Torque = dL - dt: Worked examples — Physics deep dive

We only use tools already built in the parent: angular momentum $\vec L=\vec r\times\vec p$, the cross product (which measures the perpendicular part of one vector against another), Newton's Second Law, moment of inertia $I$, and Conservation of Angular Momentum. Everything else we re-explain on the spot.

First, one new symbol: angular acceleration $α$

The parent used $ω$ (angular velocity, "how fast it spins") and $I$ (moment of inertia, "how hard it is to spin"). We now need one more.

With this defined, the parent's shortcut $τ = I α$ (valid only when $I$ is constant) is now fully spelled out: differentiate $L = I ω$ with $I$ fixed to get $τ = I d ω / d t = I α$ .

The scenario matrix

Before working problems, let us list the kinds of situation this one law produces. Each row is a "cell" — a distinct behaviour. The examples below are each tagged with the cell they cover.

Cell	What changes	Sign / degeneracy	Which relation to use
A	$ω$ grows, $I$ fixed	positive torque (speed-up)	$τ = I α$
B	$ω$ shrinks, $I$ fixed	negative torque (braking)	$τ = I α$ , watch sign
C	starts at $ω = 0$	torque creates $L$ from zero	$τ = d L / d t$ , integrate
D	$I$ changes, no torque	$τ = 0$ , $L$ conserved	$I_{1} ω_{1} = I_{2} ω_{2}$
E	$I$ changes AND $ω$ changes	must use $d (I ω) / d t$	full $τ = d L / d t$
F	force points along $r$	degenerate: $τ = 0$	$r \times F$ , $sin θ$
G	straight-line motion, off-line point	non-spinning yet $L \neq = 0$	$L = m v d$
H	real-world word problem	two-body, string+pulley	Newton + $τ = I α$
I	exam twist: vector directions	sign of $L$ , right-hand rule	$τ = r \times F$

We now cover cells A–I in order.

Reading the sign of a rotation (needed for cells A, B, I)

Everything below uses one convention. Look at the wheel below.

In the figure: the cyan circle is the wheel's rim; the amber curved arrow shows the chosen positive (CCW) sense; the small cyan circle-with-a-dot at the centre is the standard symbol for a vector pointing out of the page ( $+ \overset{z}{^}$ ) — that is the direction of $ω$ , $α$ , and $τ$ for CCW quantities. The white straight arrow on the right is a tangential drive force; notice it touches the rim sideways, not aimed at the centre, which is exactly what lets it twist the wheel.

This single rule lets us keep track of speed up versus slow down without ever getting lost.

Cell A — Positive torque, constant $I$ (speed-up)

Worked example A · A flywheel driven by a belt

A flywheel has $I = 4 kg\cdotm^{2}$ . A belt pulls tangentially at its rim (radius $R = 0.5$ m) with tension $T$ of magnitude $20$ N in the CCW sense. It starts from rest. Find the angular acceleration $α$ and the angular speed after $3$ s.

Forecast: Guess — will $ω$ after 3 s be closer to 1, 7, or 70 rad/s?

Step 1. Torque about the axle: $τ = r \times T$ , magnitude $τ = R T sin 90° = 0.5 \times 20 = + 10 N\cdotm$ , direction $+ \overset{z}{^}$ (out of page). Why this step? The belt leaves the rim tangentially, so $r ⊥ T$ , giving $sin 90° = 1$ — the whole force twists. By the right-hand rule $r \times T$ points out of the page, i.e. positive.

Step 2. $I$ is constant, so $τ = d L / d t = I α \Rightarrow α = τ / I = 10/4 = + 2.5 rad/s^{2}$ . Why this step? A rigid flywheel does not change shape, so we may pull $I$ out of $d (I ω) / d t$ , leaving $I d ω / d t = I α$ .

Step 3. Constant $α$ from rest: $ω = α t = 2.5 \times 3 = 7.5 rad/s$ . Why this step? Constant torque on constant $I$ gives constant $α$ , so the " $v = a t$ " analogue applies with $ω, α$ .

Verify: Units: $(N\cdotm) / (kg\cdotm^{2}) = kg\cdotm^{2} s^{- 2} / kg\cdotm^{2} = s^{- 2}$ ✓ (that is rad/s²). Sign positive → speeds up CCW, as a driving belt should. Answer 7.5 rad/s sits between the guesses — moderate.

Cell B — Negative torque, constant $I$ (braking)

Worked example B · Friction stops the flywheel

The same flywheel ( $I = 4$ ) now spins at $ω_{0} = 7.5$ rad/s CCW. A brake applies a constant torque of magnitude $6$ N·m opposing the motion. How long until it stops?

