Question

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.

Answer 1

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, ..., m_N situated at perpendicular distances r₁, r₂, ..., r_N 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, ..., τ_N = m_Nr_N²α

The total torque on the body is:

τ = τ₁ + τ₂ + ... + τ_N

= m₁r₁²α + m₂r₂²α + ... + m_Nr_N²α

= (m₁r₁² + m₂r₂² + ... + m_Nr_N²)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.