Relation between Torque, Moment of Inertia and Angular Acceleration

Let us consider a body on which torque x is applied, is rotating about an axis passing through a constant point and the body also has a constant angular acceleration α. All the particles of the body will have same angular acceleration a but the linear accelerations will be different. Suppose, one particle of the body has mass m₁ and the distance of it from the rotational axis is r₁. Then the linear acceleration of this particle is;
α₁ = r₁α
If \(\vec{F}\) is the force acting on the particle then;
F₁ = mass × acceleration
= \(m_{1} \vec{a}\) = m₁r₁α
The torque of this force along the rotational axis passing through O from the fig. is;
τ₁ = Force (F₁) × distance (r₁)
= m₁r₁α × r = m₁r₁²α
Similarly, if m₂, m₃,… be the masses of other particles and r₂, r₃, … the distances from the rotational axis, then the torque acting on them will be; c₂ = m₂r₂a
τ₂ = m₂r₂²α
τ₃ = m₃ r₃²α
Since, the directions of all the torque is along the same rotational axis. Hence the resultant force (torque) on the body will be the vector sum of all the torques.
τ = τ₁ + τ₂ + τ₁ + …
= m₁r₁²α + m₂r₂²α + m₃r₃²α+ ……..
= (m₁r₁² + m₂r₂² + m₃r₃²) α
τ = (Σ mr²) α
But Σ mr² is the moment of inertia I about the rotational axis. Therefore,
τ = Iα ………….. (1)
Torque = Moment of inertia × Angular Acceleration
Keeping α = 1 rad/s² in the above equation,
τ = I
Hence, the moment of inertia of a body about the rotational axis is equal to the applied torque required to generate unit angular acceleration in the body.