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Rotating rod

  • Thread starter Ene Dene
  • Start date
  • #1
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Homework Statement


The rod of length 2a swings and it rotates [itex]\dot{\phi}[/tex]. Find kinetic energy.



The Attempt at a Solution



I know how to find kinetic energy if we have rotation just in xy plane, but I'm having a problem to understand how to do it with rotation around y axis.

If we have rotation in xy plane:
[tex]x_{cm}=asin(\theta)[/tex]
[tex]y_{cm}=acos(\theta)[/tex]
[tex]\dot{x}_{cm}=acos(\theta)\dot{\theta}[/tex]
[tex]\dot{y}_{cm}=-asin(\theta)\dot{\theta}[/tex]
[tex]E_{kin}=\frac{1}{2}m(\dot{x}_{cm}^2+\dot{y}_{cm}^2)+\frac{1}{2}m\cdot\frac{1}{12}(2a)^2\dot{\theta}^2[/tex]
[tex]E_{kin}=\frac{2}{3}m\dot{\theta}^2[/tex]

For rotation around y, I assume that we have translation component in z direction, and rotation component.
[tex]z_{cm}=asin(\theta)sin(\phi)[/tex]
Is this correct? (look at the picture)
[tex]\dot{z}_{cm}=a(cos(\theta)sin(\phi)\dot{\theta}+cos(\phi)sin(\theta)\dot{\phi})[/tex]
[tex]E_{z}=\frac{1}{2}m\dot{z}_{cm}^2+\frac{1}{2}I_{cm}\dot{\phi}^2[/tex].
Rotation part of [itex]E_{z}[/itex] bothers me, maybe I have to multiply it by [itex](lsin(\theta))^2[/itex].
 

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Answers and Replies

  • #2
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Is this ok:

[tex]E=\frac{1}{2}(\dot{x}_{cm}^2+\dot{y}_{cm}^2+\dot{z}_{cm}^2)+\frac{1}{2}I_{cm}(\dot{\theta}^2+\dot{\phi}^2)[/tex]

?
 
Last edited:
  • #3
48
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I have successfully solved the problem by my self, so no need for an answer.
 

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