Torque = dL - dt: Exercises — Physics deep dive

Everything here rests on one boxed law from the parent note: $τ_{ext} = \frac{d L}{d t}$ "The net twist from outside equals how fast the spinning-amount changes." We will use its two faces:

Constant $I$ : $τ = I α$ — see Torque and Angular Acceleration (tau = I alpha).
Zero torque: $L$ = constant — see Conservation of Angular Momentum.

Symbols before we start (nothing assumed):

Definition Every character in every problem

$r$ (position vector) — the arrow from the pivot $O$ to where the force acts; its length $r$ is the distance, unit metre ( $m$ ).
$F$ (force) — a straight push or pull, unit newton ( $N$ ). See Newton's Second Law.
$T$ (tension) — the pulling force carried by a taut string, a kind of force $F$ , unit newton ( $N$ ).
$g$ (gravitational field strength) — how hard gravity pulls per kilogram near Earth, $g \approx 10 m/s^{2}$ ; a mass $M$ feels weight $M g$ downward.
$τ$ (torque) — the twisting effort, unit newton-metre ( $N\cdotm$ ). See Cross Product for $τ = r \times F$ .
$L$ (angular momentum) — the "amount of spin," unit $kg\cdotm^{2} / s$ . See Angular Momentum.
$I$ (moment of inertia) — how hard it is to spin an object, unit $kg\cdotm^{2}$ . See Moment of Inertia.
$ω$ (angular velocity) — spin rate, unit $rad/s$ .
$α$ (angular acceleration) — how fast $ω$ changes, unit $rad/s^{2}$ .

A wheel has constant moment of inertia. A steady torque of $τ = 3 N\cdotm$ acts on it. Its angular momentum is $L (t)$ . Write the equation that connects $τ$ and $L$ , and state in words what $3$ means physically.

Recall Solution 1.1

The law is $τ = \frac{d L}{d t}$ . What it says: every second, the angular momentum grows by $3 kg\cdotm^{2} / s$ . Torque is not "the spin" — it is the rate at which spin is added, exactly like force is the rate at which linear momentum is added (Newton's Second Law). After $t$ seconds starting from rest, $L (t) = 3 t$ (kg·m²/s).

Exercise 1.2 (L1)

Which of these can change an object's angular momentum about a point $O$ ? (a) A force $F$ pointing straight toward $O$ . (b) A force $F$ perpendicular to the line joining $O$ to the point of application. (c) No force at all.

Recall Solution 1.2

Only (b).

(a) points along $r$ , so $r \times F = r F sin 0^{\circ} = 0$ — no torque.
(b) is perpendicular, so $sin 9 0^{\circ} = 1$ gives the maximum torque $r F$ — it changes $L$ .
(c) no force ⇒ no torque ⇒ $L$ constant. The picture below shows why only the sideways part twists.

Figure s01 — The pivot $O$ (navy dot) with the position arrow $r$ (navy) to the point where a force acts. The orange arrow is the radial force (along $r$ ): it produces zero torque. The magenta arrow is the perpendicular force: it gives the maximum torque $r F$ . The curved arrow marks the CCW-positive sense.

Level 2 — Application

Exercise 2.1 (L2)

A disc with $I = 0.4 kg\cdotm^{2}$ (constant) is spun up from rest by a constant torque $τ = 2 N\cdotm$ . Find the angular acceleration $α$ and the angular velocity after $5 s$ .

Recall Solution 2.1

Take the driving torque as positive (it sets our CCW-positive sense). Step 1 — which face of the law? $I$ is constant, so $\frac{d L}{d t} = \frac{d ( I ω )}{d t} = I α$ . Why: pull the constant $I$ outside the derivative. Step 2 — angular acceleration: $α = \frac{τ}{I} = \frac{2}{0.4} = 5 rad/s^{2}$ (positive ⇒ speeds up in the CCW sense). Step 3 — after 5 s: constant $α$ from rest gives $ω = α t = 5 \times 5 = 25 rad/s$ .

Exercise 2.2 (L2)

A string is wrapped over a pulley of radius $R = 0.15 m$ , $I = 0.3 kg\cdotm^{2}$ . The string leaves the rim tangentially with tension $T = 8 N$ . Find the torque and the angular acceleration.

