Kinetic rotational energy of a bar hooked to a coil

  • Thread starter bznm
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  • #1
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I have solved an exercise and I'd like to know if my proceeding about finding kinetic energy is correct or not, because this is the first time that I "meet" a situation like this.

"A bar has mass M and lenght l. Its extremity A is hooked to a coil (with lenght at rest l0), its extremity B is hooked to the point O that is the origin of axes."

I have considered three coordinates: [tex]x[/tex], [tex]y[/tex] (that are the coords of the extremity A on the x-axes and y-axes) and [tex]\theta[/tex] that is the angle that the bar forms with a parallel to the y-axes and I've written
[tex]K=\frac{1}{2}I\omega^2+\frac{1}{2}Mv_{cm}^2=\frac{1}{24}Ml^2\dot\theta^2+\frac{1}{2}M(\dot y^2+\dot x^2+\frac{l^2}{2}\dot\theta^2+\dot x\dot\theta l \cos(\theta)+\dot y\dot \theta l \sin(\theta))[/tex]​
Is it correct? Did I correctly apply Koenig Theorem?
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
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By "coil" you mean "spring"?
Please provide a diagram?

It looks like you are using the moment of inertial for a rad rotating about it's center - but your description has the rod rotating about one end (extremity B).
 

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