Kinetic energy of a rotation body
Consider a rigid body of mass M rotating about an axis YY' passing through it with angular velocity ω made of n particles of mass m1, m2, m3, …, mn at distance r1, r2, r3, …, rn from the axis. Although the angular velocity is same for all the particles but the linear velocity won’t be same for all the particles. So, let v1,v2, v3, ……., vn be the linear velocities of the particles.
Then, K.E. for m1 = ½ m1v12 ………. (i)
We know, v = r. ω
Then, v1 = r1. ω
Equation (i) becomes, K.E. for m1 = ½ m1. R12. ω2
For m1, m2, m3, ……., mn
K.E. for m2 = ½ m2. R22. ω2
K.E. for m3 = ½ m3. R32. ω2
K.E. for mn = ½ mn. Rn2. ω2
For total kinetic energy,
K.E.rot = ½ m1. R12. ω2 + ½ m2. R22. ω2 + ½ m3. R32. ω2 + ……. + ½ mn. Rn2. ω2
K.E.rot = ½ ω2 (m1. R12 + m2. R22 + m3. R32 + ……. + mn. Rn2)
K.E.rot = ½ Iω2 [.: m1. R12 + m2. R22 + m3. R32 + ……. + mn. Rn2 = I]
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