Rotational Motion & Torque
Why a tiny push far from a hinge beats a huge shove close to it — the physics of turning things.
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The big idea: Forces don't just push things in straight lines — they can twist them around a pivot, and how much twisting power a force has depends on both how hard you push and how far from the pivot you push. That twisting power is torque, and once you can see torque, seesaws, wrenches, doors, and spinning wheels all become the same simple idea wearing different costumes.
Turning is different from pushing
Push a box across the floor and it slides in a straight line — simple. But push on the edge of a door, or the end of a wrench, and something different happens: the object rotates around a fixed point instead of moving in a line. That fixed point is called the pivot (or axis of rotation), and the "turning power" of a force around that pivot has its own name: torque.
Here's the key intuition to build first: the exact same force can be almost useless or incredibly powerful at turning something, depending on where you apply it. Push right next to a door's hinge and nothing happens, even if you push hard. Push at the far edge of the door, and it swings open with barely any effort. Same force, wildly different result — because torque cares about distance from the pivot just as much as it cares about the size of the push.
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The lever principle, in one sentence
Torque is force times the perpendicular distance from the pivot to where that force is applied \( (\tau = Fd) \) — so you can trade force for distance, and that trade is the secret behind every lever, wrench, and seesaw ever built.
\[ \tau = F \, d_\perp \]
Torque \(\tau\) equals force \(F\) times the perpendicular (moment-arm) distance \(d_\perp\) from the pivot to the line of action of the force. Units: newton-metres (N·m).
\[ \sum \tau = 0 \quad \text{and} \quad \sum F = 0 \]
An object is in complete static equilibrium only when both conditions hold: the net force is zero (\(\sum F = 0\), nothing slides) and the net torque about any pivot is zero (\(\sum \tau = 0\), the clockwise and counterclockwise torques balance exactly). That combination is what lets a seesaw sit level or a sign hang motionless.
🎮 Interactive: Balance the seesaw LIVE
Drag the weights and their distances from the pivot and watch the torques update live. Find how many different combinations of force and distance produce the exact same balance — that's the lever principle in action.
📝 Worked example: A 300 N child sits 2.0 m from the pivot of a seesaw. How far from the pivot must a 450 N adult sit on the other side for the seesaw to balance?
- For balance, the torques on each side must be equal: \(F_1 d_1 = F_2 d_2\).
- Plug in the child's side: \(300\text{ N} \times 2.0\text{ m} = 600\text{ N·m}\).
- Set the adult's torque equal to that: \(450\text{ N} \times d_2 = 600\text{ N·m}\).
- Solve for \(d_2\): \(d_2 = 600 / 450 = 1.33\text{ m}\).
✓ The adult must sit about 1.33 m from the pivot — closer than the child, because their weight is greater.
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Angular motion is just linear motion's rotating cousin
Every point on a spinning object shares the same angular velocity \( \omega \) (how fast the angle is changing), but points farther from the pivot cover more actual distance per second — their linear speed is \( v = \omega r \). That's why the tip of a helicopter blade moves far faster than the point near the hub, even though every point completes the same rotation in the same time. It's also why the centre of mass — the object's balance point — is what actually moves in a straight line when you throw a spinning object, while everything else swirls around it.
\[ v = \omega r \]
Linear speed \(v\) of a point on a rotating object equals its angular velocity \(\omega\) times its distance \(r\) from the pivot.
📝 Worked example: You turn a bolt with a 0.30 m wrench, pushing with 50 N perfectly perpendicular to the wrench. Then you get tired and push with the same 50 N, but now at only 30° to the wrench instead of 90°. How much torque do you lose?
- Perpendicular case: the full force counts, so \(\tau_1 = F d = 50 \times 0.30 = 15\text{ N·m}\).
- At an angle, only the component perpendicular to the wrench contributes: effective distance \(= d\sin(\theta)\), where \(\theta\) is the angle between the force and the wrench.
- Here \(\theta = 30°\), so effective distance \(= 0.30 \times \sin(30°) = 0.30 \times 0.5 = 0.15\text{ m}\).
- New torque: \(\tau_2 = 50 \times 0.15 = 7.5\text{ N·m}\).
✓ You lose exactly half your torque — dropping from 15 N·m to 7.5 N·m — just by pushing at a shallow angle instead of straight-on. That's why mechanics always tell you to push perpendicular to the wrench.
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Distance means perpendicular distance — not straight-line distance
A common mix-up: torque doesn't just care about how far the force is applied from the pivot along the object — it cares about the perpendicular distance from the pivot to the force's line of action. Push straight down the length of a wrench (along its axis) and no matter how hard you push, the torque is zero, because that force has no perpendicular distance from the pivot at all. Always ask: if I extend this force into a line, how far is the pivot from that line?
Check your understanding
1. What two quantities determine the torque of a force about a pivot?
Torque is \(\tau = Fd_\perp\) — it depends on both how strong the force is and how far (perpendicularly) it acts from the pivot. Neither one alone tells you the torque.
2. A mechanic applies 40 N of force perpendicular to a wrench, at a distance of 0.25 m from the bolt. What torque is produced?
\(\tau = Fd = 40\text{ N} \times 0.25\text{ m} = 10\text{ N·m}\).
3. Why is a door handle placed far from the hinge instead of right next to it?
Since \(\tau = Fd\), increasing \(d\) (the distance from the hinge) means you need less force \(F\) to reach the same torque needed to swing the door — that's the whole point of a lever arm.
4. Two riders sit on a spinning merry-go-round: one near the center, one near the outer edge. Both complete one full rotation in the same time. Which statement is true?
Every point on a rigid rotating object shares the same angular velocity \(\omega\), but linear speed is \(v = \omega r\) — since the outer rider has a larger radius \(r\), they travel a greater distance in the same time, meaning a higher linear speed.
✅ Key takeaways
- Torque \(\tau = Fd\) measures a force's ability to rotate something about a pivot — it depends on both force size and perpendicular distance from the pivot.
- The lever principle (\(F_1d_1 = F_2d_2\) at balance) explains why small forces far from a pivot can outmatch large forces applied close to it.
- Complete static equilibrium requires both the net force and the net torque on an object to be zero.
- Angular velocity \(\omega\) is shared by every point on a rotating rigid body, but linear speed \(v = \omega r\) grows with distance from the pivot — the centre of mass is the point that moves in a straight line overall.