Kinetic energy of a rod in a circle

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SUMMARY

The discussion focuses on deriving the Lagrangian for a uniform rod constrained to slide within a circular boundary, subtending an angle of 120 degrees. The kinetic energy (KE) of the system is primarily rotational, expressed as KE = ½ I ω², where I is the moment of inertia. The challenge lies in accurately determining the moment of inertia for the rod's new configuration using the parallel axis theorem. The participants emphasize the importance of understanding the varying velocities of the rod's ends compared to its center of mass.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with rotational dynamics and moment of inertia
  • Knowledge of the parallel axis theorem
  • Basic concepts of kinetic and potential energy
NEXT STEPS
  • Study the derivation of the Lagrangian for constrained systems
  • Learn about the parallel axis theorem in detail
  • Explore examples of rotational kinetic energy calculations
  • Investigate the dynamics of systems with multiple rods or pendulums
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Physics students, mechanical engineers, and anyone interested in advanced dynamics and Lagrangian mechanics.

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Homework Statement



Suppose that we have a uniform rod inside a circle. It's free to slide with its ends on the inside of the circle, and it subtends an angle of 120 degrees at the center of the circle. I'm looking for the Lagrangian of this system.

Homework Equations



I = ml^2/12 for the rod about its center
L = T-V

The Attempt at a Solution



Here's the problem. It seems to me that the ends of the rod (near the circle) are moving faster than the middle of the rod... which would mean that I can't use use the center of mass coordinates to find the kinetic energy. So I'm a bit confused about how to establish the correct kinetic energy - this is where I need help (I can get the potential, I think).

Can anyone help?
 
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Imagine your rod is attached to a massless 2nd rod that can rotate about its other end. You can now rotate this 2nd rod and cause your actual rod to move in a circle. This is the exact same setup as you have now with the circular boundary.

So it will only have rotational kinetic energy defined by:

[tex]KE = \tfrac12 I \omega^2[/tex]

The difficulty is finding the moment of inertia for this new setup. Although you can find it easily using the parallel axis theorem.

EDIT: It reminds me of the ride called the Sea Dragon, or Kamikaze, where you are seated in a boat that swings back and forth (and sometimes all the way around in a circle). Where the boat is your rod, and it swings around a pivot that is attached to by massless arms.
 

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