Torque = dL - dt: Question bank — Physics deep dive

Before we begin, let's re-earn every symbol and every logical step, so nothing on this page leans on memory of the parent note.

Recall The derivation in four lines (so you never need to leave this page)

These are the steps this page keeps referring to. Step 1: start from $L = r \times p$ . Step 2: differentiate with the product rule: $\frac{d L}{d t} = \frac{d r}{d t} \times p + r \times \frac{d p}{d t}$ . Step 3: the first term is $v \times m v = m (v \times v) = 0$ because any arrow crossed with itself gives zero (parallel arrows, $sin 0 = 0$ ). Step 4: in the second term $\frac{d p}{d t} = F$ (Newton), and $r \times F = τ$ . Result: $τ = \frac{d L}{d t}$ .

Formula Unpacking

$τ = \frac{d ( I ω )}{d t}$ with the chain rule — where $I α$ comes from For a fixed axis $L = I ω$ , and $I$ may itself change with time (arms moving in). Differentiate the product $I ω$ : $τ = \frac{d ( I ω )}{d t} = I α I \frac{d ω}{d t} + shape-change term ω \frac{d I}{d t} = I α + ω \frac{d I}{d t}$

Why two terms? The product rule says: rate-of-change of (something $\times$ something-else) = (first changing $\times$ second held) + (first held $\times$ second changing). Here $I$ and $ω$ are both free to change.
When is $τ = I α$ recovered? Only if the mass keeps its shape, $\frac{d I}{d t} = 0$ , killing the second term. That's the rigid-body-fixed-axis case in Torque and Angular Acceleration (tau = I alpha).
The skater with zero torque: set $τ = 0$ , so $I α = - ω d I / d t$ . As she pulls in, $d I / d t < 0$ , forcing $α > 0$ — she speeds up even though no external twist acts. That's conservation of $L$ seen through the chain rule.

True or false — justify

TRUE or FALSE: A force that points straight at the pivot

O

produces no torque about

O

True. Along that line

r

and

F

are parallel, so

θ = 0

and

r F sin θ = 0

— there is no perpendicular part to twist with.

TRUE or FALSE:

τ = I α

is always exactly equivalent to

τ = d L / d t

False. They agree only when

I

is constant. The general chain-rule form is

τ = I α + ω d I / d t

; if

I

changes (skater pulling arms in) the second term matters and

I α

alone is wrong.

TRUE or FALSE: If a system's angular momentum is conserved, its rotational kinetic energy is also conserved.

False. Take the skater at

ω_{1} = 2

rad/s with

I_{1} = 6

kg·m²; pulling her arms in to

I_{2} = 2

kg·m² keeps

L = I ω

fixed so

ω_{2} = 6

rad/s. Then

K E = \frac{1}{2} I ω^{2}

rises from

\frac{1}{2} (6) (2^{2}) = 12

J to

\frac{1}{2} (2) (6^{2}) = 36

J — a factor of 3. Her muscles supplied that energy; conserving

L

says nothing about conserving energy.

TRUE or FALSE: An object must be rotating to have nonzero angular momentum about a point.

False. A particle gliding in a straight line has

L = m v d

(

d

= perpendicular distance from

O

to the line), a fixed nonzero value even though nothing spins.

TRUE or FALSE: You may compute torque about one point and angular momentum about a different point and still expect

τ = d L / d t

False. Step 1 of the derivation starts from

L = r \times p

about one origin; both quantities must share the same point (and it must be inertial or the centre of mass), or the equation is meaningless.

TRUE or FALSE: If the net external torque on a body is zero, its angular velocity

ω

must stay constant.

False. Zero torque fixes

L = I ω

, not

ω

. If

I

changes,

ω

changes to keep

L

constant.

TRUE or FALSE: Internal forces inside a rigid body can change its total angular momentum.

False. Internal forces are Newton's-third-law pairs acting along the joining line; their torques are equal, opposite, and collinear, so they sum to zero. Only external torque changes

L

TRUE or FALSE: The term

\frac{d r}{d t} \times p

(the first term when you differentiate

L

) vanishes only for circular motion.

False. It equals

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

for any motion — the cross product of a vector with itself is always zero, regardless of path.

TRUE or FALSE: A nonzero net torque can act on a body whose angular momentum is momentarily zero.

True.

τ = d L / d t

is about the rate of change of

L

; at the instant a body starts spinning from rest

L = 0

but

τ \neq = 0

— that's exactly how spinning begins.

TRUE or FALSE: Doubling the distance from the pivot always doubles the torque of a given force.

False. Torque is

r F sin θ

; doubling

r

doubles it only if the angle

θ

between

r

and

F

is unchanged. Move the mass so

F

points more along

r

and the extra distance buys you nothing.

TRUE or FALSE: Because

τ = d L / d t

is a vector equation, a torque perpendicular to

L

changes the direction of

L

without changing its length.

True. Adding a small change-arrow

τ d t

at right angles to

L

swings

L

sideways — its length stays fixed while it rotates. This is exactly the wobble (precession) of a spinning top; planar

I α

intuition cannot capture it.

