Torque — the rotational analogue of force
Torque (moment of force) is defined as the cross product of the position vector r and the applied force F, measured from the chosen pivot point (NCERT Class 11 Physics Chapter 6, page 4):
τ = r × F
The magnitude is τ = r F sin θ, where θ is the angle between r and F. The direction follows the right-hand rule: curl fingers from r toward F, and the thumb points along τ.
Three things that control torque magnitude:
The perpendicular-distance shortcut: τ = F × d, where d = r sin θ is the perpendicular distance from the pivot to the line of action of F. Equivalently, τ = r × F⊥, where F⊥ = F sin θ is the component of force perpendicular to r. Both give the same result — pick whichever the problem makes easier.
SI unit: N·m. Note: this is dimensionally identical to the joule (J = N·m), but torque and energy are distinct physical quantities. NEET will never accept "joule" as a unit of torque.
Torque as a vector: For a fixed rotation axis (the NEET scenario), only the component of torque along the axis matters. A torque that tends to produce counterclockwise rotation is taken positive by convention.
Watch out: When multiple forces act on a body, compute the torque of each force about the same pivot, then sum algebraically. Forgetting to use the same pivot point is a silent error that produces plausible-looking wrong answers.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
What is the SI unit of torque?
Torque is defined as:
A force of 10 N acts on a rigid body. If the line of action of the force passes through the pivot point, the torque about that pivot is:
A force of 5.0 N is applied at the end of a 0.40 m wrench, perpendicular to the wrench handle. The torque about the pivot is:
A 6.0 N force acts at a point 0.50 m from the pivot, making an angle of 30° with the position vector. The magnitude of the torque is:
Two forces of equal magnitude F act on a rigid body at the same point, one at 90° to the position vector and the other at 45° to the position vector. The ratio of the torques τ₉₀ : τ₄₅ is:
A door of width w is pushed with force F at its outer edge, perpendicular to the door. If the same force is applied at w/2 from the hinge (still perpendicular), by what factor does the torque change?
Two forces act on a rod pivoted at one end: a 4.0 N force acts perpendicularly at 0.30 m from the pivot (clockwise), and a 3.0 N force acts perpendicularly at 0.80 m from the pivot (counterclockwise). The net torque about the pivot is:
Pattern: Torque calculation with multiple forces (no topic-specific PYQ pattern survived Rule 7 filtering; this worked example is constructed from the torque formula and NCERT definition to serve the lesson's scope)
Given
A uniform rod of length 1.00 m is pivoted at its left end. Two forces act on it: - Force F₁ = 8.0 N acts perpendicularly downward at 0.25 m from the pivot - Force F₂ = 5.0 N acts perpendicularly upward at 0.80 m from the pivot Find the net torque about the pivot and state the direction of rotation.
Required
Net torque (τ_net) and its rotational direction (clockwise or counterclockwise).
Concept
Each force produces a torque about the pivot: τ = r F sin θ. Since both forces are perpendicular to the rod, sin θ = 1 for both. Assign signs by convention: counterclockwise = positive, clockwise = negative. Then sum algebraically.
Formula
τ = r F sin θ Net torque: τ_net = τ₁ + τ₂ (with sign convention)
Substitution
F₁ acts downward at 0.25 m → produces clockwise rotation → negative: τ₁ = −(0.25)(8.0)(1) = −2.0 N·m F₂ acts upward at 0.80 m → produces counterclockwise rotation → positive: τ₂ = +(0.80)(5.0)(1) = +4.0 N·m
Calculation
τ_net = −2.0 + 4.0 = +2.0 N·m The integers 1 (from sin 90°) are exact values and do not limit significant figures.
Final answer
τ_net = 2.0 N·m, counterclockwise. The answer has 2 significant figures, consistent with the least precise given data (2 sig figs in 8.0 N and 5.0 N). Sin 90° = 1 is an exact value and does not constrain the sig-fig count.
Common trap
A frequent error is subtracting forces first (8.0 − 5.0 = 3.0 N) and then computing torque with some average distance. This is wrong — each force has its own lever arm, and torques must be computed individually before summing.
