Kinetic Energy of Rotating+Translating Bar

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SUMMARY

The kinetic energy of a rigid bar that is both rotating and translating can be expressed using the formula KE = (1/2)m(vC.M)2 + (1/2)(ω·LC.M), where m is the mass of the bar, vC.M is the velocity of the center of mass, ω is the rotational velocity, and LC.M is the angular momentum relative to the center of mass. This formula accounts for both translational and rotational kinetic energy components. Understanding these terms is crucial for analyzing the dynamics of systems involving rigid bodies in motion.

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enkar
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If a cart is moving in the x-direction and has a bar (not a pedulum) attached to its center that will rotate, what are the terms of the kinetic energy? I'm having a hard time figuring out what the kinetic engery of the rotating+translating bar is. Can someone break it all down into each of the terms, please?
 
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Since the bar is a rigid object(?), its kinetic energy may be written as:
[tex]KE=\frac{1}{2}m\vec{v}_{C.M}^{2}+\frac{1}{2}\vec{\omega}\cdot\vec{L}_{C.M}[/tex]

Here, we have:
m-object's mass
[tex]\vec{v}_{C.M}[/tex]-velocity of center of mass (C.M)
[tex]\vec{\omega}[/tex]-rotational velocity of object
[tex]\vec{L}_{C.M}[/tex]-angular momentum of object, computed relative yo C.M
 

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