Prerequisites
Before studying this topic, review:
- Newton’s Laws - Force and acceleration
- Moment of Inertia - Rotational mass
- Vectors - Cross product
The Hook: Why Can’t You Open a Door Pushing Near the Hinges?
Ever tried opening a heavy door by pushing near the hinges? You push with all your strength - nothing happens! Move to the handle at the edge - the same force swings it open effortlessly!
In Jawan’s iconic train scene, when Shah Rukh Khan uses a long crowbar to pry open locked doors, he always applies force at the far end. Closer to the pivot? The door won’t budge!
Watch any mechanic loosen a tight bolt - they use a longer wrench. Same force, but suddenly the bolt turns!
The mystery: Why does the same force produce different rotational effects depending on where and how you apply it?
The answer: Torque - the rotational equivalent of force!
The Core Concept
What is Torque?
Torque (symbol: $\tau$, Greek letter “tau”) is the rotational analog of force - it measures the turning effect of a force about an axis.
Force makes things move linearly. Torque makes things rotate.
Just as:
- Force = push or pull that changes linear motion
- Torque = twist or turn that changes rotational motion
The key difference: Torque depends on:
- Magnitude of force ($F$)
- Distance from axis ($r$)
- Angle between $\vec{r}$ and $\vec{F}$
Definition of Torque
Vector definition:
$$\boxed{\vec{\tau} = \vec{r} \times \vec{F}}$$where:
- $\vec{r}$ = position vector from axis to point of application of force
- $\vec{F}$ = applied force
- $\times$ = cross product
Magnitude:
$$\boxed{\tau = rF\sin\theta}$$where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$.
Alternative forms:
$$\tau = r_{\perp} F = r F_{\perp}$$where:
- $r_{\perp} = r\sin\theta$ = perpendicular distance from axis to line of action of force (lever arm or moment arm)
Interactive Demo: Visualize Torque
Explore how force, distance, and angle combine to produce rotational effects.
- $F_{\perp} = F\sin\theta$ = component of force perpendicular to $\vec{r}$
SI Unit: Newton-meter (N·m)
Direction: Given by right-hand rule (perpendicular to plane containing $\vec{r}$ and $\vec{F}$)
Memory Tricks & Patterns
Mnemonic for Torque
Memory Trick: “Twist needs Radius and Force” → $\tau = r \times F$
Remember the cross: Cross product → maximum when perpendicular ($\sin 90° = 1$), zero when parallel ($\sin 0° = 0$)
Maximum Torque Rule
Torque is maximum when force is perpendicular to position vector
$$\tau_{max} = rF \quad (\text{when } \theta = 90°)$$Torque is zero when:
- Force is zero ($F = 0$)
- Applied at axis ($r = 0$)
- Force is parallel/antiparallel to $\vec{r}$ ($\theta = 0°$ or $180°$)
Door example:
- Push perpendicular at handle → maximum torque
- Push toward hinges → zero torque (along $\vec{r}$)
- Push near hinges → small $r$ → small torque
The “Lever Arm” Shortcut
Memory trick: “Longer Lever, Larger torque”
$$\tau = r_{\perp} \times F$$where $r_{\perp}$ is the perpendicular distance from axis to the line of action of force.
This is often the fastest way to calculate torque in JEE!
Newton’s Second Law for Rotation
Just as $\vec{F} = m\vec{a}$ for linear motion:
$$\boxed{\vec{\tau} = I\vec{\alpha}}$$where:
- $\tau$ = net torque (N·m)
- $I$ = moment of inertia (kg·m²)
- $\alpha$ = angular acceleration (rad/s²)
For multiple torques:
$$\boxed{\tau_{net} = \sum \tau_i = I\alpha}$$Sign convention:
- Counterclockwise (CCW) torque: Positive (+)
- Clockwise (CW) torque: Negative (−)
| Linear Motion | Rotational Motion |
|---|---|
| $\vec{F} = m\vec{a}$ | $\vec{\tau} = I\vec{\alpha}$ |
| Force $F$ | Torque $\tau$ |
| Mass $m$ | Moment of inertia $I$ |
| Acceleration $a$ | Angular acceleration $\alpha$ |
Pattern: Replace $F \to \tau$, $m \to I$, $a \to \alpha$ - and you’ve gone from linear to rotational!
