Angular Momentum, Torque, and Central Forces

Derivation

Angular Momentum, Torque, and Central Forces

D-027 Home PU-101 Threads force · symmetry Depends on Momentum Conservation from Newton's Third Law

Statement

For a particle of position \(\vec{r}\) and momentum \(\vec{p}=m\vec{v}\), define the angular momentum about a fixed origin \(O\) as \(\vec{L}=\vec{r}\times\vec{p}\). Newton's second law implies that its time rate of change equals the net torque \(\vec{\tau}=\vec{r}\times\vec{F}\) about the same origin, \(\dfrac{d\vec{L}}{dt}=\vec{\tau}\). Consequently, whenever the net force is central — everywhere parallel or antiparallel to \(\vec{r}\), i.e. \(\vec{F}=f(r)\,\hat{r}\) — the torque vanishes and \(\vec{L}\) is conserved.

Why it matters

Angular momentum is the rotational counterpart of linear momentum, and its conservation is the single most powerful constraint on motion under gravity and electrostatics. It reduces the two-body Kepler problem to an effective one-dimensional radial problem, forces planetary orbits to lie in a fixed plane, and reproduces Kepler's second law (equal areas in equal times) as a bare kinematic identity.

The result is also the prototype of the deep link between symmetry and conservation: rotational invariance of a central potential is exactly what makes \(\vec{L}\) constant. That connection, made precise by Noether's theorem, is why angular momentum survives the passage from classical mechanics into quantum theory and field theory.

Assumptions

Newton's second law holds in an inertial frame.If the frame accelerates or rotates, \(\vec{F}=\dot{\vec p}\) acquires fictitious-force terms (centrifugal, Coriolis) and \(d\vec L/dt=\vec\tau\) fails unless those pseudo-torques are added.

The origin \(O\) is fixed (or itself inertial).Angular momentum and torque are defined about a point; measured about an accelerating origin an extra \(-m\,\vec r\times\ddot{\vec r}_O\) term appears and the clean balance is lost.

The mass is constant.For variable-mass systems \(\dot{\vec p}=m\dot{\vec v}+\dot m\,\vec v\), and one must track the angular momentum flux carried by the ejected or accreted mass.

The force is central for the conservation corollary.If \(\vec F\) has any component perpendicular to \(\vec r\) (a tangential drag, a thrust off the radial line, or a magnetic \(q\vec v\times\vec B\) force) the torque is nonzero and \(\vec L\) drifts.

Derivation

\[ \vec{L}\equiv \vec{r}\times\vec{p},\qquad \vec{p}=m\vec{v}=m\dot{\vec r} \]

Definition of angular momentum about the fixed origin \(O\). A

\[ \frac{d\vec{L}}{dt}=\frac{d}{dt}\left(\vec{r}\times\vec{p}\right)=\frac{d\vec{r}}{dt}\times\vec{p}+\vec{r}\times\frac{d\vec{p}}{dt} \]

Product rule for the derivative of a cross product; the factors must be kept in order because \(\times\) is not commutative. B

\[ \frac{d\vec{r}}{dt}\times\vec{p}=\vec{v}\times(m\vec{v})=m\,(\vec{v}\times\vec{v})=\vec{0} \]

The first term vanishes: any vector crossed with itself is zero, since \(\vec v\parallel \vec v\). A

\[ \frac{d\vec{L}}{dt}=\vec{r}\times\frac{d\vec{p}}{dt} \]

Substitute the vanished term; only the second survives. A

\[ \frac{d\vec{p}}{dt}=\vec{F}_{\text{net}} \]

Newton's second law in an inertial frame with constant mass. For an extended body this rests on the assumed result that internal third-law pair forces cancel, leaving only external forces to drive \(\vec p\). B

\[ \frac{d\vec{L}}{dt}=\vec{r}\times\vec{F}_{\text{net}}\equiv \vec{\tau} \]

Substitute Newton's law and identify the torque \(\vec\tau=\vec r\times\vec F\). This is the rotational equation of motion, valid for any force. A

\[ \vec{F}=f(r)\,\hat{r}\quad\Longrightarrow\quad \vec{\tau}=\vec{r}\times f(r)\hat{r}=r\,f(r)\,(\hat{r}\times\hat{r})=\vec{0} \]

Specialise to a central force. Since \(\vec r=r\hat r\) is parallel to \(\vec F\), their cross product vanishes identically. A

\[ \frac{d\vec{L}}{dt}=\vec{0}\quad\Longrightarrow\quad \vec{L}=\vec{r}\times\vec{p}=\text{const} \]