Similar NEET-style question
A light rod of length 2.0 m is pivoted at the centre. A 6.0 N force acts perpendicularly at one end (clockwise) and a 4.0 N force acts perpendicularly at the other end (also clockwise). Find the net torque about the pivot. *(Answer: Both torques are clockwise with lever arm 1.0 m each. τ_net = 1.0 × 6.0 + 1.0 × 4.0 = 10.0 N·m clockwise.)* ---
The moment of a force (torque) about a point is τ = r × F, where r is the position vector from the pivot to the point of force application. Magnitude |τ| = r F sin θ. SI unit: N·m.
-- NCERT Class 11 Physics, Ch. 6, p. 4For a particle: L = r x p. For a rigid body about its rotation axis: L = I omega. Vector quantity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| L | angular momentum | kg*m^2/s |
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
The position of the centre of mass equals the mass-weighted average of particle positions. For continuous bodies use integral form.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R_cm | CoM position | m |
| m_i | mass of i-th particle | kg |
| r_i | position of i-th particle | m |
Standard moments of inertia about the symmetry axis. For other axes use parallel/perpendicular axes theorems.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | mass | kg |
| R | radius | m |
| L | length | m |
| I | moment of inertia | kg*m^2 |
Moment of inertia about any axis = moment about parallel axis through CM + Md^2.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | MOI about given axis | kg*m^2 |
| I_cm | MOI about parallel CM axis | kg*m^2 |
| M | total mass | kg |
| d | perpendicular distance | m |
For planar lamina: MOI about axis perpendicular to plane = sum of MOI about two perpendicular in-plane axes through same point.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_z | MOI perp to plane | kg*m^2 |
| I_x, I_y | MOI in plane | kg*m^2 |
Rotational analogues of linear kinematic equations under constant angular acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| omega | angular velocity | rad/s |
| alpha | angular acceleration | rad/s^2 |
| theta | angular displacement | rad |
| t | time | s |
Energy of rotation about an axis. Adds to translational KE for rolling bodies.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
Cross product of position vector and force vector. Magnitude r F sin(theta).
| Symbol | Quantity | SI Unit |
|---|---|---|
| tau | torque | N*m |
| r | position from pivot | m |
| F | force | N |
| theta | angle between r and F | rad |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student answers L/2 for two-particle CoM regardless of mass ratio.
Question gives two masses on rigid rod and asks for CoM distance.
R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.
Category: Similar Terms
Student conserves rotational KE when angular momentum is conserved (or vice versa). When I changes, L = Iω is conserved but KE = ½Iω² is NOT (it depends on I and ω together).
Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).
L conservation requires zero external torque. KE conservation requires no work done — different criteria. When I changes via internal forces, L conserved, ω increases, KE increases.
Category: Similar Terms
Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).
Question gives a specific geometry and asks for I or radius of gyration.
Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.
Category: Unit Conversion
Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.
Question gives ω in rpm and asks for kinematic quantities in SI units.
Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.
Root cause: concept gap
L = Iω is conserved when external torque is zero. KE = ½Iω² is NOT conserved when I changes (since ω changes too). When skater pulls arms in, L conserved, ω increases, KE increases (work done by muscles).
Root cause: concept gap
Memorise standard moments of inertia. Solid sphere has more mass near axis (smaller MOI = 2MR²/5); hollow sphere has all mass at radius R (larger MOI = 2MR²/3).
Root cause: unit error
Convert: ω(rad/s) = (2π/60) × ω(rpm). For example, 1200 rpm = 1200 × 2π/60 = 125.66 rad/s.
The angular acceleration of a body, moving along the circumference of a circle, is
forgets conversion rpm to rad s
Treats rpm as rad/s without 2*pi/60 conversion
uses equal distribution
Default to L/2 regardless of mass ratio
uses energy conservation instead
Confusing L conservation with KE conservation
swap solid hollow coefficients
Confusing 2/5 (solid sphere) with 2/3 (hollow sphere)
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