Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this torque, I'm not sure

  1. Jul 16, 2009 #1
    When a straight rod is twisted on its' axis, causing it to spin, the top of the rod will tend to go out of center opposite to the bottom end of the rod. If this is not torque what is the force called, that causes this miss alignment referred to.

  2. jcsd
  3. Jul 16, 2009 #2


    User Avatar
    Gold Member

    I'm not sure that I understand the question. It sounds to me as if the rod is not perfectly straight and balanced.
  4. Jul 16, 2009 #3


    User Avatar
    Science Advisor

    If it is what I am thinking you are thinking, this is a dynamic balance issue. The rod is either not completely straight (which is almost always true) or there is a weight imbalance that lies off of the rotational axis. Either way, the spinning creates an imbalance which causes the end of the rod to orbit about the axis point. This is why bearings/bushings are needed in every application with a spinning shaft.
  5. Jul 16, 2009 #4
    Aha, Dynamic balance, So, am I to understand that, if the shaft was perfectly balanced it would not tend to come off of center spin axis...correct? Considering variations in speed remains constant.

    To be more clear to my original post...when I spin an entire front wheel from a bicycle it tends to rotate around an axis that is parallel to the shaft. If you hold the position of the spinning wheel, one end, compared to the other end of the spinning axis, goes in the opposite direction. Is this also dynamic balance/unbalance? Or is this torque?
  6. Jul 16, 2009 #5


    User Avatar
    Gold Member

    Sorry, man... you totally lost me on that last question. Are you starting to bring in gyroscopic effects?
  7. Jul 16, 2009 #6
    I'm obviously confused, hence the title.

    If one spins a rod on its' axis and that rod is equal in manufacture it will tend not to spin out of its' center axis. The action that causes the the orbital wobble is either due to imperfections in manufacture or is it only due to a weight, or unequal weight distribution, that is spinning tangental to the spin axis?

    I imagine that gyroscopic effect would need a weight outside of its' spin axis to be gyroscopic.

    My question is other than inperfections in design would a single spin axis tend to wobble if it were not due to weight outside of its' center axis?
  8. Jul 17, 2009 #7
    Are you just looking for a descriptive word? How about 'disequilibrium' or dynamic equilibrium?

    Once out of alignment, even a perfect rod will still continue to oscillate. It can even oscillate with several nodes. Usually this is called dynamic equilibrium, but this is not the cause but the sustained effect.
  9. Jul 17, 2009 #8
    No I'm not looking for a descriptive word, as such but I am looking for the principals involved that causes a, so designed, dynamic disequilbrium. Meaning that I would like to be able to... Set a spin axis that has a controlled wobble by placing weight of off its center point of spin.
  10. Jul 17, 2009 #9
  11. Jul 17, 2009 #10
    I'm not clear on what you mean by wobble. Try this. Place a 3 ft. length of .060 inch diameter piano wire in a drill motor. Be careful, it will whip around at the rotational velocity if the end isn't constrained. Tape a small screw ~1/3 from the motorl end. That should induce it to the first mode of oscillation.
  12. Jul 18, 2009 #11
    By a controlled wobble I guess I mean an oscillation, precession. I would imagine that I would need to place an equal weight all around the spin axis but have the plane perpendicular to some point out side of center spin axis?

    What I needed to know on the outset of my post was to determine that the center shaft contributes very little to off center rotation, It is more the so designed weight that contribute to torque...Yes?
  13. Jul 18, 2009 #12
    I'm confused by what you are after, so I read back a little.

    This is gyroscopic precession.
  14. Jul 18, 2009 #13
    Thanks for your patience, yes I must agree that this is gyroscopic precession. What I had to confirm was is the precession, more so due to the spinning shaft or the weight that is housed outside of the center axis that causes this precession. It's pretty clear now that the weight is more responsible than a perfectly balanced shaft, thanks to all that responded.

    If I understand this correctly...

    My design incorporates a shaft and a weight, much like a gyroscopes ring, but the ring is not perpendicular to the shaft. It is situated so that it touches one side of the axis and touches the opposite of the other end of the shaft. This causes the shaft to spin out of its' center point of axis, a wobble. The end points, of the shaft, that protrudes equally beyond the rings diameter on both sides, will ride in a circular groove(channel) that is housed, perpendicular to the inside wall of a cylinder. When I spin that shaft c/w the off center weight ring the shaft will experience a torque and run on the upper part of the channel while the opposite end of the shaft will run on the bottom of that channel causing the shaft and weight assembly to rotate in the channel as well as spin on its axis.
    Last edited: Jul 18, 2009
  15. Jul 19, 2009 #14


    User Avatar
    Science Advisor

  16. Jul 19, 2009 #15
    Thanks for the link and terminologies involved. ....Nutation....
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Is this torque, I'm not sure
  1. Torque calculation (Replies: 3)

  2. Torque on a shaft (Replies: 3)

  3. Torque amplification (Replies: 5)