A vector with zero derivative is constant in magnitude and direction; \(\vec L\) is conserved throughout the motion. A

Result

\[ \frac{d\vec{L}}{dt}=\vec{\tau}=\vec{r}\times\vec{F},\qquad \vec{F}\ \text{central}\ \Longrightarrow\ \vec{L}=\vec{r}\times\vec{p}=\text{const} \]

Reading. The net torque about a point is the time rate of change of angular momentum about that same point — the exact rotational analogue of \(\vec F=d\vec p/dt\). When the force always points along the line to the origin it can spin nothing up, so the angular momentum vector is frozen: fixed in magnitude, and fixed in direction, which pins the orbit to a single plane.

Units check. \([\vec L]=[\vec r][\vec p]=\mathrm{m}\cdot(\mathrm{kg\,m\,s^{-1}})=\mathrm{kg\,m^2\,s^{-1}}=\mathrm{J\,s}\). \([\vec\tau]=[\vec r][\vec F]=\mathrm{m}\cdot\mathrm{N}=\mathrm{N\,m}=\mathrm{kg\,m^2\,s^{-2}}\). Then \([d\vec L/dt]=\mathrm{kg\,m^2\,s^{-1}}/\mathrm{s}=\mathrm{kg\,m^2\,s^{-2}}\), matching \([\vec\tau]\). Dimensionally consistent.

Limiting cases

Free particle (\(\vec F=\vec 0\)): the torque is zero, so \(\vec L\) is constant even though the particle moves in a straight line — \(\vec L=\vec r\times\vec p\) equals \(2m\) times the (constant) areal velocity about \(O\).
Purely radial motion (\(\vec v\parallel\vec r\)): \(\vec L=\vec r\times m\vec v=\vec 0\); a mass falling straight toward the origin carries no angular momentum and stays on the line.
Circular orbit (\(\vec v\perp\vec r\)): \(L=mvr\) is maximal for given \(r,v\); the central force supplies exactly the centripetal acceleration and does no work.
Kepler / equal areas: the areal velocity \(dA/dt=\tfrac12|\vec r\times\vec v|=L/(2m)\) is constant, recovering Kepler's second law directly.
Small non-central perturbation (\(\vec F=f(r)\hat r+\vec g\)): \(d\vec L/dt=\vec r\times\vec g\) is small, so \(\vec L\) precesses slowly rather than staying exactly fixed.

Breaks when

Non-central forces act. A tangential component — atmospheric drag \(-b\vec v\), a rocket thrust off the radial line, or sliding friction — gives \(\vec\tau=\vec r\times\vec F\neq\vec 0\), so \(\vec L\) decays or grows. The orbital decay of low satellites is exactly this.
The chosen origin is non-inertial. Working about a rotating or accelerating point introduces Coriolis and centrifugal pseudo-forces; \(d\vec L/dt=\vec\tau\) then omits their pseudo-torques and gives the wrong answer unless corrected.
Velocity-dependent field forces. The magnetic force \(q\vec v\times\vec B\) is generally not central, so mechanical \(\vec L\) alone is not conserved — one must include the angular momentum stored in the electromagnetic field.
Relativistic or radiative regimes. When \(v\to c\), \(\vec p=\gamma m\vec v\) and the Newtonian bookkeeping must be replaced; radiating charges also carry angular momentum away in their fields (analogous to orbital decay by gravitational waves in general relativity).

Failure modes

Origin amnesia: computing \(\vec L\) about one point and \(\vec\tau\) about another. Both must reference the same origin, or the balance is meaningless.
Sign/order errors in the cross product: writing \(\vec p\times\vec r\) instead of \(\vec r\times\vec p\), which flips the sign of \(\vec L\).
Assuming central \(\Rightarrow\) constant speed: \(\vec L\) is conserved but \(|\vec v|\) is not — the particle speeds up near perihelion. Only \(L=mr^2\dot\phi\) and (for conservative \(f\)) energy are constant.
Confusing "no torque" with "no force": a central force can be enormous yet exert exactly zero torque; students wrongly conclude the force is negligible.
Using \(L=mvr\) when \(\vec v\not\perp\vec r\): the correct magnitude is \(L=mvr\sin\theta\); only the velocity component perpendicular to \(\vec r\) contributes.
Forgetting the vanishing first term: failing to note \(\vec v\times\vec v=\vec 0\) and carrying a spurious \(m\vec v\times\vec v\) term through the algebra.

