1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Dipoles and angular velocity

  1. Sep 6, 2008 #1
    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
     

    Attached Files:

  2. jcsd
  3. Sep 6, 2008 #2
    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.
     
  4. Sep 6, 2008 #3
    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
  5. Sep 6, 2008 #4
    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
     
  6. Sep 7, 2008 #5

    alphysicist

    User Avatar
    Homework Helper

    Hi StephenDoty,

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

    [tex]
    U = - \vec p \cdot\vec E
    [/tex]
    and so the potential energy at a particular orientation depends on the dot product of those two vectors. Does that help?
     
  7. Sep 7, 2008 #6
    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
     
  8. Sep 7, 2008 #7

    alphysicist

    User Avatar
    Homework Helper

    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?
     
  9. Sep 7, 2008 #8
    max kinetic energy at that final point. So what do I set .5Iw^2 to?
     
  10. Sep 7, 2008 #9

    alphysicist

    User Avatar
    Homework Helper

    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 [itex]\theta_0[/itex]) to the final point where it points along the [itex]y[/itex]-axis?
     
  11. Sep 7, 2008 #10
    change in PE = - P dot E
     
  12. Sep 7, 2008 #11

    alphysicist

    User Avatar
    Homework Helper

    No, the initial potential energy is found from calculating [itex]-\vec p\cdot \vec E[/itex] when the dipole is in its first orientation, and the final potential energy is found from calculating [itex]-\vec p\cdot \vec E[/itex] when the dipole is at the second point. The change is the final potential energy minus the initial potential energy.
     
  13. Sep 7, 2008 #12
    ok how do I find the angular velocity from -Pinitial dot E to -Pfinal dot E???
     
  14. Sep 7, 2008 #13

    alphysicist

    User Avatar
    Homework Helper

    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.
     
  15. Sep 7, 2008 #14
    so
    -PEcos(theta) = .5 I w^2?
     
  16. Sep 7, 2008 #15

    alphysicist

    User Avatar
    Homework Helper

    No, since energy is conserved you have:

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

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

    Remember that the initial angle is [itex]\theta_0[/itex]; what is the angle to use for the final potential energy?
     
  17. Sep 8, 2008 #16
    so
    (PEcos(pi/2)- PEcos(theta 0) = .5Iw^2??
     
  18. Sep 8, 2008 #17

    alphysicist

    User Avatar
    Homework Helper

    I don't believe the final angle is [itex]\pi/2[/itex]. The angle between the dipole and the electric field is zero degrees at the final point.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?