Torque = dL - dt — Physics notes

WHAT is it?

The key word is same point — $\vec\tau$ and $\vec L$ must be measured about the same origin (and an inertial or non-accelerating origin, or the centre of mass).

WHY is it true? (Derive from scratch)

We start from Newton's second law and build the rotational law — no memorising.

Step 1 — Start with angular momentum of one particle. $\vec L = \vec r \times \vec p$ Why this step? This is the definition; everything follows from differentiating it.

Step 2 — Differentiate with respect to time (product rule for cross products): $\frac{d\vec L}{dt} = \frac{d\vec r}{dt}\times \vec p \;+\; \vec r \times \frac{d\vec p}{dt}$ Why this step? $\vec L$ depends on time through both $\vec r$ and $\vec p$ , so both get differentiated.

Step 3 — Kill the first term. $\frac{d\vec r}{dt}\times \vec p = \vec v \times (m\vec v) = m(\vec v\times\vec v) = 0$ Why this step? The cross product of any vector with itself is zero ( $\sin 0° = 0$ ). So velocity-with-momentum contributes nothing.

Step 4 — Use Newton's law on the second term. $\frac{d\vec p}{dt} = \vec F \quad\Rightarrow\quad \vec r\times\frac{d\vec p}{dt} = \vec r\times\vec F = \vec\tau$ Why this step? Newton's 2nd law gives $\vec F = d\vec p/dt$ ; and $\vec r\times\vec F$ is the definition of torque.

Result: $\frac{d\vec L}{dt} = \vec\tau$

HOW does this scale to a rigid body? (System of particles)

For many particles, sum over all of them: $\vec L_{\text{tot}}=\sum \vec L_i$ , so $\frac{d\vec L_{\text{tot}}}{dt}=\sum_i \vec\tau_i = \vec\tau_{\text{ext}} + \underbrace{\vec\tau_{\text{int}}}_{=0}$

For a rigid body spinning about a fixed axis with $\vec L = I\vec\omega$ (where $I$ is the moment of inertia about that axis): $\vec\tau_{\text{ext}} = \frac{d(I\vec\omega)}{dt} = I\frac{d\vec\omega}{dt} = I\vec\alpha \quad(\text{if } I \text{ constant})$ This is the famous $\tau = I\alpha$ — the rotational $F=ma$ . But $\tau = dL/dt$ is more general: it still holds when $I$ changes (e.g. a spinning skater pulling arms in).

The most important consequence

Worked Examples

Worked example 1 — Ice skater pulls arms in

A skater spins at $\omega_1 = 2\ \text{rad/s}$ with $I_1 = 6\ \text{kg·m}^2$ . She pulls her arms in, dropping $I$ to $I_2 = 2\ \text{kg·m}^2$ . Find $\omega_2$ .

Step 1: No external torque (friction negligible) $\Rightarrow L$ conserved. Why? This lets us use $\tau=dL/dt=0$ instead of tracking complicated internal muscle forces. Step 2: $I_1\omega_1 = I_2\omega_2 \Rightarrow \omega_2 = \dfrac{6\times 2}{2} = 6\ \text{rad/s}$ . Why? $L$ is the conserved quantity, not $\omega$ . She spins 3× faster. Sanity check: Kinetic energy increases ( $\tfrac12 I\omega^2$ goes $12\to36$ J) — supplied by her muscles doing work. $L$ stays the same; energy need not.

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine you're on a spinning office chair holding heavy books with your arms out. When you yank the books close to your chest, you suddenly spin faster — whee! Nobody pushed you. The "spinning-amount" (angular momentum) likes to stay the same, so when you make yourself smaller (less spread out), you must spin faster to keep it balanced. A torque is the only thing that can change that spinning-amount, and a torque is just a twist applied off-centre — like pushing a door near the handle, not near the hinge.

Active Recall — Flashcards

What is the fundamental rotational form of Newton's second law?

\vec\tau_{\text{ext}} = d\vec L/dt

(net external torque = rate of change of angular momentum about the same point).

In deriving

\tau=dL/dt

, why does the term

\frac{d\vec r}{dt}\times\vec p

vanish?

Because it equals

\vec v\times m\vec v = m(\vec v\times\vec v) = 0

— cross product of a vector with itself is zero.

Why do internal torques cancel for a system of particles?

Internal forces are Newton's-3rd-law pairs acting along the joining line; their torques are equal, opposite, and collinear, so they sum to zero.

When is

\tau = I\alpha

valid but

\tau=dL/dt

always valid?

\tau=I\alpha

needs

I

= constant;

\tau=dL/dt=d(I\omega)/dt

handles changing

I

too.

A skater drops

I

from 6 to 2 kg·m² while spinning at 2 rad/s. New

\omega

I_1\omega_1=I_2\omega_2 \Rightarrow \omega_2 = 12/2 = 6

rad/s.

Does a particle moving in a straight line have angular momentum about an off-line point?

Yes:

L = mvd

where

d

is the perpendicular distance; it's constant since torque is zero.

If net external torque is zero, what is conserved?

Angular momentum

\vec L

(constant in magnitude and direction).

Is kinetic energy conserved when angular momentum is conserved?

Not necessarily — internal work (e.g. skater's muscles) can change KE while

L

stays fixed.

Connections

Angular Momentum — the quantity $\vec L=\vec r\times\vec p$ this law governs.
Newton's Second Law — the linear parent $\vec F=d\vec p/dt$ we derived from.
Moment of Inertia — the $I$ in $L=I\omega$ .
Conservation of Angular Momentum — the $\tau=0$ special case.
Cross Product — the maths behind why only perpendicular force twists.
Torque and Angular Acceleration (tau = I alpha) — the constant- $I$ special case.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jaise linear motion mein force momentum ko change karta hai ( $F = dp/dt$ ), waise hi rotation mein torque angular momentum ko change karta hai: $\tau = dL/dt$ . Yeh koi naya alag rule nahi hai — yeh wahi Newton ka second law hai, bas ghuma ke (rotated) likha gaya hai. Force ka rotational twin hai torque, momentum ka twin hai angular momentum $L$ , mass ka twin hai moment of inertia $I$ .

Derivation simple hai: $L = r \times p$ likho, time ke saath differentiate karo. Ek term $v \times mv$ ban jaata hai jo zero ho jaata hai (kyunki same vector ka cross product zero hota hai), aur doosra term $r \times F = \tau$ ban jaata hai. Bas, ho gaya — $dL/dt = \tau$ . Cross product isliye aata hai kyunki sirf force ka woh part jo $r$ ke perpendicular hai wahi cheez ko ghuma sakta hai; pivot ki taraf seedha push karne se kuch ghoomega nahi.

Sabse important baat: agar external torque zero hai, to $L$ constant rehta hai. Yeh hai conservation of angular momentum. Ice skater jab apne haath andar khींchta hai, uska $I$ kam ho jaata hai, isliye $\omega$ badh jaata hai taaki $I\omega$ same rahe — isiliye woh tezi se ghoomne lagta hai. Yaad rakho: $\tau = I\alpha$ sirf tab chalega jab $I$ constant ho; lekin $\tau = dL/dt$ hamesha chalega, isliye yahi asli (fundamental) formula hai. Aur ek trap: $L$ conserve hone ka matlab energy conserve hona nahi hai — skater ke muscles kaam karte hain to KE badh sakti hai.

Torque = dL - dt

WHAT is it?

WHY is it true? (Derive from scratch)

HOW does this scale to a rigid body? (System of particles)

The most important consequence

Worked Examples

Common Mistakes (Steel-manned)

Active Recall — Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rotational Mechanics

Connections