Torque and Rotational Motion
Since the year 2000, NASA and Roscosmos have operated the International Space Station at an orbital speed of 7.66 km/s and altitude of 408 km, completing 15.5 orbits per day with a period of 92.68 minutes. Crew members experience approximately 5×10⁻⁶ g — not because gravity is absent, but because the station and crew are in continuous circular motion whose mathematics this lesson unpacks.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Why is it easier to open a door by pushing near the handle than near the hinges? What quantity is different in each case?
Warm-up — what is the rotational equivalent of force called?
Know — Define and Calculate Torque
- Understand torque as the rotational equivalent of force
- Apply $\tau = rF_\perp = rF\sin\theta$ correctly
Understand — Analyse Lever Arms and Angles
- Identify the perpendicular distance from pivot to line of action
- Determine when torque is maximised or zero
Can Do — Solve Rotational Equilibrium Problems
- Apply the principle that net torque equals zero for balance
- Solve seesaw, door and wrench problems
Torque is measured in joules.
A force directed through the pivot produces a large torque.
Core Content
The rotational equivalent of force
Torque Diagram
Torque Detailed
Sit on a spinning office chair and extend your arms outward: you slow down. Pull them inward: you spin faster. The chair rotates continuously — and the rate of rotation changes depending on how mass is distributed around the axis. The quantity that causes rotation to change is torque, and it depends on three things:
- The magnitude of the applied force $F$
- The distance $r$ from the pivot (axis of rotation) to where the force is applied
- The angle $\theta$ between the force vector and the lever arm
$\tau = rF_\perp = rF\sin\theta$
$\tau$ = torque (N m) · $r$ = distance from pivot (m) · $F$ = applied force (N) · $\theta$ = angle between force and lever arm
Key insight: Only the perpendicular component of force ($F_\perp = F\sin\theta$) contributes to torque. The component parallel to the lever arm ($F_\parallel = F\cos\theta$) pulls along the arm and does not cause rotation.
The HSC syllabus uses the notation $\tau = rF_\perp = rF\sin\theta$. You must be able to identify $r$, $F$ and $\theta$ in any diagram, and calculate torque for any angle. The unit of torque is the newton-metre (N m) — do not confuse this with joules, even though the units look the same. Torque is not energy.
Torque $\tau = rF\sin\theta$ (N m), where $r$ is the pivot-to-force distance and $\theta$ is the angle between the force and the lever arm. Only $F_\perp = F\sin\theta$ causes rotation; torque is NOT energy (unit is N m, not J).
Pause — copy the highlighted torque formula and the unit rule into your book before moving on.
A force of 40 N is applied perpendicular to a lever 0.5 m from the pivot. The torque is:
How angle and distance affect rotation
We just saw how torque is defined as $\tau = rF\sin\theta$. That raises a question: which angle gives the most (or least) rotation effect? This card answers it → 90° gives maximum torque; 0° or 180° gives zero.
From $\tau = rF\sin\theta$, we can identify three special cases:
| Angle $\theta$ | $\sin\theta$ | Torque | Physical situation |
|---|---|---|---|
| $90°$ (force perpendicular to lever arm) | $1$ | $\tau_{\max} = rF$ | Pushing a door at right angles — maximum effectiveness |
| $0°$ or $180°$ (force parallel to lever arm) | $0$ | $\tau = 0$ | Pushing directly toward or away from the hinge — door does not rotate |
| $\theta$ between $0°$ and $90°$ | $0 < \sin\theta < 1$ | $0 < \tau < rF$ | Pushing a door at an angle — only the perpendicular component works |
Two ways to increase torque:
- Increase the force $F$ — push harder
- Increase the perpendicular distance $r$ — push farther from the pivot
Door handles are placed far from the hinges because this maximises $r$ and therefore maximises torque for the same pushing force. If you tried to open a door by pushing near the hinges, even a very large force would produce little rotation because $r$ is small.
