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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Derive an expression for the kinetic energy of a rotating body with uniform angular velocity. - Physics

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प्रश्न

Derive an expression for the kinetic energy of a rotating body with uniform angular velocity. 

व्युत्पत्ती
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उत्तर

  1. 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
  2. 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.
  3. 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 
  4. 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.
  5. 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`
  6. 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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Rotational Dynamics
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Rotational Dynamics - Short Answer II

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