Forecast: Longer or shorter than the 3 s it took to spin up?

Step 1. The brake opposes CCW motion, so $τ$ points into the page: $τ = - 6 N\cdotm$ (negative). Why this step? Positive $ω$ (out of page); a torque that reduces it must point the opposite way (into page), hence the negative sign — this is the whole point of the sign convention.

Step 2. $α = τ / I = - 6/4 = - 1.5 rad/s^{2}$ . Why this step? $I$ constant again, so $τ = I α$ . Negative $α$ means $α$ opposes $ω$ — deceleration.

Step 3. Stop when $ω = 0$ : $0 = ω_{0} + α t \Rightarrow t = - ω_{0} / α = - 7.5/ (- 1.5) = 5 s$ . Why this step? Same kinematics, now decelerating; solve for the time the speed hits zero.

Verify: $t = 5$ s is positive (good — a negative time would mean an impossible answer). It takes longer to stop than to spin up because the brake torque (6) is weaker than the drive torque (10). Units: $(rad/s) / (rad/s^{2}) = s$ ✓.

Cell C — Torque creates $L$ from zero (starting from rest, integrate)

This is the trap "if $ω = 0$ there is no $L$ , so torque does nothing." Wrong — torque builds $L$ .

Worked example C · A time-varying torque from rest

A disc ( $I = 2 kg\cdotm^{2}$ ) is at rest. A torque $τ (t) = 3 t N\cdotm$ (growing with time, all along $+ \overset{z}{^}$ ) is applied. Find $L$ and $ω$ at $t = 4$ s.

Forecast: Because $τ$ grows, do you expect $ω$ to grow linearly or faster than linearly?

Step 1. Use the fundamental form directly: $\frac{d L}{d t} = τ (t) = 3 t$ . Why this step? $τ$ is not constant, so $α$ is not constant either — we cannot use $ω = α t$ . But $τ = d L / d t$ always holds, and we can integrate it.

Step 2. Integrate from rest ( $L (0) = 0$ ): $L (4) = \int_{0}^{4} 3 t d t = \frac{3}{2} t^{2}_{0}^{4} = \frac{3}{2} \times 16 = 24 kg\cdotm^{2} / s .$ Why this step? The area under the torque-versus-time curve is the accumulated angular momentum — the "impulse" idea for rotation.

Step 3. $ω = L / I = 24/2 = 12 rad/s$ . Why this step? With constant $I$ , $L = I ω$ , so dividing recovers the speed.

Verify: Units of $L$ : $(N\cdotm) (s) = kg\cdotm^{2} s^{- 1}$ ✓. $ω$ grew from zero — torque manufactured all of it. Because $τ \propto t$ , $L \propto t^{2}$ : faster than linear, matching the forecast.

Cell D — $I$ changes, no external torque ( $L$ conserved)

Cell E — Both $I$ and $ω$ change (must use full $d (I ω) / d t$ )

Here $τ = I α$ is illegal — this cell exists to force the fundamental form.

Worked example E · A spinning platform gaining mass

A turntable's moment of inertia grows linearly as sand falls on it: $I (t) = (2 + t) kg\cdotm^{2}$ . It is driven so that its speed also grows: $ω (t) = (1 + t) rad/s$ . Find the torque needed at $t = 2$ s.

Forecast: Will the required torque be smaller or larger than the naive $I α$ estimate?

Step 1. Write $L (t) = I (t) ω (t) = (2 + t) (1 + t)$ . Why this step? Both factors depend on time, so we cannot pull either out; we must form $L$ first.

Step 2. Expand and differentiate: $L = 2 + 3 t + t^{2}$ , so $τ = \frac{d L}{d t} = 3 + 2 t$ . Why this step? $τ = d L / d t$ is the only correct relation when $I$ varies; the product rule automatically includes both the $I d ω / d t$ and $ω d I / d t$ effects.

Step 3. At $t = 2$ : $τ = 3 + 2 (2) = 7 N\cdotm$ . Why this step? Plug the instant of interest into the derivative.

Verify: The naive (wrong) $I α$ would give $I (2) \cdot \overset{ω}{˙} = 4 \times 1 = 4$ N·m — too small, because it ignores the $ω d I / d t = 3 \times 1 = 3$ term. Correct total $4 + 3 = 7$ ✓ matches. This is exactly why the parent warns "trust the fundamental form."

Cell F — Force along $r$ : the degenerate zero-torque case

Worked example F · Pushing straight at the pivot

A rod is hinged at $O$ . You push a point on the rod at distance $r = 0.4$ m with force $F$ of magnitude $15$ N, but the force points directly toward $O$ (along $r$ ). Find the torque.

