Derive an expression for torque in terms of the moment of inertia.

Solution:

Consider a rigid body rotating about a given axis with a uniform angular acceleration α, under the action of torque.

Let the body consist of particles of masses m₁, m₂, m₃, . . … , m_n at a distance r₁, r₂, r₃, . . … r_n respectively from the axis of rotation

If a₁, a₂, a₃, . . … , a_n are the respective linear acceleration of the particles, then,

a₁ = r₁α, a₂ = r₂α, a₃ = r₃α…..

Force on particles of mass m is

f₁ = m₁a₁ = m₁r₁α

Moment of this force about the axis of rotation f₁ × r₁ = (m₁r₁α) × r₁ = m₁ r₁²α

Similarly, the moment of forces on other particles about the axis of rotation is

m₂ r₂²α, m₃ r₃²α ———-m_n r_n²α

Torque acting on the body

τ = m₁ r₁²α + m₂ r₂²α + m₃ r₃²α ———-m_n r_n²α

So, 

τ = Iα

where m₁ r₁² = moment of inertia of the body about the given axis of rotation