Kinetic Energy of Rotating+Translating Bar

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The discussion focuses on calculating the kinetic energy of a bar that rotates while attached to a cart moving in the x-direction. The total kinetic energy (KE) of the system is expressed as the sum of translational and rotational components. The translational kinetic energy is given by the formula 1/2 * m * v_C.M^2, where m is the mass of the bar and v_C.M is the velocity of its center of mass. The rotational kinetic energy is represented by 1/2 * ω · L_C.M, where ω is the angular velocity and L_C.M is the angular momentum of the bar relative to its center of mass. The discussion emphasizes understanding these components to accurately calculate the overall kinetic energy of the rotating and translating bar.
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:
KE=\frac{1}{2}m\vec{v}_{C.M}^{2}+\frac{1}{2}\vec{\omega}\cdot\vec{L}_{C.M}

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