Torque and Angular Acceleration — τ = Iα

Question

A wheel of moment of inertia $I = 2 \text{ kg·m}^2$ starts from rest and reaches an angular velocity of $\omega = 10 \text{ rad/s}$ in $t = 5 \text{ s}$. Find the torque required, assuming constant angular acceleration.

Solution — Step by Step

Why This Works

The equation $\tau = I\alpha$ is the rotational form of Newton’s second law. Every linear quantity has a rotational twin: force $\leftrightarrow$ torque, mass $\leftrightarrow$ moment of inertia, linear acceleration $\leftrightarrow$ angular acceleration.

The key insight is that moment of inertia $I$ tells us how hard it is to change the spin state of a body. A wheel with $I = 2 \text{ kg·m}^2$ is harder to spin up than one with $I = 0.5 \text{ kg·m}^2$ — you’d need 4× the torque to achieve the same $\alpha$.

This problem is a direct application — no integration needed because $\alpha$ is constant throughout. That makes it a favourite for NCERT exercises and CBSE board short-answer questions.

Alternative Method

We can use the work-energy theorem for rotation as a cross-check.

The angular displacement in 5 s (starting from rest, constant $\alpha = 2 \text{ rad/s}^2$):

Work done by torque = gain in rotational kinetic energy:

Both methods give the same answer — good habit to cross-verify in JEE Main.

Common Mistake