Torque and Angular Momentum: Why the Skater Spins Faster (NEET & JEE)
Why does a spinning ice skater whirl faster the moment she pulls her arms in, even though nobody pushes her? And why does a door open easily when you push far from the hinge but barely budge when you push near it? Both puzzles are answered by one pair of ideas: torque and angular momentum. This topic matters for both NEET and JEE, since it sits inside Class 11 Rotational Motion.
Quick Summary
Torque is the turning effect of a force: τ = r × F × sinθ, where r is the distance from the axis to the point of application and θ is the angle between r and F.
Angular momentum of a particle is L = m × v × r × sinθ; for a rigid body about a fixed axis, L = Iω, where I is the moment of inertia and ω the angular velocity.
The rotational form of Newton's second law is τ = dL/dt, which becomes τ = Iα for a rigid body (α is angular acceleration).
If the net external torque on a system is zero, its angular momentum stays constant.
Building the idea from what you know
In straight-line motion, a net force changes momentum: F = dp/dt. Rotation has an exact parallel. Force is replaced by torque, momentum by angular momentum, mass by moment of inertia, and velocity by angular velocity. If you trust F = dp/dt, then τ = dL/dt is simply the same statement in rotational language.
Why does torque depend on distance and angle? Think of the door. A force pointing straight at the hinge, or along the line of the door, cannot open it. Only the component of force perpendicular to the line from the axis to the point of application does any turning. That component is F sinθ, and multiplying it by the lever arm r gives τ = rF sinθ. Equivalently, torque is force times the perpendicular distance from the axis to the line of action of the force. In vector form, τ = r × F, a cross product, so its direction is given by the right-hand rule and is perpendicular to the plane containing r and F.
Angular momentum follows the same logic. Linear momentum is p = mv. For a particle at position r from a chosen point, angular momentum is L = r × p, so its size is L = mvr sinθ. Notice that this depends on the point you choose. A particle moving in a perfectly straight line has a non-zero L about any point not on its path, equal to mv times the perpendicular distance from that point to the path.
Now the central fact. Differentiate L = r × p with respect to time. You get dL/dt = r × F, which is τ. So if no external torque acts, dL/dt = 0 and L cannot change. This is conservation of angular momentum, and it is a consequence of the rotational form of Newton's law, not a separate rule.
Worked example: the skater
A skater spins with her arms stretched out. Her moment of inertia is I₁ = 4.0 kg·m² and her angular velocity is ω₁ = 3.0 rad/s. She pulls her arms in, and her moment of inertia drops to I₂ = 1.5 kg·m². Find her new angular velocity and the change in kinetic energy. Ignore friction with the ice.
The ice exerts no meaningful torque about her vertical axis, and gravity and the normal force act along that axis, so net external torque is zero. Angular momentum is conserved.
Initial angular momentum: L₁ = I₁ω₁ = 4.0 × 3.0 = 12 kg·m²/s.
Set L₂ = L₁: 1.5 × ω₂ = 12, so ω₂ = 8.0 rad/s.
Initial kinetic energy: ½ × 4.0 × 3.0² = 18 J.
Final kinetic energy: ½ × 1.5 × 8.0² = 48 J.
Her kinetic energy rose by 30 J. Angular momentum was conserved, but energy was not, because her muscles did work pulling the arms inward against the outward tendency of the arms to fly away. That is a very common exam trap, and we return to it below.
A second quick example: torque on a wrench
A 20 N force acts on a wrench at a point 0.50 m from the bolt, at 30° to the wrench handle. The torque is τ = rF sinθ = 0.50 × 20 × sin30° = 0.50 × 20 × 0.5 = 5.0 N·m. Had the force been perpendicular, the torque would be 10 N·m, the maximum possible for this r and F.
Mistakes students actually make
Using r × F as if it were r times F. Forgetting sinθ, or using the angle between the wrong pair of vectors, is the most common error. The angle is between r (from axis to application point) and F.
Conserving kinetic energy along with angular momentum. In the skater problem, L is conserved but kinetic energy is not, as the numbers above show. Conserve a quantity only when you can justify it.
Ignoring the choice of origin. Both τ and L depend on the point about which you compute them. Torque and angular momentum must be taken about the same point in τ = dL/dt.
Saying a particle in straight-line motion has zero angular momentum. It does not, unless the chosen point lies on its line of motion.
Treating zero net force as zero net torque. A couple, two equal and opposite forces with different lines of action, has zero net force but a non-zero torque.
How it is asked in NEET and JEE
Typical shapes include: finding the torque of a force given in vector form using the cross product (JEE especially); a person on a rotating platform walking inward or outward, or a child jumping onto a merry-go-round, where you apply conservation of L; a particle in projectile or circular motion where you compute L about a given point; and statements asking which of torque, angular momentum, kinetic energy are conserved in a given process. Questions on a disc with a lump of clay dropped on it are conservation of L with a changed I.
Key Insight: Before applying conservation of angular momentum, ask one question: is there any external torque about the same point or axis you will use for L? A force can be large and still produce zero torque about your chosen axis if its line of action passes through that axis. Choosing the axis cleverly, for example through the point where an unknown force acts, often removes that force from the torque equation and solves the problem in one line.
Frequently Asked Questions
Is torque the same as work?
No. Torque and work (energy) share the unit N·m, but torque is a vector with direction along the axis, while work is a scalar. The work done by a torque turning a body through angle θ is τθ, a different quantity altogether. Joule is reserved for energy, and torque is written as N·m.
Why does the skater speed up if no torque acts?
Her angular momentum Iω must stay constant. Pulling her arms in lowers I, so ω must rise to keep the product the same. The extra kinetic energy comes from the work her muscles do, not from any external torque.
When is angular momentum conserved?
When the net external torque about the chosen point or axis is zero. This can happen because no external force acts, or because every external force passes through that point, as with a planet orbiting under a central gravitational force.
What is the direction of angular momentum?
It follows the right-hand rule: curl the fingers of your right hand in the direction of rotation and your thumb points along L. For a wheel spinning anticlockwise as seen from above, L points upward along the axis.
Is τ = Iα always valid?
It holds for a rigid body rotating about a fixed axis, or about an axis through the centre of mass, where I is taken about that same axis. For more general motion you return to the fundamental law τ = dL/dt.
Stuck on a rotation problem or want to see the vector cross product worked out step by step? Drop your doubt in the comments, and share this with a fellow aspirant who still mixes up torque and energy.
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