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Rotational dynamics: Rotating rod with two attached balls about a non principal axis

  • Thread starter Nacho_rc11
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Homework Statement:

The object here is a rod with two point particles attached, one to each side of the rod.
This rod has an inclination (α degrees) with respect to the vertical axis (z), but it rotates around the z axis.
Prove that Angular Momentum is not conserved with time, express Angular Momentum as a matter of time, find Torque with respect to a principal axis.

Relevant Equations:

First time here, don't know what to put
Firstly I deduced that in this situation the moment of inertia I, is not going to be parallel to w.
And I calculated it as a matter of the angle, for the rod and the two point particles attached (with a mass 'm'), and the total moment of Inertia ended up being:
I=((R²*sin²α)/2)*(M/6 + m)
Being M the mass of the rod and m the mass of each ball.
Now I'm stuck, could anyone help
 

Answers and Replies

  • #2
BvU
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Hello nacho, :welcome: !
moment of inertia I, is not going to be parallel to w
that would be difficult: one (##\vec \omega##) is a vector (with a direction and a magnitude) and thte other is a number.

To 'Prove that Angular Momentum is not conserved with time' all you have to do is to 'express Angular Momentum as a matter function of time' so your 'don't know what to put 'is answered: you need the definition of angular momentum ##\vec L##
 
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  • #3
Hello nacho, :welcome: !
that would be difficult: one (##\vec \omega##) is a vector (with a direction and a magnitude) and thte other is a number.

To 'Prove that Angular Momentum is not conserved with time' all you have to do is to 'express Angular Momentum as a matter function of time' so your 'don't know what to put 'is answered: you need the definition of angular momentum ##\vec L##
Sorry, I meant that the angular momentum was not going to be parallel to 'w', I made a mistake.
Anyway, my question is how could I express mathematically this angular momentum as a function of time.
Thanks
 
  • #4
BvU
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you need the definition of angular momentum ##\vec L##
write it down and work it out ...
 
  • #5
haruspex
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one (##\vec \omega##) is a vector (with a direction and a magnitude) and thte other is a number.
In a problem like this, the moment of inertia needs to be treated as a tensor.
But if unfamiliar with that, the easier way is to consider the two masses separately to start with. Each has a linear momentum, and these both lie in a tilted plane. The overall angular momentum is a vector normal to that plane.
 
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  • #6
BvU
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Have we lost @Nacho_rc11 ?

Can I ask for the context of this exercise ? Is it in an introductory part for mechanics of rotation, or in a more advanced course ?

I can imagine a simple case where one of the particles is on the z axis and it is clear that ##\vec L = \vec r\times \vec p ## is describing a cone.

Or are we supposed to have the z axis through the center of the rod ?
 
  • #7
haruspex
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I can imagine a simple case where one of the particles is on the z axis and it is clear that ##\vec L = \vec r\times \vec p ## is describing a cone.

Or are we supposed to have the z axis through the center of the rod ?
Really? If one particle is on the axis, that contributes nothing, surely. So that leaves us a single particle moving in a circle. Its instantaneous velocity would give it angular momentum about a chosen axis, but not an inherent angular momentum.
I would say the axis of rotation of the rod is through its centre.
 
  • #8
kuruman
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I would say the axis of rotation of the rod is through its centre.
I agree. Also, this looks like an application of Euler's equations for rigid body rotation.
 
  • #9
BvU
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@Nacho_rc11 ? (Not seen since Thursday ...)

What is the full problem statement ?
 
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