Recall Solution 2.2

Step 1 — torque: the string is tangential so the tension $T$ (a force $F$ ) is perpendicular to $r$ , $θ = 9 0^{\circ}$ : $τ = R T sin 9 0^{\circ} = 0.15 \times 8 = 1.2 N\cdotm .$ Take this driving torque as positive. Step 2 — angular acceleration: $I$ constant ⇒ $τ = I α$ : $α = \frac{τ}{I} = \frac{1.2}{0.3} = 4 rad/s^{2} .$

Level 3 — Analysis

Exercise 3.1 (L3)

A particle of mass $m = 2 kg$ moves in a straight line at constant velocity $v = 3 m/s$ . The line passes a distance $d = 4 m$ from a fixed point $O$ . Find its angular momentum about $O$ , and explain why it stays constant even though nothing spins.

Recall Solution 3.1

Step 1 — the magnitude: $L = m v d = 2 \times 3 \times 4 = 24 kg\cdotm^{2} / s$ . Why $d$ and not $r$ ? $L = r \times m v = m v r sin θ$ , and $r sin θ$ is exactly the perpendicular distance $d$ from $O$ to the line — see the figure. Step 2 — why constant: no force ( $F = 0$ ) ⇒ $τ = r \times F = 0$ ⇒ $\frac{d L}{d t} = 0$ . The surprise: angular momentum needs no circle. As the particle slides along, $r$ grows but $θ$ shrinks so that $r sin θ = d$ stays locked. The product never changes.

Figure s02 — The fixed point $O$ (navy) and a straight path (violet) a perpendicular distance $d$ (orange) away. Two snapshots of the particle: at the near point (magenta) $r$ is short but the angle to $v$ is wide; at the far point (navy) $r$ is long but the angle is narrow. The product $r sin θ = d$ — and hence $L = m v d$ — is the same in both, so $L$ is constant.

Exercise 3.2 (L3)

A merry-go-round ( $I = 100 kg\cdotm^{2}$ ) spins freely at $ω_{0} = 1.5 rad/s$ . A child of mass $m = 30 kg$ (treat as a point) drops vertically onto its edge at radius $R = 2 m$ . Find the new angular velocity.

Recall Solution 3.2

Step 1 — which face? The child adds mass, changing $I$ . There is no external horizontal torque (the drop is vertical), so $L$ is conserved. Step 2 — child's added inertia: a point mass on the rim has $I_{child} = m R^{2} = 30 \times 2^{2} = 120 kg\cdotm^{2}$ . Step 3 — conserve $L$ : $I_{0} ω_{0} = (I_{0} + I_{child}) ω_{1}$ $ω_{1} = \frac{100 \times 1.5}{100 + 120} = \frac{150}{220} \approx 0.682 rad/s .$ The ride slows down — spreading mass out raises $I$ , so $ω$ drops to keep $L$ fixed.

Level 4 — Synthesis

Exercise 4.1 (L4)

A mass $M = 1 kg$ hangs from a string wound over a pulley of radius $R = 0.1 m$ and moment of inertia $I = 0.02 kg\cdotm^{2}$ . The mass is released and falls. Find the linear acceleration $a$ of the falling mass. (Take $g = 10 m/s^{2}$ , the gravitational field strength defined in the glossary.)

The free-body picture below fixes every sign before we compute.

Figure s03 — Left: the pulley (violet disc) with the string leaving the rim; the tension $T$ (magenta) pulls tangentially, giving a torque $T R$ that spins the pulley in the chosen positive sense (curved orange arrow). Right: the hanging mass $M$ with weight $M g$ (orange, down) and string tension $T$ (magenta, up); down is chosen positive so $M g - T = M a$ . The link $a = R α$ joins the two diagrams.

Recall Solution 4.1

Two objects, so we write two laws and link them. Sign choices: for the mass, down is positive; for the pulley, the sense that the falling string turns it is positive. These two choices are made consistent by the link $a = R α$ (both positive together). Step 1 — mass (Newton): weight $M g$ down (positive), tension $T$ up (negative): $M g - T = M a .$ Step 2 — pulley ( $τ = d L / d t = I α$ , $I$ constant): the tension is the only turning force; its torque $+ T R$ drives the positive sense: $+ T R = I α .$ Step 3 — link (string doesn't slip): the rim speed equals the mass speed, so $a = R α \Rightarrow α = a / R$ (same sign — both positive). Step 4 — substitute: $T R = I \frac{a}{R} \Rightarrow T = \frac{I a}{R ^{2}}$ . Put into Step 1: $M g - \frac{I a}{R ^{2}} = M a \Rightarrow a = \frac{M g}{M + I / R ^{2}} .$ Step 5 — numbers: $I / R^{2} = 0.02/0.01 = 2 kg$ , so $a = \frac{1 \times 10}{1 + 2} = \frac{10}{3} \approx 3.33 m/s^{2} .$ Less than $g$ : the pulley's inertia "holds back" the fall.

