Why total kinetic energy is the sum of translational and rotational kinetic energies?

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  • #1
Afo
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Homework Statement:
Why is total kinetic energy always equal to the sum of rotational and translational kinetic energies?
Relevant Equations:
1/2 I W^2
KE = 1/2 m v^2 + 1/2 I W^2
Why is the total energy energy equal to the sum of translational kinetic energy and rotational kinetic energy? I understand the derivation KE = 1/2 I w^2 for a rigid object rotating around an axis:

sum 0.5 * m_n * (v_T)^2 = sum 0.5 * m_n * (wr_n)^2 = 0.5 * w^2 * sum m_n r_n^2 = 0.5 * I * w^2

But I don't understand how the KE of a rolling without slipping object is simply the addition of them. I think the proof is pretty complicated since the tangential velocity and translational velocity makes an angle which varies.
 
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  • #2
hmmm27
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From an arithmetic point of view, translational KE, rotational KE and PE are what's used for simple kinematics exercises.

Conservation of Energy principle is the "proof".

Example of a ball coming down a ramp:
1) for a non-frictionless interaction, the PE at the top of the ramp is equal to the translational KE at the bottom ; translational only, since there is no rotation.
2) for a rolling ball, the PE at the top of the ramp is equal to the added KE's at the bottom.

The ball in the second example will not have as high a translational velocity as the first, when they hit the bottom of the ramp, but will of course have a higher rotational velocity.

My apologies if this is too simplistic an explanation.
 
  • #3
Orodruin
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Write the total kinetic energy as an integral over the mass distribution. Then split the position vector into the center of mass vector and the displacement from the center of mass. Then start simplifying. The cross terms between the center of mass and displacement will cancel out and leave the center of mass translational energy and the rotational energy relative to the center of mass.

It is important that the rotational energy is relative to the center of mass or the statement is not necessarily true.
Conservation of Energy principle is the "proof".
No, it is not.
 
  • #4
PeroK
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Homework Statement:: Why is total kinetic energy always equal to the sum of rotational and translational kinetic energies?
Relevant Equations:: 1/2 I W^2
KE = 1/2 m v^2 + 1/2 I W^2

Why is the total energy energy equal to the sum of translational kinetic energy and rotational kinetic energy? I understand the derivation KE = 1/2 I w^2 for a rigid object rotating around an axis:

sum 0.5 * m_n * (v_T)^2 = sum 0.5 * m_n * (wr_n)^2 = 0.5 * w^2 * sum m_n r_n^2 = 0.5 * I * w^2

But I don't understand how the KE of a rolling without slipping object is simply the addition of them. I think the proof is pretty complicated since the tangential velocity and translational velocity makes an angle which varies.
It's clearly true in the centre of mass frame. What happens when you add the same linear velocity to every point mass? That transforms to a reference frame where the centre of mass has that velocity.

Hint: using the definition of centre of mass should cause some terms in the equation to vanish.
 
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  • #5
Afo
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Ok, thanks I got it.
 

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