Mechanical System: Kinetic Energy of a Rotating Bar with Gravity

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The discussion focuses on calculating the kinetic energy of a mechanical system consisting of a homogeneous bar subjected to gravity, with one end hinged at a fixed point on the y-axis. The y-axis is rotating with an angular velocity ω(t), which influences the kinetic energy of the system. The kinetic energy can be determined by combining the rotational energy about the center of mass and the translational energy of the mass moving with the center of mass's velocity. The parallel axis theorem may also be applied for more accurate calculations. Understanding these principles is crucial for analyzing the dynamics of the rotating bar system.
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Consider a mechanical system (subject to gravity) consisting of an homogeneous bar, lying in the xy -plane, of mass m, length \ell and negligible section, with an extreme hinged in a fixed point P of the y axis. Now, suppose that the y-axis is rotating with angular velocity \omega = \omega(t). What is the kinetic energy of the system?
 
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Hi Goklayeh! :smile:

(have an omega: ω :wink:)

KE of rigid body = rotational energy (about c.o.m.) plus energy of total mass with same velocity as c.o.m.

(or use the parallel axis theorem)
 

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