How does rotation influence the motion of objects like a thrown pen?

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Rotation of objects like a thrown pen is complex, involving both spin and wobble. When a pen is thrown, it rotates around its longitudinal axis while also experiencing a wobble due to mass imbalances and external forces, leading to a non-returning position after a complete rotation. The angular momentum of the pen consists of two components: the orbital momentum from its trajectory and the spin momentum from its rotation. Accurate calculations require breaking down the system into simpler components and integrating them to understand the overall motion. This rigid body problem illustrates the intricacies of rotational dynamics in real-world scenarios.
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Can someone explain to me, how does rotation work?

I thought before that rotation of an object could be described by an axis through the mass center of the object, and an angular velocity (this would be equal to describing the rotation as an angular momentum). This means that the rotation would cause the object to turn around 360 degrees in a certain time, and come back to the same possition after a whole turn.

But if I throw a pen up in the air, it won't be as simple as that. It will rotate fast around one axis (the one going through both ends of the pen), but it will also wobble. So after it has turned one time around the axis, it will not be back in the same possition, since it has also wobbled a little bit. So, how do you describe rotation, and how do you use it in calculations?
 
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Your model description is one of a simplistic system. The pen in question is not symmetrical in surface area or mass dispersion. When thrown in the air it is not rotating along its mass centers, nor are the frictional effects of air interaction symmetrical. Your pen is both rotating along its longitudinal center and its lateral centers. Initially the spins have axix induced by the external forces of the throw, not the centers of mass. The mass imbalances are causing direction changes interacting with gyroscopic forces of the spinning pen. If the pen had sufficient time to fall without additional forces of friction it would stabilize into a mass balanced centered single axis spin, although it would not be the pen's visual center by any means. As far as calculations go, you must analyze the system into smaller more simplistic ones then integrate the small systems into an overall complex system.
 
TriKri said:
Can someone explain to me, how does rotation work?

I thought before that rotation of an object could be described by an axis through the mass center of the object, and an angular velocity (this would be equal to describing the rotation as an angular momentum). This means that the rotation would cause the object to turn around 360 degrees in a certain time, and come back to the same possition after a whole turn.

But if I throw a pen up in the air, it won't be as simple as that. It will rotate fast around one axis (the one going through both ends of the pen), but it will also wobble. So after it has turned one time around the axis, it will not be back in the same possition, since it has also wobbled a little bit. So, how do you describe rotation, and how do you use it in calculations?
This is a rigid body problem. Basically the pen will rotates on itself and in the same time its center of mass would describe a parabola. The angular momentum of the pen is in this case the sum of 2 angular momentums : the orbital one (due to the translation motion of the pen into the air) and the spin one (due to its rotation on itself). So it is a common problem (1st year university) to solve. (if you are asking the equations of motion. From it you can determinate what would be the velocity of any point of the pen at any given time).
You might like to read http://en.wikipedia.org/wiki/Rigid_body and http://en.wikipedia.org/wiki/Rigid_body_dynamics.
Have a nice time.
P.S.: I misunderstood something, but I think I get it now. You mean it would rotates around 2 axis so just do the calculations for an axis and do the same for the other axis. The problem remains solved I believe.
 
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fluidistic said:
The angular momentum of the pen is in this case the sum of 2 angular momentums : the orbital one (due to the translation motion of the pen into the air) and the spin one (due to its rotation on itself). So it is a common problem (1st year university) to solve.

Don't know if it is first year thing, cause if these two motions alone the posed returning of starting point would be there. I see at least three motions in here...

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