 #1
Afo
 17
 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.
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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