Kinetic Energy of Rigid Object Swinging in Vertical Plane

AI Thread Summary
The discussion centers on the total kinetic energy of a rigid object swinging in a vertical plane, specifically addressing the effects of a stretchable cord versus a non-stretchable one. When the cord is stretchable, there is confusion about whether to include translational kinetic energy associated with the center of mass in the total kinetic energy equation. Some participants argue that only the vertical component of the center of mass's velocity should be included, while others suggest incorporating both horizontal and vertical components. The conversation also touches on the application of the parallel axis theorem in calculating moment of inertia and its relevance to the kinetic energy formulation. Clarification is sought on when to account for translational kinetic energy in rotational systems.
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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