Exercise 4.2 (L4)

A spinning space station is a ring of $I_{1} = 5000 kg\cdotm^{2}$ turning at $ω_{1} = 0.4 rad/s$ . Crew extend two arms, raising $I$ to $I_{2} = 8000 kg\cdotm^{2}$ . Find $ω_{2}$ and the change in rotational kinetic energy. Where did the energy go?

Recall Solution 4.2

Step 1 — no external torque (deep space) ⇒ $L$ conserved: $ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{5000 \times 0.4}{8000} = 0.25 rad/s .$ Step 2 — kinetic energies ( $K E = \frac{1}{2} I ω^{2}$ ): $K E_{1} = \frac{1}{2} (5000) (0.4)^{2} = 400 J, K E_{2} = \frac{1}{2} (8000) (0.25)^{2} = 250 J .$ Step 3 — change: $Δ K E = 250 - 400 = - 150 J$ . Energy dropped. Where it went: the crew's muscles did negative work resisting the outward-flung arms; energy left the rotation and went into the crew/heat. $L$ stayed fixed; energy need not.

Level 5 — Mastery

Exercise 5.1 (L5)

A star of moment of inertia $I_{1} = 1.0 \times 1 0^{40} kg\cdotm^{2}$ rotates once every $30$ days. It collapses to a neutron star with $I_{2} = 2.0 \times 1 0^{38} kg\cdotm^{2}$ . With no external torque, find its new rotation period. Then explain, using $τ = d L / d t$ (not $I α$ ), why we could not have solved this with $τ = I α$ .

Figure s04 — Left: the large slow star ( $I_{1}$ big, $ω_{1}$ small, long arrow for period $T_{1}$ ). Right: the tiny neutron star ( $I_{2}$ small, $ω_{2}$ large, short arrow for period $T_{2}$ ). The band between them is labelled $I ω = L$ = constant: as $I$ shrinks the spin $ω$ shoots up so the product stays fixed — no torque needed.

Recall Solution 5.1

Step 1 — convert spin to $ω$ . Angular velocity is $ω = \frac{2 π}{T}$ where $T$ is the period. Initial period $T_{1} = 30 days = 30 \times 86400 = 2.592 \times 1 0^{6} s$ . Step 2 — no external torque ⇒ $L$ conserved ⇒ $I_{1} ω_{1} = I_{2} ω_{2}$ . Writing $ω = 2 π / T$ , the $2 π$ cancels: $\frac{I _{1}}{T _{1}} = \frac{I _{2}}{T _{2}} \Rightarrow T_{2} = T_{1} \frac{I _{2}}{I _{1}} .$ Step 3 — numbers: $T_{2} = 2.592 \times 1 0^{6} \times \frac{2.0 \times 1 0 ^{38}}{1.0 \times 1 0 ^{40}} = 2.592 \times 1 0^{6} \times 0.02 = 5.184 \times 1 0^{4} s \approx 0.6 days .$ So the period drops from 30 days to about 14.4 hours — the collapse spins it up dramatically, the astrophysical version of the skater pulling arms in. Step 4 — why not $τ = I α$ ? During the collapse $I$ is changing enormously. The formula $τ = I α$ secretly assumes $I$ is constant (so it can be pulled outside $d / d t$ ). The honest law $τ = \frac{d ( I ω )}{d t}$ has an extra $ω \frac{d I}{d t}$ piece. Here $τ_{ext} = 0$ , so instead of chasing $α$ we use the whole product $I ω = L$ staying constant. That is exactly what makes $τ = d L / d t$ more general.

Exercise 5.2 (L5)

During the collapse in 5.1 the process takes $Δ t$ and the star's $L$ stays constant at $L = I_{1} ω_{1}$ . A student claims "the internal gravitational squeeze exerts a huge torque to spin it up." Refute this using $τ_{ext} = d L / d t$ , and state what actually causes the speed-up.

Recall Solution 5.2

Step 1 — apply the law to the whole star. Gravity here is internal (star acting on itself). By the same pair-cancellation argument that kills internal torques in the parent derivation, internal gravity contributes zero net torque about the spin axis. Step 2 — so $τ_{ext} = 0 \Rightarrow d L / d t = 0$ . $L$ never changes. There is no torque spinning it up. Step 3 — the real cause. $L = I ω$ is fixed. Collapse shrinks $I$ , so $ω$ must rise to keep the product constant. The speed-up is not from a torque adding spin — it is from redistributing the same spin onto a smaller $I$ . "Torque adds $L$ ; shrinking $I$ redistributes existing $L$ ." Different mechanisms entirely.

Exercises — Torque = dL - dt

Level 1 — Recognition

Exercise 1.1 (L1)