dipole is at the second point. The change is the final potential energy minus the initial potential energy.

12. Sep 7, 2008

### StephenDoty

ok how do I find the angular velocity from -Pinitial dot E to -Pfinal dot E???

13. Sep 7, 2008

### alphysicist

The problem statement gives you E, q, D, and theta_0 so those should be in your answer. Using those quantities only, what is (-Pinitial dot E)? and what is (-Pfinal dot E)?

Once you have those two, you can subtract them to find out how much the potential energy is decreasing, which will equal how much the kinetic energy is increasing. At that point you can use the kinetic energy formula to relate angular velocity to the rotational kinetic energy.

14. Sep 7, 2008

### StephenDoty

so
-PEcos(theta) = .5 I w^2?

15. Sep 7, 2008

### alphysicist

No, since energy is conserved you have:

\begin{align} \Delta PE + \Delta KE &= 0 \nonumber\\ (PE_f - PE_i) + (KE_f - KE_i) &= 0 \nonumber \end{align}

Since it starts from rest, $KE_i$ will be zero, but your equation will still have three nonzero terms. Your last post is correct that $KE_f=\frac{1}{2}I\omega_f^2$. So what do you get for $PE_f$ and $PE_i$?

Remember that the initial angle is $\theta_0$; what is the angle to use for the final potential energy?

16. Sep 8, 2008

### StephenDoty

so
(PEcos(pi/2)- PEcos(theta 0) = .5Iw^2??

17. Sep 8, 2008

### alphysicist

I don't believe the final angle is $\pi/2$. The angle between the dipole and the electric field is zero degrees at the final point.