Spot the error

Find the flaw: "The skater spins faster after pulling her arms in, so angular momentum increased."

Angular momentum is conserved (

I ω

constant); it did not increase. She spins faster precisely because

I

dropped and

ω

had to rise to keep

L

fixed.

Find the flaw: "No net force acts on a straight-line-moving particle, so its angular momentum about any point is zero."

No force means

L

is constant, not zero. Its constant value is

L = m v d

, which is nonzero for any point off the line of motion.

Find the flaw: "Because

F = ma

, the rotational law must be

τ = I α

— full stop."

The true analogue of

F = d p / d t

τ = d L / d t

τ = I α

is the special case with constant

I

, just as

F = ma

F = d p / d t

with constant

m

Find the flaw: "Torque and force point the same direction, since torque is just rotational force."

Torque is

r \times F

; by the right-hand rule the cross product points perpendicular to both

r

and

F

(along the axis of rotation), generally not along

F

at all.

Find the flaw: "A torque acting for a while adds kinetic energy equal to the angular momentum it delivered."

Torque delivers angular momentum (

\int τ d t = Δ L

), which has different units from energy. Energy comes from work

\int τ d θ

; the two are separate accountings.

Find the flaw: "The string tension gives no torque on the pulley because tension pulls along the string, not around the wheel."

The string leaves the rim tangentially, so

r

(axle-to-rim) is perpendicular to

T

θ = 90°

sin 90° = 1

, and

τ = R T

is maximal, not zero.

Find the flaw: "

τ = d (I ω) / d t

just equals

I α

because

ω

is what's changing."

Only if

I

is held fixed. The product rule gives

I α + ω d I / d t

; ignoring the second term is exactly the mistake that breaks the skater and collapsing-star cases.

Why questions

Why must the reference point for

τ = d L / d t

be inertial or the centre of mass?

In an accelerating frame each particle feels a fake force

- m a_{O}

, whose total torque is

- (M R_{cm}) \times a_{O}

; this vanishes when

O

is inertial (

a_{O} = 0

) or when

O

is the centre of mass (

R_{cm} = 0

), and only then does the clean law survive.

Why does the cross product, not ordinary multiplication, appear in torque and angular momentum?

Only the component of the force perpendicular to

r

can twist about

O

. The cross product's

sin θ

automatically discards the useless parallel part and keeps the twisting part.

Why is

τ = d L / d t

called "more fundamental" than

τ = I α

It follows directly from Newton's law with no assumption about

I

, so it covers changing-

I

situations (skaters, collapsing stars) where the dropped term

ω d I / d t

would otherwise be lost.

Why can

τ = d L / d t

describe a wobbling top while

τ = I α

cannot?

Because

d L / d t

is a full vector rate: a sideways torque changes

L

's direction (precession), a genuinely 3D effect. The scalar

I α

only tracks speeding up or slowing down about one fixed axis.

Why does angular momentum stay constant for a free particle even though it "moves away" from

O

r

lengthens, the angle to the velocity shrinks so that the perpendicular distance

d

stays fixed;

L = m v d

therefore never changes without a torque.

Why can a spinning system's kinetic energy rise while its angular momentum stays put?

Since

ω = L / I

, we get

K E = L^{2} / (2 I)

, so halving

I

at fixed

L

doubles

K E

; a threefold drop in

I

(like the skater's

6 \to 2

) triples it. The extra energy comes from work done pulling the mass inward.

Why do we say internal torques cancel "in pairs" rather than "on average"?

Each Newton's-third-law force pair is equal, opposite, and shares the same line of action, so each pair contributes exactly zero torque individually — no averaging or approximation is involved.

Edge cases

What is the torque if the force is applied exactly at the pivot (

r = 0

Zero:

τ = r F sin θ = 0

because

r = 0

. A push right at the hinge cannot rotate the door no matter how hard you push.

What happens to

ω

I \to 0

while

L

is conserved?

ω = L / I

blows up toward infinity — a mathematical warning that mass can never fully collapse to the axis; real bodies have a minimum

I

For a particle moving straight toward the point

O

, what is its angular momentum?

Zero. The line of motion passes through

O

, so the perpendicular distance

d = 0

and

L = m v d = 0

— equivalently

r ∥ v

makes the cross product vanish.

If two equal internal forces act along the joining line, what net linear momentum and net torque do they contribute?

Both are zero: the forces cancel for linear momentum, and being equal-opposite-collinear their torques cancel too, which is why internal interactions never change total

p

or total

L

At the exact instant a body starts from rest under a torque, what are

L

ω

, and

d L / d t

L = 0

and

ω = 0

, but

d L / d t = τ \neq = 0

— the body has no spin yet but is acquiring it, the whole point of

τ = d L / d t

What is the torque when the force is parallel to

r

but the body is already spinning fast?

Still zero:

sin θ = 0

kills it regardless of the current spin. Existing

ω

has no bearing on whether a given force produces torque.

In a genuinely 3D case where

τ

is perpendicular to

L

, does the spin rate change?

No — the length of

L

(hence the spin rate) is unchanged; only its direction swings. This is steady precession, invisible to any planar

I α

treatment.