Torque and angular momentum

A body's angular velocity changes only when something twists it, and building the quantity that does the twisting completes the rotational half of mechanics.

What is carried in: the moment of inertia $I = \sum m_{i} r_{i}^{2}$ and the rotational kinetic energy $\frac{1}{2} I ω^{2}$ from the previous lesson, together with $\sum \vec{F} = d \vec{p} / d t$ and the third law.

Torque

Push on a door at the handle and it swings; push with the same force near the hinge and almost nothing happens; push straight at the hinge and nothing happens at all. What matters is not the force but the combination of the force, the distance from the axis and the direction of the push. That combination is the torque:

τ = r F \sin φ = F_{⊥} r = F d

where $φ$ is the angle between the position vector from the axis and the force. The three forms are the same number read three ways: the distance times the perpendicular component of the force, or the force times the perpendicular distance from the axis to the line of the force, which is called the moment arm. Torque is measured in newton metres. Deliberately not in joules, though the units are identical, because torque is not energy.

A sign convention is needed, and the usual one takes anticlockwise as positive. In three dimensions torque is a vector, $\vec{τ} = \vec{r} \times \vec{F}$ , pointing along the axis by the right hand rule, but for the fixed axis problems in this course a sign is enough.

Example. A spanner 0.30 m long is pulled with 150 N. What torque does it apply to the bolt when the force is perpendicular to the spanner, and when it is at 60° to it?

Perpendicular: $τ = (0.30) (150) = 45$ N m. At 60°: $τ = (0.30) (150) \sin 60^{\circ} = 39.0$ N m, thirteen per cent less for the same effort. This is why a torque wrench must be pulled square, and why a longer spanner is worth more than a stronger arm: doubling $r$ doubles the torque exactly.

Now you. A force of 80 N is applied at 40° to a lever arm 0.45 m long. What is the torque, and what perpendicular force at the same point would match it?

Answer

$τ = (0.45) (80) \sin 40^{\circ} = 23.1$ N m. A perpendicular force would need to be $23.1 / 0.45 = 51.4$ N, which is $80 \sin 40^{\circ}$ , as it must be.

The rotational second law

Now derive rather than assert. Take one particle of mass $m_{i}$ at distance $r_{i}$ from the axis, in a rigid body. Only the tangential component of the force on it changes its speed, and Newton's second law along the tangent gives $F_{i, t} = m_{i} a_{i, t} = m_{i} r_{i} α$ , using $a_{t} = α r$ and the fact that $α$ is common to the whole rigid body. Multiply both sides by $r_{i}$ :

τ_{i} = F_{i, t} r_{i} = m_{i} r_{i}^{2} α

Sum over every particle. The internal torques cancel in pairs, because internal forces are equal, opposite and act along the line joining the particles, so they have the same moment arm and opposite signs. What survives is

\sum τ_{ext} = I α

the rotational counterpart of $\sum F = m a$ , with torque in place of force and moment of inertia in place of mass. Everything about it must be taken about the same axis: a torque is meaningless until the axis is named.

Example. A 2.0 kg mass hangs from a light cord wound round a uniform pulley of mass 1.0 kg and radius 0.10 m, which turns freely on its axle. Find the acceleration of the mass and the tension in the cord.

Two free bodies, two equations. For the hanging mass, taking down as positive, $m g - T = m a$ . For the pulley, the only torque is the cord's, $T R = I α$ , with $I = \frac{1}{2} M R^{2}$ and $α = a / R$ since the cord does not slip. The second becomes $T R = \frac{1}{2} M R^{2} (a / R)$ , so $T = \frac{1}{2} M a$ . Substituting,

a = \frac{m g}{m + M / 2} = \frac{(2.0) (9.81)}{2.0 + 0.5} = 7.85 {m s}^{- 2}

and $T = \frac{1}{2} (1.0) (7.85) = 3.92$ N. The pulley's mass matters: a massless pulley would have given $a = g$ and $T = 0$ . Notice that the radius cancelled entirely, so a heavy pulley of any size behaves the same, and that what enters is $M / 2$ , the pulley's mass discounted by its shape factor.