Forecast: Zero, or $0.4 \times 15 = 6$ N·m?

Step 1. Torque magnitude $τ = ∣ r \times F ∣ = r F sin θ$ , where $θ$ is the angle between $r$ and $F$ . Why this step? The cross product keeps only the part of the force perpendicular to $r$ ; $sin θ$ measures that perpendicular fraction.

Step 2. Force points toward $O$ , so $F$ is anti-parallel to $r$ : $θ = 180°$ , $sin 180° = 0$ . Why this step? A push aimed at the hinge has no sideways "twisting" component — geometrically it cannot rotate anything. (Same reasoning kills the $v \times p$ term in the parent's derivation.)

Step 3. $τ = 0.4 \times 15 \times 0 = 0 N\cdotm$ . Why this step? Zero perpendicular component ⇒ zero torque ⇒ no change in $L$ .

Verify: A radial (central) force gives zero torque — the same fact that makes planetary orbits conserve angular momentum. Answer 0, not 6, catching the forecast trap.

Cell G — Straight-line motion, off-line point (non-spinning, yet $L \neq = 0$ )

In the figure: the cyan horizontal line is the car's straight path; the amber dot at the origin is the observation point $O$ ; the two white arrows are the position vector $r$ drawn to two different car positions; the amber vertical double-arrow marks the fixed perpendicular distance $d = 3$ m. The key visual: although $r$ swings and lengthens as the car moves, its perpendicular projection onto $O$ — the quantity $r sin θ$ — is always the same $d$ , which is why $L$ never changes.

Worked example G · A car driving past you

A $1200$ kg car moves in a straight line at constant $v = 10$ m/s. Its path passes at a perpendicular distance $d = 3$ m from you (point $O$ ). What is its angular momentum about $O$ , and does it change?

Forecast: Zero (it is not going around you), or something nonzero?

Step 1. For straight-line motion, $L = m v d$ with $d$ the perpendicular ("closest approach") distance from $O$ to the line. Why this step? $L = r \times m v$ has magnitude $m v (r sin θ)$ , and $r sin θ$ is exactly that perpendicular distance $d$ — see the figure's right triangle.

Step 2. $L = 1200 \times 10 \times 3 = 36000 kg\cdotm^{2} / s$ . Why this step? Plug in; the value is fixed because $m$ , $v$ , and $d$ are all constant along the straight path.

Step 3. Does $L$ change? We must check the net torque about $O$ , and specifically its component along the $L$ -axis (here vertical, $\pm \overset{z}{^}$ ). Gravity ( $m g$ down) and the road's normal force ( $N$ up) are real, nonzero forces, and each can have a nonzero torque about $O$ . But both point vertically (in the plane of the page here), so $r \times F$ for each lies in the horizontal plane — it has zero component along $\overset{z}{^}$ . Along the axis of $L$ , the net torque is zero, so $L_{z}$ (our $36000$ ) does not change. Why this step? "No horizontal force" is the honest statement, not "no force at all." Vertical forces do exert torque, but only about horizontal axes; they cannot alter the vertical angular momentum. If the car also travels at constant height, those vertical torques stay balanced and never tip it.

Verify: As the car moves, $r$ shrinks then grows and $θ$ changes, but the product $r sin θ = d$ stays exactly $3$ m (look at the figure — every position projects to the same $d$ ). So $L_{z}$ is constant at $36000$ . Angular momentum does not require circular motion.

Cell H — Real-world word problem (string + pulley, two bodies)

In the figure: the cyan circle is the pulley (moment of inertia $I$ , radius $R$ ); the white vertical line on its right is the string; the amber arrow up at the rim is the tension $T$ that produces the torque; the cyan rectangle below is the hanging block; on the block, the small amber arrow up is tension and the white arrow down is gravity $m g$ ; the amber arrow beside the block marks its downward acceleration $a$ . Watch how the tension acts tangentially at the rim — that is what makes $sin 90° = 1$ in the torque.

Worked example H · Falling block turns a pulley

A block of mass $m = 2$ kg hangs from a light string wrapped around a pulley of moment of inertia $I = 0.1 kg\cdotm^{2}$ and radius $R = 0.2$ m. Released from rest, the block falls. Find its downward acceleration $a$ . (Take $g = 10 m/s^{2}$ .)

Forecast: More or less than free-fall $g = 10$ ? By roughly how much?

Step 1. Block (linear, Newton's Second Law): taking down as positive, $m g - T = ma$ , where $T$ is the string tension magnitude. Why this step? Two vertical forces on the block — gravity down, tension up.