Torque is maximised when $\theta = 90°$ ($\tau_{\max} = rF$) and is zero when the force is parallel to the lever arm ($\theta = 0°$ or $180°$). To increase torque: increase $F$ or increase $r$ (push farther from the pivot).
Add the highlighted angle rules to your notes before the check below.
Pushing a door perpendicular to its surface (at $90°$) produces the maximum torque.
A force directed directly toward the pivot produces a large torque.
Torque can be increased by pushing further from the pivot.
Calculating torque at different angles
We just saw that torque depends on angle, with maximum at 90°. That raises a question: how do we apply this to a real calculation with different angles? This card answers it → always identify $r$, $F$ and $\theta$, then substitute into $\tau = rF\sin\theta$.
A mechanic applies a force of 80 N to a wrench of length 0.25 m. Calculate the torque when: (a) the force is perpendicular to the wrench, (b) the force is at 30° to the wrench, (c) the force is directed along the wrench toward the nut.
- Given. $F = 80\ \text{N}$, $r = 0.25\ \text{m}$
- Find. Torque $\tau$ for three angles.
- Method. Use $\tau = rF\sin\theta$ for each case.
- $(a)\ \theta = 90°: \tau = 0.25 \times 80 \times \sin(90°) = 0.25 \times 80 \times 1 = \mathbf{20\ \text{N m}}$
- $(b)\ \theta = 30°: \tau = 0.25 \times 80 \times \sin(30°) = 0.25 \times 80 \times 0.5 = \mathbf{10\ \text{N m}}$
- $(c)\ \theta = 0°: \tau = 0.25 \times 80 \times \sin(0°) = 0.25 \times 80 \times 0 = \mathbf{0\ \text{N m}}$
- Answer. (a) 20 N m, (b) 10 N m, (c) 0 N m. Pushing along the wrench produces no rotation because the force is directed through the pivot.
Always identify $r$, $F$ and $\theta$ before substituting into $\tau = rF\sin\theta$. When $\theta = 0°$: $\tau = 0$ (force through the pivot); when $\theta = 90°$: $\tau = rF$ (maximum).
Pause — write the highlighted problem-solving steps into your book before the check below.
A mechanic applies 80 N to a wrench 0.25 m from the nut at 30° to the wrench handle. What is the torque?
Balancing torques for zero rotation
We just saw how to calculate torque from $\tau = rF\sin\theta$. That raises a question: what happens when multiple torques act together — can they cancel? This card answers it → in rotational equilibrium $\sum\tau = 0$; clockwise torques exactly balance anticlockwise ones.
When an object is in rotational equilibrium, the sum of all torques acting on it is zero:
$\sum \tau = 0$
Clockwise torques balance anticlockwise torques.
A uniform seesaw of length 3.0 m is pivoted at its centre. A child of mass 30 kg sits 1.2 m from the pivot on the left side. Where must a second child of mass 25 kg sit on the right side to balance the seesaw?
- Given. $m_1 = 30\ \text{kg}$ at $r_1 = 1.2\ \text{m}$; $m_2 = 25\ \text{kg}$; $g = 9.8\ \text{m/s}^2$
- Find. Distance $r_2$ for balance.
- Method. For balance, clockwise torque = anticlockwise torque. Both children sit perpendicular to the seesaw, so $\theta = 90°$ and $\sin\theta = 1$.
- $\tau_{\text{left}} = \tau_{\text{right}}$
- $r_1 F_1 = r_2 F_2$
- $r_1 m_1 g = r_2 m_2 g$
- $r_2 = \dfrac{r_1 m_1}{m_2} = \dfrac{1.2 \times 30}{25} = \dfrac{36}{25} = \mathbf{1.44\ \text{m}}$
- Answer. The 25 kg child must sit 1.44 m from the pivot on the right side. Notice that $g$ cancels — the balance position is independent of gravitational field strength.
