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Torque Free Precession

  1. Dec 2, 2005 #1
    What is torque free precession, and why does it occur? I'm surprised that it predicts that a solid cylinder rotating in space would "wobble" a bit. Why is it not possible to have a cylinder rotating around a constant axis?

    I understand how it comes from Euler's equations, but I'm not too sure I understand the equations in the first place.
  2. jcsd
  3. Dec 2, 2005 #2
    Essentially, torque expresses how a force F results in a rotation of the body it is acting on. That is why the torque is defined as a vector product.

    Torque governs the dynamics of rotating inertia so you really need it.
    I am sure that you know Newton's second law F=ma

    Now, one can translate this in angular form by inserting torque and looking at the tangential force-component (this component yields the circular motion) :
    [tex]F_t = ma_t[/tex]
    [tex]torque = F_tr = ma_tr[/tex]
    realizing that
    [tex]a_t = \alpha r[/tex]

    [tex]torque = m(\alpha r)r = (mr^2) \alpha[/tex]

    The nutation movement is determined by the normal force component and this is what the Euler equations (well, in an easy language) express.
    Also keep in mind that in the Euler equations, the moment of inertia is very important. So, the mass-distribution throughout the object's volume is a very important parameter that determines the dynamics of the object.


    ps : as a piece of advice, when studying these equations, be sure to know what variables you have and how they influence the object's motion. Just like what i told you about the moment of inertia I.
    I know this sounds a bit vague but i do not know how well you know this theory...
  4. Dec 2, 2005 #3
    Ohh yes and torque free rotations actually mean free rotations. This means, there is no force acting onto an already rotating object, that can change the direction of the angular impuls momentum L (you know dL/dt = torque = 0 this L is constant in time)

    See for example this or this

  5. Dec 2, 2005 #4
    I appreciate your response, but I think you might have misunderstood what I'm asking.

    I'm familiar with torque, the moment of inertia tensor, and the angular velocity and angular momentum vectors.

    My question deals with a phenomenon predicted by euler's equations called torque free precession. See http://en.wikipedia.org/wiki/Precession
    Why does this occur? Why isn't a plate stable rotating solely around it's axis of symmetry?
  6. Dec 2, 2005 #5
    Woops we posted at about the same time. I'll look at those links. Thanks!
  7. Dec 2, 2005 #6
    Okay, perhaps I'm misunderstanding all together. It seems to me now that rotating purely about the axis of symmetry is only one of many stable configurations, and other configurations where w precesses around L are also stable configurations. Is this correct?
  8. Dec 3, 2005 #7
    If I understand this correctly (it's been a year since we covered it in my mechanics class), you're only going to see this effect when your angular momentum vector does not point exactly along one of the axes of symmetry. Note how the [tex]\Omega[/tex] term (following the notation here) only has an effect on the equations of motion when either [tex]\omega_1[/tex] or [tex]\omega_2[/tex] are non-zero, in addition to [tex]\omega_3[/tex] being non-zero.

    I think that the real source of confusion here is that we're looking at the situation from two different refrence frames: that of the object which is rotating and an external inertial refrence frame. In the inertial frame everything happens the way you expect it to, but if you look at things from the refrence frame of the object things start to get weird.

    A simple example of this is the case of a cylinder. Imagine that you're out in space, define a set of axis such that z is up towards your head, x is to your right, and y is straight out in front. Align your cylinder straight up along z, then rotate it 45 degrees into the x direction and bring it to a stop so that it's now pointing at a diagonal. Give the cylinder a spin by pushing forward on the top end and pulling back on the bottom end. Assuming you can impart identical impulse forces to both ends this will give it an angular momentum along the z axis.

    This rotation will be stable, you could sit there and watch that cylinder spin about the z axis for all eternity and nothing would change. Consider what it looks like from the refrence frame of the cylinder though. Define the [tex]e_3[/tex] axis as being along the major axis of the cylinder, then define [tex]e_1[/tex] and [tex]e_2[/tex] perpindicular to [tex]e_3[/tex]. Imagine watching these axes, glued onto your cylinder, spin around in front of you; and now imagine the angular momentum vector pointing straight up. You should be able to see (a visual aid is helpful here, I used an empty soda can) that the component of the angular momentum along the [tex]e_3[/tex] axis remains constant, while the [tex]e_1[/tex] and [tex]e_2[/tex] axes pick up more or less of a projection due to their rotation.

    It is that effect which the Euler equations are describing and which is termed torque-free precession.
    Last edited: Dec 3, 2005
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