Total KE = Sum of Translational & Rotational KE: Proving the Equation

AI Thread Summary
The discussion focuses on understanding why the total kinetic energy of an object is the sum of its translational and rotational kinetic energies. It highlights the derivation of rotational kinetic energy and emphasizes the importance of considering the center of mass when analyzing rolling objects. The conservation of energy principle is mentioned as a key proof for this relationship, particularly in examples involving objects rolling down ramps. The conversation also touches on the complexity of the proof due to varying angles between tangential and translational velocities. Ultimately, the discussion clarifies that the total kinetic energy can be accurately expressed when accounting for the center of mass frame.
Afo
Messages
17
Reaction score
5
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.
 
Last edited:
Physics news on Phys.org
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.
 
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.
hmmm27 said:
Conservation of Energy principle is the "proof".
No, it is not.
 
Afo said:
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.
 
Last edited:
Ok, thanks I got it.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top