QCC Notes
CLASS 11 · PHYSICSJEE MAIN × NEETहिंदी
§ 6.3NCERT Class 11 · Physics · Chapter 6

Torque & Angular Momentum

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 9 min read

Just as force causes translational acceleration and linear momentum measures translational inertia in motion, Torque (Moment of Force) and Angular Momentum (Moment of Momentum) are their exact rotational counterparts. They constitute the cornerstone of rigid body dynamics.

1. Torque (Moment of Force) and Couple

When a force F acts on a particle located at position vector r relative to a reference origin O, the torque τ about O is defined as:

τ = r × F = (r F sin θ) n̂

Alternatively, the scalar magnitude can be written in two illuminating physical forms:

  • τ = r⊥ F, where r⊥ = r sin θ is the lever arm (perpendicular distance from origin to the line of action of force).
  • τ = r F⊥, where F⊥ = F sin θ is the component of force perpendicular to the position vector r.

Units & Dimensions: SI Unit is Newton-meter (N·m). Dimensional formula: [ML²T⁻²]. (Caution: Although dimensionally identical to Joule, torque is a vector perpendicular to the plane of motion, while work is a scalar; torque is never expressed in Joules).

The Concept of a Couple

A couple consists of a pair of equal and opposite parallel forces whose lines of action do not coincide:

  • Net translational force is zero: Fnet = F − F = 0 (no translational acceleration).
  • Net torque is independent of the choice of origin: τcouple = Force × perpendicular distance between force lines (arm of couple).
  • Everyday examples: Turning a water tap, opening a bottle cap, rotating a steering wheel with two hands.

2. Angular Momentum of a Particle & a System

The angular momentum L of a particle of mass m moving with linear velocity v (and linear momentum p = m v) about origin O is:

L = r × p = m (r × v) = (m v r sin θ) n̂

In scalar magnitude: L = p r⊥ = m v r⊥ (Momentum × Perpendicular distance from line of motion to origin).

Units & Dimensions: SI unit is kg·m²/s or Joule-second (J·s). Dimensions: [ML²T⁻¹] (same as Planck's constant h).

Angular Momentum of a Rigid Body About a Fixed Axis

For a rigid body rotating with angular velocity ω about a fixed axis:

L = I ω   (Rotational analogue of p = m v)

where I is the Moment of Inertia of the body about that axis.

3. Fundamental Relation: τ = dL/dt and Angular Momentum Conservation

Differentiating angular momentum with respect to time:

dL/dt = d(r × p)/dt = (dr/dt × p) + (r × dp/dt)

Since dr/dt = v and p = m v, the term (v × m v) = 0. Substituting Newton's second law dp/dt = Fext:

τext = dL/dt = I α   (Rotational analogue of F = dp/dt = m a)

Principle of Conservation of Angular Momentum

If the net external torque acting on a system is zero (τext = 0):

dL/dt = 0 ⇒ L = constant ⇒ I1 ω1 = I2 ω2

When the moment of inertia changes due to internal reconfiguration of mass, angular velocity automatically adjusts in inverse proportion:

  • Ballet Dancer / Figure Skater: Outstretched arms ⇒ high I, low ω. Drawing arms close to body ⇒ I drops sharply, ω surges into a rapid spin.
  • Springboard Diver: Curls body into a compact tuck position to reduce I and perform multiple somersaults before extending to enter water cleanly.
  • Planetary Motion (Kepler's 2nd Law): Gravitational force is central (directed along r), so r × Fg = 0. Hence L = m r v = const. As the planet nears the sun at perihelion (r decreases), its orbital speed v increases.

4. Work Done by Torque and Rotational Power

When a torque τ rotates a body through an infinitesimal angular displacement dθ:

dW = τ dθ ⇒ W = ∫ τ dθ   and   P = dW/dt = τ (dθ/dt) = τ ω

This is identical to the translational work-power relations dW = F dx and P = F v.

Figure 6.3 — Conservation of Angular Momentum: The Spinning Skater

[STATE 1: Arms Extended] [STATE 2: Arms Tucked Close] O / O | / <-- Large Radius R /| <-- Small Radius r | / | | | / / High Moment of Inertia (I1) Low Moment of Inertia (I2) Slow Angular Speed (ω1) FAST Angular Speed (ω2) ======================================================================== L1 = I1 * ω1 === I2 * ω2 = L2
Kinetic Energy Trap: Notice that rotational kinetic energy Krot = L² / (2I). Because I decreases while L is constant, Krot actually INCREASES! Where does the extra kinetic energy come from? The skater's muscles do positive chemical work pulling the arms inward against centrifugal reaction.

Step-by-Step Solved Numerical Examples

Solved Example 1: Angular Momentum of a Projectile About Projection Point

Problem: A projectile of mass m is launched with velocity u at an angle θ with the horizontal. Find the magnitude of its angular momentum about the point of projection when it reaches its maximum height H.