Rotational Equilibrium
Conditions for Equilibrium
A body is in rotational equilibrium when it’s not rotating or rotating with constant angular velocity.
Condition: Net torque about any axis = 0
$$\boxed{\sum \vec{\tau} = 0}$$Or component-wise:
$$\sum \tau_{CW} = \sum \tau_{CCW}$$For complete equilibrium (both translational + rotational):
- Translational equilibrium: $\sum \vec{F} = 0$ (no linear acceleration)
- Rotational equilibrium: $\sum \vec{\tau} = 0$ (no angular acceleration)
For equilibrium problems:
- Choose a convenient axis (usually where you have unknown forces - their torque becomes zero!)
- Write torque equation: $\sum \tau = 0$
- Choose positive direction (CCW is standard)
- Solve for unknowns
Pro tip: In equilibrium, $\sum \tau = 0$ about any axis. So pick the axis that makes calculation easiest!
Important Cases of Torque
Case 1: Force Applied Perpendicular at Distance $r$
$$\tau = rF$$Example: Opening a door by pushing perpendicular at handle
Case 2: Force at Angle $\theta$
$$\tau = rF\sin\theta$$Example: Pulling a door at an angle
Case 3: Multiple Forces
$$\tau_{net} = \tau_1 + \tau_2 + \tau_3 + \cdots$$(with appropriate signs for direction)
Case 4: Continuous Force Distribution
$$\tau = \int r dF$$Example: Torque due to gravity on a rod - integrate along length
Torque Due to Gravity
For a rigid body, gravity acts at the centre of mass.
$$\boxed{\tau_g = r_{cm} \times Mg}$$where $r_{cm}$ is the position of COM from the axis.
Magnitude:
$$\tau_g = r_{cm} \cdot Mg \cdot \sin\theta$$This simplifies many problems! Treat entire weight as acting at COM.
Common Mistakes to Avoid
Mistake: Thinking larger force always means larger torque
Truth: Torque depends on both force AND distance: $\tau = rF\sin\theta$
Example:
- 10 N at 2 m perpendicular: $\tau = 20$ N·m
- 100 N at 0.1 m perpendicular: $\tau = 10$ N·m
- Smaller force wins!
Mistake: $\tau = r F$ when force is not perpendicular
Correct: $\tau = r_{\perp} F$ or $\tau = r F_{\perp}$ or $\tau = rF\sin\theta$
Example: Force at 30° to position vector:
- Wrong: $\tau = rF$
- Correct: $\tau = rF\sin 30° = 0.5rF$
Mistake: Mixing up clockwise and counterclockwise signs
Fix: Choose a convention at the start (CCW = +, CW = −) and stick to it throughout the problem!
Check: Net torque direction matches angular acceleration direction
Mistake: Calculating torque for force through axis or parallel to $\vec{r}$
Remember - Torque is ZERO when:
- Force passes through axis ($r = 0$)
- Force parallel to $\vec{r}$ ($\sin\theta = 0$)
- Force is zero ($F = 0$)
Example: Tension in a string pulling radially on a pulley produces zero torque!
Worked Examples
Level 1: Foundation (NCERT)
A force of 10 N is applied perpendicular to a wrench at a distance of 0.3 m from the bolt. Find the torque.
Solution:
Force is perpendicular, so $\theta = 90°$:
$$\tau = rF\sin\theta = 0.3 \times 10 \times \sin 90°$$ $$\tau = 0.3 \times 10 \times 1 = \boxed{3 \text{ N·m}}$$A force of 20 N is applied at a distance of 0.5 m from the axis at an angle of 30° to the position vector. Calculate the torque.
Solution:
$$\tau = rF\sin\theta = 0.5 \times 20 \times \sin 30°$$ $$\tau = 0.5 \times 20 \times 0.5 = \boxed{5 \text{ N·m}}$$A wheel of moment of inertia 2 kg·m² experiences a net torque of 8 N·m. Find the angular acceleration.
Solution:
Using $\tau = I\alpha$:
$$\alpha = \frac{\tau}{I} = \frac{8}{2} = \boxed{4 \text{ rad/s}^2}$$Level 2: JEE Main
A uniform seesaw of length 4 m and mass 20 kg is balanced on a pivot at its center. A child of mass 30 kg sits 1.5 m from the left end. Where should a 40 kg child sit from the right end to balance the seesaw?
Solution:
Choose pivot as axis (weight of seesaw produces zero torque about center).