Step 2. Pulley (rotational): the falling block unwinds the string on the right, spinning the pulley CW as drawn — take that spin sense as positive. The tension acts tangentially at the rim, so $τ = r \times T$ has magnitude $τ = R T sin 90° = R T$ and points along $+$ (the CW spin axis), i.e. it drives the rotation. Thus $τ = I α$ gives $R T = I α$ with both $τ$ and $α$ positive in this chosen sense. Why this step? $I$ is constant, so $τ = d L / d t = I α$ . Fixing a positive spin sense keeps every sign consistent with the rest of the page; tension is the only torque (the axle force acts at $r = 0$ , so its torque is zero).

Step 3. Rolling-string constraint: $a = R α \Rightarrow α = a / R$ , with the same positive sense (block falling ↔ pulley spinning our positive way). Why this step? The string does not slip, so the rim's tangential acceleration equals the block's fall rate, and the two positive directions are tied together.

Step 4. Substitute $α = a / R$ into $R T = I α$ : $T = I a / R^{2}$ . Put into Step 1: $m g - \frac{I a}{R ^{2}} = ma \Rightarrow a = \frac{m g}{m + I / R ^{2}} .$ Why this step? Eliminate $T$ and $α$ to solve for the one unknown $a$ .

Step 5. Numbers: $I / R^{2} = 0.1/0.04 = 2.5 kg$ , so $a = \frac{2 \times 10}{2 + 2.5} = \frac{20}{4.5} \approx 4.44 m/s^{2} .$

Verify: $a < g$ ✓ — the pulley's inertia holds the block back, as forecast. If $I \to 0$ (massless pulley), $a \to g$ ✓, the sanity limit. Units: $I / R^{2}$ has $kg\cdotm^{2} / m^{2} = kg$ , so the denominator is a mass ✓.

Cell I — Exam twist: direction of $L$ via right-hand rule

Worked example I · Which way does

$L$ point? A particle of mass $m = 0.5$ kg is at position $r = (2, 0, 0)$ m and moves with velocity $v = (0, 3, 0)$ m/s. Find $L$ about the origin, including its direction.

Forecast: Will $L$ point along $+ \overset{z}{^}$ (out of page) or $- \overset{z}{^}$ ?

Step 1. $L = r \times m v = m (r \times v)$ . Why this step? Definition of angular momentum; the Cross Product gives both magnitude and direction, so it answers "which way" as well as "how much."

Step 2. Compute $r \times v$ using the right-hand rule $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ : $r \times v = (2 \overset{x}{^}) \times (3 \overset{y}{^}) = 6 (\overset{x}{^} \times \overset{y}{^}) = 6 \overset{z}{^} .$ Why this step? Only the $x$ -component of $r$ and the $y$ -component of $v$ survive; their product times $\overset{x}{^} \times \overset{y}{^}$ gives a vector along $+ \overset{z}{^}$ .

Step 3. $L = 0.5 \times 6 \overset{z}{^} = 3 \overset{z}{^} kg\cdotm^{2} / s$ . Why this step? Multiply by the mass; the direction is $+ \overset{z}{^}$ , i.e. out of the page — CCW, matching our sign convention.

Verify: Magnitude $= m v d = 0.5 \times 3 \times 2 = 3$ ✓ (here $d = 2$ since $r ⊥ v$ ). Direction $+ \overset{z}{^}$ : point right-hand fingers from $r$ (along $+ x$ ) curling toward $v$ (along $+ y$ ); the thumb points out of the page ✓. Had $v$ been $(0, - 3, 0)$ , the same rule would give $- \overset{z}{^}$ (into the page) — always let the right-hand rule, not intuition, fix the sign.

Recall Quick self-test across the matrix

Which cell forbids using $τ = I α$ ? ::: Cell E — $I$ changes, so you must use $τ = d (I ω) / d t$ . A radial force gives what torque? ::: Zero (Cell F): $sin θ$ with $θ = 0°$ or $180°$ is zero. Straight-line motion, off-line point — is $L$ zero? ::: No (Cell G): $L = m v d \neq = 0$ and it stays constant along the $\overset{z}{^}$ -axis. Skater/collapsing ring shrinks $I$ by 4× with no torque — new $ω$ ? ::: 4× faster (Cell D), since $I ω$ is fixed. Torque acting on something at rest does what? ::: Builds $L$ from zero (Cell C) — integrate $d L / d t = τ$ .

Back to the parent: Torque = dL/dt · related: Torque and Angular Acceleration (tau = I alpha).