Rotational equilibrium: $\sum\tau = 0$ (clockwise torques = anticlockwise torques). For two weights on a balanced beam: $r_1 m_1 g = r_2 m_2 g$, so $r_1 m_1 = r_2 m_2$ ($g$ cancels). Heavier objects sit closer to the pivot.
Add the highlighted equilibrium condition to your notes before the check below.
A 30 kg child sits 1.2 m from the pivot on a seesaw. A 25 kg child sits on the other side. What distance from the pivot balances the seesaw?
Why torque matters for rotating systems
We just saw torque applied to static equilibrium problems. That raises a question: does torque also matter for objects already moving in circles? This card answers it → net torque causes angular acceleration; centripetal force passes through the centre and produces zero torque.
In circular motion, a net torque causes a change in angular velocity (angular acceleration). This is the rotational equivalent of Newton's second law:
$\tau_{\text{net}} = I\alpha$
net torque = moment of inertia × angular acceleration
Beyond syllabus: The moment of inertia $I$ depends on how mass is distributed relative to the axis. The syllabus focuses on the definition and calculation of torque ($\tau = rF\sin\theta$), not on rotational dynamics. However, understanding that torque causes rotation (just as force causes acceleration) helps connect this topic to uniform circular motion.
Centripetal force ($F = mv^2/r$) maintains circular motion by pulling toward the centre. It does not cause angular acceleration because it acts through the centre of rotation, producing zero torque ($r = 0$ for the centre). Torque is required only to start, stop or change the speed of rotation.
Net torque causes angular acceleration: $\tau_{\text{net}} = I\alpha$. Centripetal force acts through the pivot (zero torque). Torque is needed only to start, stop or change rotational speed — not to maintain uniform circular motion.
Add the highlighted rotational analogy to your notes before the check below.
Three of these statements about torque are correct. Pick the odd one out.
Errors to avoid in torque problems
We just saw the connections between torque, circular motion and angular acceleration. That raises a question: what mistakes do students most commonly make on torque questions? This card answers it → torque is not a force, $r$ is the lever-arm distance, and the unit is N m not J.
Three key torque rules: (1) torque ≠ force — it is the turning effect; (2) $r$ is measured along the lever arm, not along the force direction; (3) units are N m, never joules, even though both are kg m² s⁻².
Pause — write the highlighted unit rule and the three distinctions into your book before moving on.
Complete the sentence. For rotational equilibrium, the sum of all torques must equal _____.
Practise resolving and analysing motion.
- A projectile is launched at 22 m/s, 34° above the horizontal. Find the horizontal and vertical components of the initial velocity.
- At a point in flight, the horizontal velocity is 15 m/s and the vertical velocity is 11 m/s. Find the resultant velocity (magnitude and direction).
- A ball is thrown horizontally at 11 m/s from a cliff 20 m high. Calculate the time of flight and the horizontal distance travelled.
A ball is thrown horizontally at 11 m/s from a 20 m cliff. Approximately how long does it take to hit the ground?
Explain the independence of horizontal and vertical motion.
Explain why a projectile launched horizontally and one dropped from the same height hit the ground simultaneously. Use the concept of independence of horizontal and vertical motion in your answer.
Misconceptions — final check
Copy into your books
Key Definitions
- Torque: rotational effect of a force
- Lever arm: perpendicular distance from pivot to line of force
- Rotational equilibrium: $\sum\tau = 0$
Torque Formula
- $\tau = rF_\perp = rF\sin\theta$
- Units: N m (not joules)
- Max at $\theta = 90°$; zero at $\theta = 0°$
Equilibrium Rule
- $\sum\tau = 0$ for balance
- $r_1 m_1 = r_2 m_2$ for seesaws
- $g$ cancels in balance problems
Key Principles
- Only $F\sin\theta$ causes rotation
- More distance from pivot → more torque
- Centripetal force → zero torque at centre
Which set of statements correctly describes torque?
