Kinetic Energy of Rigid Object Swinging in Vertical Plane

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SUMMARY

The discussion centers on the total kinetic energy of a rigid object swinging in a vertical plane, specifically a bar attached to a stretchable cord. The kinetic energy is expressed as K = \frac{I}{2}\omega^2 + \frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2) for a stretchable cord, while some participants argue that only K = \frac{I}{2}\omega^2 + \frac{M}{2}\dot{y}_c^2 is necessary. The confusion arises from whether to include translational kinetic energy based on the changing center of mass and the application of the parallel axis theorem. Understanding the moment of inertia relative to different axes is crucial for resolving these discrepancies.

PREREQUISITES
  • Understanding of kinetic energy formulas, specifically K = \frac{I}{2}\omega^2.
  • Familiarity with the parallel axis theorem in physics.
  • Knowledge of rigid body dynamics and rotational motion.
  • Concept of moment of inertia and its calculation.
NEXT STEPS
  • Study the application of the parallel axis theorem in various rigid body scenarios.
  • Learn about the relationship between translational and rotational kinetic energy in complex systems.
  • Explore the effects of different types of constraints (e.g., stretchable vs. rigid) on kinetic energy calculations.
  • Investigate the implications of changing the axis of rotation on moment of inertia and kinetic energy.
USEFUL FOR

Physics students, mechanical engineers, and anyone studying dynamics of rigid bodies in motion will benefit from this discussion.

KFC
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I am quite confusing about the total kinetic energy of an rigid object swing in a vertical plane. Imagine a bar hanging to the ceiling with a massless but stretchable cord. The bar has mass M. Because the cord is stretchable (not very hard but like a spring with very small spring constant), the center of the bar (x_c, y_c) changed.

Well if the cord is NOT stretchable, since the bar is swinging, the kinetic energy is K = \frac{I}{2}\omega^2. In this case, the center of the bar is also changing from place to place, but we don't account \frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2) into the total kinetic energy. But for the stretchable cord, the center of the bar is also changing but in other way, in this case, what will the total kinetic energy look like? Will it be

K = \frac{I}{2}\omega^2 + \frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2)

But someone tell me it is not necessary to include the term about x_c, only the following is enough

K = \frac{I}{2}\omega^2 + \frac{M}{2}\dot{y}_c^2

This is very confusing! For rotation, when should we include the translation kinetic energy? and when could we use the rotation kinetic energy only for the whole system?

By the way, if we use a mass instead, what would be different?
 
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KFC said:
K = \frac{I}{2}\omega^2 + \frac{M}{2}(\dot{x}_c^2 + \dot{y}_c^2)

Hi KFC! :smile:

You're getting confused about whether I is measured relative to an axis through the ceiling or through the centre of mass.

From the PF Library:
The parallel axis theorem:

The Moment of Inertia of a body about an axis is

I = (I_C\,+\,md^2)

where m is the mass, d is the distance from that axis to the centre of mass, and I_C is the Moment of Inertia about the parallel axis through the centre of mass.

This is the same as your formula if you put x'2 + y'2 = v2 = w2d2. :smile:
But someone tell me it is not necessary to include the term about x_c, only the following is enough

K = \frac{I}{2}\omega^2 + \frac{M}{2}\dot{y}_c^2

Sorry, don't understand that. :confused:
 
Sometimes I think it is useful to think of mv2/2 for each point first.
 

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