Left child: Distance from center = $2 - 1.5 = 0.5$ m (left side)
Right child: Let distance from center = $x$ (right side)
Torque balance (taking CCW as positive):
$$\tau_{left} + \tau_{right} = 0$$ $$30g(0.5) - 40g(x) = 0$$ $$15g = 40gx$$ $$x = \frac{15}{40} = 0.375 \text{ m from center}$$Distance from right end: $2 - 0.375 = \boxed{1.625 \text{ m}}$
A uniform ladder of length 5 m and mass 10 kg leans against a smooth wall at an angle of 60° to the ground. The ground is rough. Find the normal force from the wall.
Solution:
Forces:
- Weight $W = 10g$ at center (2.5 m from bottom)
- Normal from ground: $N_1$ (upward)
- Normal from wall: $N_2$ (horizontal)
- Friction from ground: $f$ (horizontal)
Taking torque about bottom point (eliminates $N_1$ and $f$):
Torque by weight:
- Perpendicular distance from bottom = $2.5\cos 60° = 1.25$ m
- $\tau_W = 10g \times 1.25$ (clockwise)
Torque by $N_2$:
- Perpendicular distance = $5\sin 60° = 5\sqrt{3}/2$ m
- $\tau_{N_2} = N_2 \times 5\sqrt{3}/2$ (counterclockwise)
Equilibrium:
$$N_2 \times \frac{5\sqrt{3}}{2} = 10g \times 1.25$$ $$N_2 = \frac{10g \times 1.25 \times 2}{5\sqrt{3}} = \frac{25g}{5\sqrt{3}} = \frac{5g}{\sqrt{3}}$$ $$N_2 = \frac{5 \times 10}{\sqrt{3}} = \frac{50}{\sqrt{3}} \approx \boxed{28.9 \text{ N}}$$A uniform disc of mass 2 kg and radius 0.5 m is free to rotate about a horizontal axis through its center. A rope is wrapped around it with masses 1 kg and 3 kg hanging from its ends. Find: a) Angular acceleration of the disc b) Tensions in the rope
Solution:
Moment of inertia of disc: $I = \frac{MR^2}{2} = \frac{2 \times (0.5)^2}{2} = 0.25$ kg·m²
For 3 kg mass (downward positive):
$$3g - T_1 = 3a \quad \text{...(1)}$$For 1 kg mass (upward positive):
$$T_2 - 1g = 1a \quad \text{...(2)}$$For disc (torque equation):
$$\tau_{net} = I\alpha$$ $$T_1 R - T_2 R = I\alpha$$Since $a = R\alpha$:
$$R(T_1 - T_2) = I \frac{a}{R}$$ $$T_1 - T_2 = \frac{Ia}{R^2} \quad \text{...(3)}$$Adding (1) and (2):
$$3g - T_1 + T_2 - g = 4a$$ $$2g + (T_2 - T_1) = 4a$$From (3): $T_2 - T_1 = -\frac{Ia}{R^2}$
$$2g - \frac{Ia}{R^2} = 4a$$ $$2g = 4a + \frac{Ia}{R^2} = a\left(4 + \frac{I}{R^2}\right)$$ $$a = \frac{2g}{4 + I/R^2} = \frac{2 \times 10}{4 + 0.25/0.25} = \frac{20}{4 + 1} = \boxed{4 \text{ m/s}^2}$$Angular acceleration:
$$\alpha = \frac{a}{R} = \frac{4}{0.5} = \boxed{8 \text{ rad/s}^2}$$From (1): $T_1 = 3g - 3a = 30 - 12 = \boxed{18 \text{ N}}$
From (2): $T_2 = g + a = 10 + 4 = \boxed{14 \text{ N}}$
Level 3: JEE Advanced
A uniform rod of mass $M$ and length $L$ is hinged at one end and released from horizontal position. Find: a) Initial angular acceleration b) Linear acceleration of free end c) Reaction at hinge at initial moment
Solution:
a) Initial angular acceleration:
Torque about hinge (by weight acting at COM):
$$\tau = Mg \times \frac{L}{2} \times \sin 90° = \frac{MgL}{2}$$Moment of inertia about hinge:
$$I = \frac{ML^2}{3}$$Angular acceleration:
$$\alpha = \frac{\tau}{I} = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L} = \boxed{\frac{3g}{2L}}$$b) Linear acceleration of free end:
$$a = L\alpha = L \times \frac{3g}{2L} = \boxed{\frac{3g}{2}}$$(Greater than $g$! Free end accelerates faster than free fall!)
c) Reaction at hinge:
At initial moment, rod is horizontal.
