Physics · System Of Particles And Rotational Motion · NEET
Yes. NCERT says τ = Iα is the rotational form of Newton's second law F = ma. Force F becomes torque τ (the turning effect), mass m becomes moment of inertia I (how hard the body is to spin up), and linear acceleration a becomes angular acceleration α. So every step you did in F = ma problems, you repeat here with rotation words. This is why remembering one equation gives you both.
Torque is the cause, angular acceleration is the effect. Torque (τ, unit N m) is how strongly you twist the body. Angular acceleration (α, unit rad/s^2) is how fast its spinning speed changes as a result. They are linked by τ = Iα, so α = τ / I. A large torque does not always give large α — if I is big, the same torque gives a small α.
In straight-line motion, mass tells you how hard it is to speed something up. In rotation, the SAME body can be easy or hard to spin depending on where its mass sits relative to the axis. Moment of inertia I captures both the mass and how far that mass is from the axis (I = sum of m r^2). So I is the true 'rotational mass' and it replaces m in the equation.
Use α = τ / I. Step 1: find the torque, usually τ = F × R (force times its distance from the axis). Step 2: find the moment of inertia I of the body about that axis (use a standard formula, e.g. hollow cylinder I = MR^2, solid cylinder I = MR^2/2). Step 3: divide, α = τ / I. The answer is in rad/s^2, never m/s^2.
Only if I stays the same. From α = τ / I, angular acceleration is directly proportional to torque but inversely proportional to moment of inertia. So doubling the torque doubles α only when the body's I is unchanged. If you also spread the mass farther from the axis (larger I), α can stay small even with big torque.
α is the angular acceleration — the rate at which angular velocity ω changes, α = dω/dt, measured in rad/s^2. It is an axial vector pointing along the rotation axis (right-hand rule), not along the radius or tangent. When a torque acts on a fixed-axis body, α is what the torque produces, exactly like a produced by F.
A rope wound on a hollow cylinder of mass 3 kg and radius 40 cm is pulled with a force of 30 N. The angular acceleration of the cylinder is:
A solid cylinder of mass 2 kg and radius 4 cm rotates about its axis at 3 rpm. The torque required to stop it within 2 revolutions is:
A constant torque of 100 N·m turns a wheel of moment of inertia 300 kg·m² about an axis through its centre. Starting from rest, its angular velocity after 3 s is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The core formula is τ = Iα, where τ is the net torque about the axis, I is the moment of inertia, and α is the angular acceleration. Rearranged: α = τ / I.
Torque τ is in N m, moment of inertia I is in kg m^2, and angular acceleration α is in rad/s^2. Check: N m = (kg m^2)(rad/s^2), which balances because radian is dimensionless.
τ = Iα is the rotational equivalent of F = ma. Torque replaces force, moment of inertia replaces mass, and angular acceleration replaces linear acceleration.
Almost every NEET dynamics-of-rotation problem is solved in three steps: find torque (usually τ = F × R), find I from a standard formula, then get α = τ / I. Often you then use rotational kinematics (ω = ω0 + αt or ω^2 = ω0^2 + 2αθ) to finish.
Yes. If the force passes through the axis, its perpendicular distance R is zero, so τ = 0 and α = 0. A force only produces angular acceleration when it has a torque about the axis.