Now you. A 3.0 kg mass hangs from a cord round a uniform pulley of mass 2.0 kg. Find the acceleration and the tension.

Answer

$a = (3.0) (9.81) / (3.0 + 1.0) = 7.36$ m s⁻², and $T = \frac{1}{2} (2.0) (7.36) = 7.36$ N.

Angular momentum

Define, for a rigid body turning about a fixed axis,

L = I ω

in kg m² s⁻¹, and for a single particle, more generally, $L = m v r \sin φ$ , the momentum times its moment arm about the chosen point. Then differentiating $L = I ω$ for a rigid body of fixed shape gives $d L / d t = I α$ , and the rotational second law becomes

\sum τ_{ext} = \frac{d L}{d t}

which is the exact counterpart of $\sum \vec{F} = d \vec{p} / d t$ and is the more general statement, because it survives when $I$ itself changes.

The immediate corollary is the one that matters. If no external torque acts about an axis, the angular momentum about that axis is constant. For a body that can change shape, $I_{1} ω_{1} = I_{2} ω_{2}$ , so pulling mass inward, which cuts $I$ , must raise $ω$ in the same proportion.

This is a genuinely independent conservation law, not a consequence of momentum conservation. A system can have zero momentum and enormous angular momentum: two equal masses whirling about their common centre in opposite directions have $\vec{P} = 0$ and $L \neq 0$ .

Example. A skater spinning at 2.0 rad s⁻¹ with arms outstretched has a moment of inertia of 3.5 kg m². Pulling their arms in reduces it to 1.2 kg m². What is the new spin rate, and what happened to the kinetic energy?

The ice exerts no vertical torque worth mentioning, so $L$ is conserved: $ω_{2} = (3.5 / 1.2) (2.0) = 5.83$ rad s⁻¹, nearly three times faster. The kinetic energy was $\frac{1}{2} (3.5) (4.0) = 7.00$ J and is now $\frac{1}{2} (1.2) (34.0) = 20.4$ J. Energy is not conserved and has risen by 13.4 J.

That is not a paradox and it is the interesting part of the example. The skater did that work. Their arms were moving in circles and needed a centripetal force to hold them there; pulling them inward means exerting that inward force through a real inward displacement, which is positive work. The general result is that $K = L^{2} / 2 I$ at fixed $L$ , so shrinking $I$ raises $K$ , and the energy comes from whoever does the shrinking. Letting the arms back out returns it.

Now you. A turntable of moment of inertia 0.80 kg m² spins freely at 4.0 rad s⁻¹. A 0.50 kg lump of putty is dropped onto it at 0.40 m from the axis and sticks. What is the new angular velocity?

Answer

The putty adds $m r^{2} = (0.50) (0.16) = 0.080$ kg m², so $I_{2} = 0.880$ kg m². Conservation of angular momentum gives $ω_{2} = (0.80) (4.0) / 0.880 = 3.64$ rad s⁻¹. Energy fell from 6.40 J to 5.82 J, the difference going into the impact, exactly as in a perfectly inelastic linear collision.

Where the law shows itself

Angular momentum conservation is not confined to skaters. A diver leaves the board with a fixed $L$ , tucks to cut $I$ by a factor of three or four, completes the somersaults quickly, then opens out to slow the rotation for a clean entry, and no torque is available in mid air to help.

A collapsing star is the extreme case. A star of radius $7 \times 10^{8}$ m turning once in 25 days that collapses to a neutron star 10 km across cuts its radius by 70000, so $I$ falls by the square of that, $4.9 \times 10^{9}$ , and $ω$ rises by the same factor. The 25 day period becomes about $4.4 \times 10^{- 4}$ s. That estimate is crude, because a real collapse sheds mass and the core is not uniform, but it explains at a stroke why pulsars spin hundreds of times a second, which is otherwise an absurd rate for an object heavier than the Sun.

