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Question
Derive an expression for the kinetic energy of a rotating body with uniform angular velocity.
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Solution
- Consider a rigid object rotating with a constant angular speed ω about an axis perpendicular to the plane of the paper.
A body of N particles - For theoretical simplification, let us consider the object to be consisting of N particles of masses m1, m2, …..mN at respective perpendicular distances r1, r2, …..rN from the axis of rotation.
- As the object rotates, all these particles perform UCM with the same angular speed ω, but with different linear speeds,
v1 = r1ω, v2 = r2ω,…., vN = rNω - Translational K.E. of the first particle is
(K.E.)1 = `1/2m_1v_1^2 = 1/2m_1r_1^2omega^2`
Similar will be the case of all the other particles. - The rotational K.E. of the object is the sum of individual translational kinetic energies.
Thus,
Rotational K.E. = `1/2m_1r_1^2omega^2 + 1/2m_2r_2^2omega^2..... + 1/2m_Nr_N^2omega^2`
∴ Rotational K.E. = `1/2(m_1r_1^2 + m_2r_2^2..... + m_Nr_N^2)omega^2` - But I = `Σ_{i = 1}^N m_ir_i^2 = m_1r_1^2 + m_2r_2^2...... + m_Nr_N^2`
∴ Rotational K.E. = `1/2"I"omega^2`
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