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प्रश्न
Obtain an expression for torque acting on a body rotating with uniform angular acceleration.
Obtain an expression for torque acting on a rigid body rotating with constant angular acceleration.
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उत्तर
Consider a rigid body rotating about an axis passing through the point O and perpendicular to the plane of the figure. Suppose that a torque `vecτ` on the body produces uniform angular acceleration `vecα` along the axis of rotation.
The body can be considered as made up of N particles with masses m1, m2, ..., mN situated at perpendicular distances r1, r2, ..., rN respectively from the axis of rotation. `vecα` is the same for all the particles as the body is rigid. Let `vecF_1, vecF_2, ..., vecF_N` be the external forces on the particles.
The torque `vecτ_1`, on the particle of mass m1, is:
`vecτ_1 = vecr_1 × vecF_1`
∴ τ1 = r1F1 sin θ
where θ is the smaller of the two angles between `vecr_1 and vecF_1`.
Since, in this case, θ = 90°
∴ τ1 = r1F1
Now, F1 = m1a1 = m1r1α
where a1 = r1α is the tangential acceleration of the particle.
∴ τ1 = r1(m1r1α)
= m1r12α
Similarly, τ2 = m2r22 α, ..., τN = mNrN2α
The total torque on the body is:
τ = τ1 + τ2 + ... + τN
= m1r12α + m2r22α + ... +mNrN2α
= (m1r12 + m2r22 + ... +mNrN2)α
= `(sum_(i = 1)^N m_ir_i^2)α`
∴ τ = I α
where I = `(sum_(i = 1)^N m_ir_i^2)` is the moment of inertia of the body about the axis of rotation.
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