L = (m u³ sin²θ cosθ) / (2g)
Step 1: Identify Velocity at Maximum Height:
At maximum height, the vertical velocity component is zero (vy = 0). The projectile has only horizontal velocity: v = u cos θ.

Step 2: Find the Perpendicular Distance (Lever Arm):
The line of action of the velocity at the summit is horizontal at height H above the ground.
Thus, the perpendicular distance from the projection point (origin) to the velocity vector line is simply r⊥ = H.
Recall maximum height formula: H = (u² sin²θ) / (2g).

Step 3: Calculate Angular Momentum:
L = m v r⊥ = m (u cos θ) [ (u² sin²θ) / (2g) ]
L = (m u³ sin²θ cosθ) / (2g) directed perpendicularly into the plane of trajectory.
Solved Example 2: Turntable with Dropped Mass (Angular Momentum Conservation)

Problem: A uniform horizontal disc of mass M = 2 kg and radius R = 0.2 m is rotating freely about its central vertical axis at 60 rpm. A small lump of wax of mass m = 0.5 kg is dropped gently onto the disc and sticks at a distance of 0.1 m from the axis. Calculate the new rotational speed of the system in rpm.

New Speed: 53.33 rpm
Step 1: Initial Moment of Inertia:
For the uniform circular disc: I1 = ½ M R² = ½(2)(0.2)² = 0.04 kg·m².
Initial angular speed: N1 = 60 rpm.

Step 2: Final Moment of Inertia:
When wax sticks at r = 0.1 m:
Iwax = m r² = (0.5)(0.1)² = 0.005 kg·m².
Total final moment of inertia: I2 = I1 + Iwax = 0.04 + 0.005 = 0.045 kg·m².

Step 3: Apply Conservation of Angular Momentum:
Since dropping the wax exerts no net external torque along the vertical axis:
I1 N1 = I2 N2
(0.04)(60) = (0.045) N2 ⇒ N2 = 2.4 / 0.045 = 53.33 rpm.

5. Solved NEET & JEE Practice Questions

Q1. A force F = 2î − 3ĵ + 4k̂ N acts at a point having position vector r = 3î + 2ĵ + 3k̂ m. The torque acting on the body about the origin is:

A. 17î − 6ĵ − 13k̂ N·m
B. −17î + 6ĵ + 13k̂ N·m
C. 17î + 6ĵ − 13k̂ N·m
D. −17î − 6ĵ − 13k̂ N·m

Answer: A
Explanation: τ = r × F =
î[(2)(4) − (3)(−3)] − ĵ[(3)(4) − (3)(2)] + k̂[(3)(−3) − (2)(2)]
= î[8 + 9] − ĵ[12 − 6] + k̂[−9 − 4] = 17î − 6ĵ − 13k̂ N·m.
Q2. When a mass rotates in a plane about a fixed point, its angular momentum is directed along:

A. The radius vector
B. The tangent to the orbit
C. A line perpendicular to the plane of rotation
D. At 45° to the plane of rotation

Answer: C
Explanation: Angular momentum is defined as L = r × p. By definition of cross product, L is perpendicular to both r and p, which means it points along the axis of rotation perpendicular to the orbital plane.
Q3. If the Earth suddenly shrinks to half of its present radius without any change in its mass, the duration of one day will become:

A. 6 hours
B. 12 hours
C. 24 hours
D. 48 hours

Answer: A
Explanation: Modeling Earth as a solid sphere: I = (2/5) M R². Shrinking radius to R' = R/2 makes I' = (2/5) M (R/2)² = I / 4.
Since no external torque acts: L = I ω = I (2π/T) = constant ⇒ T ∝ I.
Therefore, T' = T / 4 = 24 / 4 = 6 hours.
Q4. A particle moves with constant velocity along a straight line not passing through the origin. Its angular momentum about the origin:

A. Increases with time
B. Decreases with time
C. Remains constant in magnitude and direction
D. Is zero

Answer: C
Explanation: Angular momentum is L = p r⊥ = m v d, where d is the fixed perpendicular distance from the origin to the straight-line trajectory. Because velocity v and perpendicular distance d remain constant, L remains constant throughout the motion.

6. Frequently Asked Questions (FAQs)

Can a body have zero angular momentum about one point and non-zero angular momentum about another point?
Yes, absolutely. Angular momentum L = r × p depends explicitly on the choice of the origin through the position vector r. For a particle moving along a straight line, its angular momentum is identically zero about any point located on that straight line (rperp = 0), but non-zero about any origin situated off the line (rperp > 0).
Why does torque have the same dimensional formula as work and energy, yet they represent completely different physical concepts?
Both work (W = F · s) and torque (τ = r × F) have dimensions of [ML²T⁻²]. However, work is a scalar product measuring energy transfer along displacement (units: Joules), whereas torque is a vector cross product representing rotational effort perpendicular to displacement (units: N·m, never Joules).
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