COM acceleration: $a_{cm} = \frac{L}{2} \alpha = \frac{L}{2} \times \frac{3g}{2L} = \frac{3g}{4}$ (downward)
Vertical forces:
$$Mg - R = M a_{cm} = M \times \frac{3g}{4}$$ $$R = Mg - \frac{3Mg}{4} = \boxed{\frac{Mg}{4}}$$Horizontal component: $R_x = 0$ (no horizontal acceleration of COM initially)
Total reaction: $\boxed{R = \frac{Mg}{4} \text{ (vertical)}}$
A uniform cube of side $a$ and mass $m$ rests on a rough horizontal surface (coefficient of friction $\mu$). A horizontal force $F$ is applied at height $h$ from the base. Find: a) Condition for sliding b) Condition for toppling c) Which occurs first if $h = 3a/4$?
Solution:
a) Condition for sliding:
Maximum friction: $f_{max} = \mu N = \mu mg$
For sliding: $F > \mu mg$
$$\boxed{F > \mu mg}$$b) Condition for toppling:
Cube topples about the edge opposite to the applied force.
Torque about edge:
- By $F$: $\tau_F = F \times h$ (clockwise)
- By weight: $\tau_W = mg \times \frac{a}{2}$ (counterclockwise, perpendicular distance from edge to COM)
For toppling:
$$F \times h > mg \times \frac{a}{2}$$ $$\boxed{F > \frac{mga}{2h}}$$c) For $h = 3a/4$:
Sliding condition: $F > \mu mg$
Toppling condition: $F > \frac{mga}{2 \times 3a/4} = \frac{2mg}{3}$
Comparison:
- If $\mu > \frac{2}{3}$: Toppling occurs first (at $F = \frac{2mg}{3}$)
- If $\mu < \frac{2}{3}$: Sliding occurs first (at $F = \mu mg$)
For $h = 3a/4$:
- Toppling threshold: $\frac{2mg}{3}$
- Sliding threshold: $\mu mg$
If $\mu = 0.5 < 2/3$: Sliding first If $\mu = 0.8 > 2/3$: Toppling first
A solid cylinder of radius $R$ and mass $m$ is placed on a truck. The truck accelerates with acceleration $a$. What maximum acceleration can the truck have without the cylinder slipping? Coefficient of friction = $\mu$.
Solution:
For cylinder (translational motion):
$$f = ma \quad \text{...(1)}$$where $f$ is friction force.
For cylinder (rotational motion about COM):
Torque by friction:
$$\tau = fR = I\alpha$$For solid cylinder: $I = \frac{mR^2}{2}$
$$fR = \frac{mR^2}{2} \alpha \quad \text{...(2)}$$For pure rolling (no slipping):
$$a = R\alpha \quad \text{...(3)}$$From (3): $\alpha = \frac{a}{R}$
Substituting in (2):
$$fR = \frac{mR^2}{2} \times \frac{a}{R} = \frac{maR}{2}$$ $$f = \frac{ma}{2} \quad \text{...(4)}$$Maximum friction available:
$$f_{max} = \mu mg$$For no slipping:
$$\frac{ma}{2} \leq \mu mg$$ $$\boxed{a \leq 2\mu g}$$Maximum acceleration: $a_{max} = 2\mu g$
Real-World Applications
Wrenches and Spanners: Longer handle → larger $r$ → more torque for same force
Doors: Handle at edge → maximum torque; push near hinge → almost no rotation
Seesaws and Balances: Torque balance determines equilibrium position
Steering Wheels: Large diameter → easier to turn (more torque for same force)
Tightrope Walking: Long pole increases moment of inertia → resists toppling torque
Gyroscopes in Phones: Detect orientation changes via torque measurements
Quick Revision Box
| Concept | Formula | Key Point |
|---|---|---|
| Torque | $\vec{\tau} = \vec{r} \times \vec{F}$ | Turning effect of force |
| Magnitude | $\tau = rF\sin\theta$ | Maximum when $\perp$ |
| Lever arm | $\tau = r_\perp F$ | Perpendicular distance |
| Newton’s 2nd Law | $\tau = I\alpha$ | Rotational analog |
| Equilibrium | $\sum \tau = 0$ | No angular acceleration |
| Direction | Right-hand rule | CCW = +, CW = − |
Memory aid:
- Force → linear motion → $F = ma$
- Torque → rotational motion → $\tau = I\alpha$
Decision Tree: Solving Torque Problems
Step 1: Identify all forces and their points of application
Step 2: Choose an axis of rotation
- For equilibrium: Choose axis through unknown forces (their torque = 0)
- For dynamics: Usually axis through hinge/pivot
Step 3: Calculate torque for each force
- Use $\tau = r_\perp F$ (easiest) or $\tau = rF\sin\theta$
- Assign signs: CCW = +, CW = −
Step 4: Apply equation
- Equilibrium: $\sum \tau = 0$
- Dynamics: $\sum \tau = I\alpha$
Step 5: Solve for unknowns
Pro tip: For equilibrium, try different axes - pick the one that eliminates most unknowns!
