Equation of motion for a rigid body

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The discussion focuses on deriving the equation of motion for a rigid body subjected to an external force acting at a point not coinciding with its center of mass. The participants emphasize separating the motion into translational and rotational components, with the translational motion described by F=ma and the rotational motion expressed as torque, τ = r x F. They clarify that the angular momentum is related to the inertia tensor and angular velocity, leading to the equation r x F = Iα, but note complications arise when considering non-inertial frames. An additional term related to the angular velocity and inertia tensor is introduced, highlighting the complexity of time derivatives in rotating frames. The conversation concludes with confirmation that F=ma remains valid regardless of the force's application point.
Gavroy
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hi

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.
 
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Hi Gavroy! :smile:
Gavroy said:
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.

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)
 
tiny-tim said:
r x F = Iα
That isn't quite correct. What is valid is that angular momentum is the product of the inertia tensor and angular velocity: \vec L = \mathbf I\,\vec{\omega}.

Now differentiate both sides. The left hand side is easy: It's torque, which is given by \vec {\tau} = \vec r \times \vec F. 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
\vec r \times \vec F = \mathbf I \frac{d\vec{\omega}}{dt} + \vec{\omega}\times (\mathbf I\,\vec{\omega})
 
D H said:
\vec r \times \vec F = \mathbf I \frac{d\vec{\omega}}{dt} + \vec{\omega}\times (\mathbf I\,\vec{\omega})

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?
 
That extra term results from operating in a rotating frame. For any vector quantity \vec q, the time derivatives of that vector from the perspectives of inertial and rotating observers are related via
<br /> \left(\frac {d\vec q}{dt}\right)_{\text{inertial}} =<br /> \left(\frac {d\vec q}{dt}\right)_{\text{rotating}} +\quad<br /> \vec{\omega} \times \vec q
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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