Rigid body kinetic energy+ constraints (upper level classical mechanics)

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SUMMARY

This discussion focuses on calculating the kinetic energy of a homogeneous cylinder of radius a rolling inside a cylindrical surface of radius R, where R > a. The kinetic energy formula used is T = (mv^2_{CM})/2 + (I_1ω^2_x + I_2ω^2_y + I_3ω^2_z)/2, with moments of inertia I_1 = I_2 = (m/4)(h^2/3 + a^2) and I_3 = (ma^2)/2. The key constraint is that the rolling cylinder does not slide against the cylindrical surface, necessitating the use of angular velocities to relate the motion of the center of mass and the cylinder itself.

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  • Understanding of rigid body dynamics
  • Familiarity with kinetic energy equations in classical mechanics
  • Knowledge of moments of inertia for various shapes
  • Concept of rolling without slipping
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  • #31
darkxponent said:
okay

NOW let's visualise:

fisrt i forgot 'Wc' . Just forget that it exists in the question.

i only know Wcm and Vcm:

NOW according to pure rolling Vcm = Wcm * a;

now the only twist is that this cyllinder is rolling not on a plane but in cyllindrical surface

We Find time period(Tc) of the cyllinder = 2*pi*(R-a)/Vcm

THIS GIVES Wc as::

Wc = 2*pi/Tc

= Vcm/(R-a)

=Wcm *a/(R-a)
I understand everything if I assume that Vcm = Wcm * a. Could you explain a bit more how do you get it?
You considered a cylinder of radius a rolling over a plane (with rolling and not slipping?)? In those case I don't understand the meaning of \omega _{CM}. I mean the cylinder rotates around an axis that passes through the center of mass over a plane. Since the center of mass does not rotate, omega of center of mass is worth 0... What am I misunderstanding?
 
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  • #32
i took Wcm is not the angular velocity of centre of mass. It is angular velocity of the cyllinder about its centre of mass and similarly Wc is the angular velocity if the cyllinder about the centre of the cyllindrical surface
 

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