Kinetic Energy of Rotation

At any instant, the general motion of a rigid body consists of the translation of reference point P in the body and the rotation about ‘some axis’ passing through P. If the angular velocity of rotation of the body is then the velocity v_i of ith particle with respect to inertial frame fixed in space at O, at any instant, is given by



where vp is the translational velocity of P relative to O, and is the velocity of ith particle relative to P, arising due to its rotation about P. Note that a rotating from (X Y Z) is fixed with the body at point P.

Hence, the total kinetic energy of the rigid body in frame O is,

K = Σ_i ½ m_i (v_i . v_i)



(i) The first term on the right side above expresses the translational kinetic energy of the point P.


    
(ii) The second term represents the rotational energy of the body.



where we used the formula, (a × b) . (c . d) = (a . c) (b . d) – (a . d) (b . c)



Expanding the above bracket and writing in relation, we find after rearranging the terms (try it), that



If XYZ-axes are the principal axes, then products of inertia vanish, and we get



(iii) The third term on the right side of relation (iii) represents a mixing between translational and rotational motions:



If point P happens to be CM of the rigid body, then Σ m_i r_i = 0, and hence K_int = 0.

Thus, if we choose center-of-mass as the reference point within the body then the total kinetic energy can be neatly separated into pure translational and rotational parts:

K = K_trans + Krot



where, I_x, I_y and I_z now represent the moments of inertia about principal axes through CM of the body.

In the particular case where remains fixed in space, we get the special case of plane rotation so that,

K_rot = ½ I_ω denotes the moment of inertia about axis of rotation.

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