Kinetic Energy of a Rotating Body
Suppose the body is rotating about an axis through O with a constant angular velocity ω. A particle A of mass m, at a distance r1 from O describes its own circular path. If v1 is its linear velocity along the tangent to the path, at the instant shown then v1=r1ω and the kinetic energy of A=(1/2)m1v12=(1/2)m1v12ω2
It follows that the kinetic energy of the whole body is the sum of the kinetic energy of its component particles. If these have masses m1, m2, m3, etc. and are distributed at distances r1, r2, r3, etc. from O then since all the particles have the same angular velocity (the body is rigid).
The quantity ∑miri represents the sum of the mr2 values depends on the mass and distribution of and is taken as a measure of the moment of inertia of the body about the axis in question. It is denoted by the symbol I.
Therefore, the rotational K.E. of the body is