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 4(3 marks) 1. A force of 50 N is applied to a door handle 0.80 m from the hinges. Calculate the torque when the force is (a) perpendicular to the door, (b) at 60° to the door surface. Explain which pushing direction is most effective for opening the door.
1 mark: correct torque for (a) · 1 mark: correct torque for (b) · 1 mark: explanation using $\sin\theta$
AnalyseBand 5(4 marks) 2. A metre ruler is pivoted at the 50 cm mark. A 200 g mass is hung at the 20 cm mark. Where must a 150 g mass be placed to balance the ruler horizontally? Show all working and state the principle used.
1 mark: principle stated · 1 mark: correct $r_1$ from pivot · 1 mark: correct method · 1 mark: position on ruler
EvaluateBand 6(5 marks) 3. Evaluate the statement: "A force always produces rotation if it is applied away from the pivot." Use the torque formula to identify a situation where a non-zero force applied away from the pivot produces zero torque. Explain the physical meaning of this result.
1 mark: identifies statement as incorrect · 1 mark: references $\tau = rF\sin\theta$ · 1 mark: identifies $\theta = 0°$ as the zero-torque case · 1 mark: physical explanation · 1 mark: example
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (3 marks):
(a) $\tau = rF\sin\theta = 0.80 \times 50 \times \sin(90°) = 0.80 \times 50 \times 1 = \mathbf{40\ \text{N m}}$ (1 mark)
(b) $\tau = 0.80 \times 50 \times \sin(60°) = 0.80 \times 50 \times 0.866 = \mathbf{34.6\ \text{N m}}$ (1 mark)
The perpendicular push is most effective because $\sin(90°) = 1$ is the maximum value of $\sin\theta$. Any angle less than $90°$ reduces the perpendicular component and therefore reduces torque (1 mark).
Q2 (4 marks):
Principle: For rotational equilibrium, the sum of clockwise torques equals the sum of anticlockwise torques ($\sum \tau = 0$) (1 mark).
Distance of 200 g mass from pivot: $r_1 = 50 - 20 = 30\ \text{cm} = 0.30\ \text{m}$
Let $r_2$ be the distance of the 150 g mass on the opposite side.
$r_1 m_1 g = r_2 m_2 g \Rightarrow r_2 = \dfrac{r_1 m_1}{m_2} = \dfrac{0.30 \times 0.200}{0.150} = \dfrac{0.060}{0.150} = \mathbf{0.40\ \text{m}}$ (2 marks for method + answer)
Position: $50 + 40 = \mathbf{90\ \text{cm}}$ mark (or 40 cm to the right of the pivot) (1 mark).
Q3 (5 marks):
The statement is incorrect (1 mark). From $\tau = rF\sin\theta$, torque depends on the angle $\theta$ between the force and the lever arm, not just on the magnitude of the force or its distance from the pivot (1 mark).
When $\theta = 0°$ or $180°$ (force directed along the lever arm, either toward or away from the pivot), $\sin\theta = 0$ and therefore $\tau = 0$ (1 mark).
Physical meaning: A force directed through or away from the pivot pulls along the lever arm but does not tend to rotate the object (1 mark). For example, pushing a door directly toward its hinges produces no rotation regardless of how hard you push (1 mark).
Five timed questions on torque and rotational motion. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaA quick arcade game to wind down — keep your momentum going before the next lesson.
At the start you were asked about the International Space Station — orbiting at 7.66 km/s, 408 km altitude, completing 15.5 orbits per day since 2000.
The ISS demonstrates circular motion on a planetary scale. Its angular velocity is $\omega = 2\pi/T = 2\pi/5561 \approx 1.13 \times 10^{-3} \text{ rad/s}$. Its linear orbital speed $v = 7.66 \text{ km/s}$ is related to this by $v = \omega r$. The 92.68-minute period, the altitude, and the speed are all inter-related through the circular motion equations you have now learned. Has your understanding of these quantities changed?