Mechanical System: Kinetic Energy of a Rotating Bar with Gravity

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SUMMARY

The discussion focuses on calculating the kinetic energy of a homogeneous bar subjected to gravity, with one end hinged at a fixed point on the y-axis, which is rotating with an angular velocity ω(t). The kinetic energy (KE) of the system is determined by combining the rotational energy about the center of mass (c.o.m.) and the translational energy of the total mass moving with the same velocity as the c.o.m. The parallel axis theorem is also applicable for this calculation, providing a method to find the KE in this mechanical system.

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  • Understanding of rotational dynamics
  • Familiarity with the parallel axis theorem
  • Knowledge of kinetic energy calculations for rigid bodies
  • Basic principles of angular velocity
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  • Learn about the parallel axis theorem in detail
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Mechanical engineers, physics students, and anyone interested in the dynamics of rotating systems will benefit from this discussion.

Goklayeh
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Consider a mechanical system (subject to gravity) consisting of an homogeneous bar, lying in the [tex]xy[/tex] -plane, of mass [tex]m[/tex], length [tex]\ell[/tex] and negligible section, with an extreme hinged in a fixed point [tex]P[/tex] of the [tex]y[/tex] axis. Now, suppose that the [tex]y[/tex]-axis is rotating with angular velocity [tex]\omega = \omega(t)[/tex]. 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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