Practice Problems
Foundation Level
A force of 50 N is applied at the edge of a door 0.8 m wide at an angle of 60° to the door. Find the torque about the hinges.
A disc of moment of inertia 0.5 kg·m² is subjected to a torque of 2 N·m. Find the angular acceleration.
A uniform rod of length 2 m and mass 5 kg is pivoted at its center. Masses of 2 kg and 3 kg are hung from its ends. Is it in equilibrium?
JEE Main Level
A uniform beam of length 6 m and mass 20 kg is supported at its ends. A person of mass 60 kg stands 2 m from one end. Find the reactions at the supports.
A wheel of radius 0.4 m and moment of inertia 2 kg·m² is rotating at 10 rad/s. A tangential force of 5 N acts on it for 4 seconds. Find the final angular velocity.
A thin rod of length $L$ and mass $M$ is hinged at one end. What horizontal force applied at the free end will keep it inclined at 30° to vertical in equilibrium?
JEE Advanced Level
A solid sphere and a hollow sphere of same mass and radius roll down an incline. Compare the forces of friction acting on them.
A uniform rod AB of mass $m$ and length $L$ is placed on a smooth table. Two forces each of magnitude $F$ act at A and B perpendicular to the rod in opposite directions. Find: a) Linear acceleration of COM b) Angular acceleration about COM
A disc of mass $M$ and radius $R$ has a string wrapped around it with a mass $m$ hanging. The disc is free to rotate about a fixed horizontal axis through its center. Find: a) Acceleration of mass $m$ b) Tension in string c) Angular acceleration of disc
Connection to Other Topics
Prerequisites you should know:
- Newton’s Laws - $\tau$ is rotational analog of $F$
- Moment of Inertia - Needed for $\tau = I\alpha$
- Vectors - Cross product for torque
What’s next:
- Angular Momentum - $\tau = \frac{dL}{dt}$
- Rolling Motion - Torque by friction enables rolling
- Work in Rotation - $W = \tau \theta$
Cross-chapter links:
- Equilibrium of Rigid Bodies - Static equilibrium uses $\sum \tau = 0$
- Circular Motion - Centripetal force can create torque
- Gravitation - Torque by gravitational forces
Teacher’s Summary
Torque is the rotational analog of force - it’s the “twist” that causes angular acceleration, just as force causes linear acceleration
Three ways to calculate torque:
- $\tau = rF\sin\theta$ (when you know the angle)
- $\tau = r_\perp F$ (perpendicular distance - often fastest!)
- $\tau = rF_\perp$ (perpendicular component of force)
Torque is zero when:
- Force through axis ($r = 0$)
- Force parallel to $\vec{r}$ ($\sin\theta = 0$)
- No force ($F = 0$)
Newton’s Second Law for rotation: $\tau = I\alpha$ (just like $F = ma$)
Equilibrium strategy: Choose axis through unknown forces → their torque = 0 → easier calculation!
JEE loves: Ladder problems, seesaw/beam equilibrium, pulley with disc/cylinder, toppling vs sliding
Sign convention matters: CCW = +, CW = − (or vice versa, but be consistent!)
“Force says ‘move,’ torque says ’turn’ - master torque and you control rotation!”
This topic is a goldmine - 3-5 questions (12-20 marks) in JEE involve torque. Practice equilibrium and $\tau = I\alpha$ problems until they become second nature!