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Equation of motion for a rigid body

  1. May 31, 2012 #1

    let me says you have an arbitrary rigid body and a force F acts on this body at some point A, which is not the center of mass that is called C.

    how do you get the equation of motion?

    my idea was to separate out translation and rotation:

    maybe the whole body moves by F=ma
    and the rotation is given by: D=r x F where r is the distance between A and C.

    but i am not sure at all about this.

    by the way: i am rather looking for a general idea that would give me the equation of motion.
  2. jcsd
  3. May 31, 2012 #2


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    Hi Gavroy! :smile:
    yes, F = ma,

    and r x F = Iα

    where I is the moment of inertia about an axis through the centre of mass and parallel to r x F

    (but if that axis is not a principal axis of the body, I needs to be the inertia tensor)
  4. May 31, 2012 #3

    D H

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    That isn't quite correct. What is valid is that angular momentum is the product of the inertia tensor and angular velocity: [itex]\vec L = \mathbf I\,\vec{\omega}[/itex].

    Now differentiate both sides. The left hand side is easy: It's torque, which is given by [itex]\vec {\tau} = \vec r \times \vec F[/itex]. The right hand side is a bit tougher. In which frame? Which derivative? The inertia tensor is time varying from the perspective of an inertial frame. The inertia tensor for a rigid body is constant in a frame fixed with respect to the body, but now you have the problem of take time derivatives in a non-inertial frame. It's easier than making the inertia tensor time varying, but it is not that simple r x F = Iα. Instead you get
    [tex]\vec r \times \vec F = \mathbf I \frac{d\vec{\omega}}{dt} + \vec{\omega}\times (\mathbf I\,\vec{\omega})[/tex]
  5. Jun 1, 2012 #4
    interesting, thank you. can you give me a hint where this extra term comes from or how i can get this one? (i am referring to the cross prouct (angular velocity cross inertia tensor dot angular velocity)

    by the way: where i right about the translation given by F=ma? i am a little sceptical now, because if this equation is right it would not make a difference (referring to the translation) whether the force acts on the center of mass or an arbitrarily chosen other point of the body?
  6. Jun 1, 2012 #5

    D H

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    That extra term results from operating in a rotating frame. For any vector quantity [itex]\vec q[/itex], the time derivatives of that vector from the perspectives of inertial and rotating observers are related via
    \left(\frac {d\vec q}{dt}\right)_{\text{inertial}} =
    \left(\frac {d\vec q}{dt}\right)_{\text{rotating}} +\quad
    \vec{\omega} \times \vec q[/tex]
    This is called the transport theorem. Wiki reference: http://en.wikipedia.org/wiki/Rotati...tives_in_the_two_frames[url]. And yes, F=ma.
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