The same law explains why a helicopter needs a tail rotor. Spinning the main rotor one way requires the engine to exert a torque on it, and the third law returns an equal torque to the fuselage, which would otherwise spin the other way. The tail rotor supplies an external torque to cancel it.

Central forces, and Kepler's second law

Now the result that reaches furthest. A central force is one directed always along the line joining a body to a fixed point: the tension in a string held at the centre, the electrostatic attraction of a nucleus, and, though it has not been introduced yet, gravity.

The torque of a central force about that centre is exactly zero, because $τ = r F \sin φ$ and $φ$ is zero or 180° by definition. So the angular momentum of a body moving under any central force whatever is conserved. Three consequences follow at once, with no knowledge of the force law needed.

The motion is confined to a plane, since $\vec{L}$ is a fixed vector and the position and velocity must stay perpendicular to it. The body speeds up as it approaches the centre and slows as it recedes, since $m v r$ is fixed. And the rate at which the line from the centre to the body sweeps out area is constant: in a short time $d t$ the body sweeps a thin triangle of area $d A = \frac{1}{2} r (v \sin φ d t)$ , so

\frac{d A}{d t} = \frac{r v \sin φ}{2} = \frac{L}{2 m}

which is constant. That statement, that a planet sweeps out equal areas in equal times, is Kepler's second law, published in 1609 from Tycho Brahe's observations of Mars. It is derived here without ever saying that gravity is an inverse square force, or indeed anything about gravity at all: the law holds for any central force, and what it really reports is that the Sun's pull on a planet points at the Sun.

The Earth's orbit shows the effect in numbers. At perihelion in early January the Earth is $1.471 \times 10^{11}$ m from the Sun and moving at 30290 m s⁻¹; at aphelion in early July it is $1.521 \times 10^{11}$ m away and moving at 29294 m s⁻¹. The products $v r$ agree to four figures, as conservation of angular momentum demands, and the three per cent speed difference is why the northern winter half of the year is about seven days shorter than the summer half.

Gyroscopes, and an honest limit

Everything above treats angular momentum as a signed number about one fixed axis. In three dimensions it is a vector, and the interesting behaviour is that a torque changes the direction of $\vec{L}$ rather than its size.

A spinning top leaning over does not fall. Gravity supplies a horizontal torque about the pivot, perpendicular to the top's angular momentum, so $d \vec{L} / d t$ is perpendicular to $\vec{L}$ and the vector turns without changing length. The top precesses, sweeping a cone at a rate $Ω = τ / L$ , which is slower the faster it spins.

The honest limit is that this course does not develop the vector treatment properly. In three dimensions the relation between $\vec{L}$ and $\vec{ω}$ is not a simple multiplication by a number: a body has three different moments of inertia about three principal axes, and $\vec{L}$ and $\vec{ω}$ are generally not parallel. That is why a badly balanced wheel shakes its bearings, and why a book spun about its intermediate axis tumbles chaotically while the same book spun about either of the other two is stable. Those results need machinery beyond this course, and asserting the simple formula outside its range is where most confusion about gyroscopes comes from.

The equation nobody can solve

One thing rotation has quietly failed to deliver is a solved problem of the kind the earlier lessons produced. Applying $τ = I α$ to a pendulum, a rod swinging on a pivot, gives

I \ddot{θ} = - m g d \sin θ

and that equation has no solution in elementary functions. The $\sin θ$ makes it nonlinear, and nonlinear differential equations are, with rare exceptions, unsolvable in closed form.

The escape is the most productive approximation in physics. For small angles $\sin θ \approx θ$ , the equation becomes linear, and its solution is a sine wave. The same escape works for a mass on a spring, a molecule vibrating, an atom in a crystal and a circuit oscillating, because every potential energy curve looks like a parabola near its minimum. Making that argument properly, solving the resulting equation, and finding out exactly how wrong the approximation is, is the next lesson.