Discussion

The derivation is almost entirely kinematic: only Step 5 injects physics (Newton's law), and everything else is the geometry of the cross product. This is why the torque law \(d\vec L/dt=\vec\tau\) is so robust — it holds for any force whatsoever, central or not. Centrality is a separate, stronger condition that makes the right-hand side vanish. Keeping these two facts distinct is the key to using angular momentum correctly.

Conservation of \(\vec L\) has two independent pieces. Constancy of its direction confines the motion to the plane through \(O\) perpendicular to \(\vec L\), collapsing a three-dimensional problem to two dimensions. Constancy of its magnitude, \(L=mr^2\dot\phi\), then serves as a first integral: it lets one eliminate \(\dot\phi\) from the energy equation and reduce the dynamics to a single radial coordinate moving in an effective potential \(V_{\text{eff}}(r)=V(r)+L^2/(2mr^2)\), where the second term is the centrifugal barrier. This reduction is the backbone of the entire theory of orbits.

Structurally, this is Noether's theorem in miniature. A central potential \(V(r)\) depends only on the distance to \(O\) and is therefore invariant under all rotations about \(O\); the conserved quantity generated by that continuous rotational symmetry is precisely \(\vec L\). Each rotation axis contributes one component, so full spherical symmetry conserves all three components of \(\vec L\), whereas a merely axially-symmetric potential conserves only \(L_z\). In the Hamiltonian formulation the components of \(\vec L\) generate rotations through the Poisson brackets \(\{L_i,L_j\}=\varepsilon_{ijk}L_k\), an algebra that carries over verbatim (up to a factor of \(i\hbar\)) into the quantum commutators \([\hat L_i,\hat L_j]=i\hbar\,\varepsilon_{ijk}\hat L_k\).

Common misconceptions. Angular momentum is not "the momentum of rotation" reserved for spinning bodies — a particle moving in a straight line has nonzero \(\vec L\) about any off-line origin. Conservation of \(\vec L\) does not mean the orbit is circular or the speed constant; it means one specific vector is frozen while \(r\), \(\dot r\), and \(|\vec v|\) all change. And "central" is a statement about direction, not strength: a force can be huge and still exert zero torque.

Worked examples

Elliptical orbit: relate perihelion and aphelion speeds.

A comet orbits the Sun. At perihelion \(r_p=8.0\times10^{10}\,\mathrm{m}\) its speed is \(v_p=5.4\times10^{4}\,\mathrm{m\,s^{-1}}\); at aphelion \(r_a=5.2\times10^{12}\,\mathrm{m}\). Find \(v_a\). Gravity is central, so \(\vec L\) is conserved. A

\[ L=m v_p r_p = m v_a r_a \quad\Longrightarrow\quad v_a=v_p\,\frac{r_p}{r_a} \]

At both apsides \(\vec v\perp\vec r\), so \(L=mvr\) with no \(\sin\theta\) factor; the mass cancels. Symbols first. A

\[ v_a=(5.4\times10^{4})\times\frac{8.0\times10^{10}}{5.2\times10^{12}}=(5.4\times10^{4})(1.538\times10^{-2})\ \mathrm{m\,s^{-1}} \]

Insert numbers with units. A

\[ v_a\approx 8.3\times10^{2}\ \mathrm{m\,s^{-1}} \]

Reading. The comet is about 65 times slower at aphelion, exactly the ratio \(r_a/r_p\); the product \(vr\) stays fixed because the pull toward the Sun exerts no torque.

Torque from a non-central (tangential) force.

A \(0.50\,\mathrm{kg}\) puck slides on a frictionless table, held at fixed radius \(r=0.40\,\mathrm{m}\) from a pivot by a radial guide, with tangential speed \(v_0=3.0\,\mathrm{m\,s^{-1}}\). A tangential force \(F=0.60\,\mathrm{N}\) (perpendicular to \(\vec r\)) acts for \(t=2.0\,\mathrm{s}\). Find the final angular momentum and speed. B

\[ \tau=rF,\qquad \frac{dL}{dt}=\tau\ \Rightarrow\ L_f=L_0+\tau\,t=mv_0 r + rF\,t \]

Here \(\vec\tau\neq\vec0\) because the force is tangential; integrate the torque law over constant \(\tau\). Symbols first. B

\[ L_0=(0.50)(3.0)(0.40)=0.60\ \mathrm{kg\,m^2\,s^{-1}},\quad \tau=(0.40)(0.60)=0.24\ \mathrm{N\,m} \]

Evaluate the initial angular momentum and the torque. A

\[ L_f=0.60+(0.24)(2.0)=1.08\ \mathrm{kg\,m^2\,s^{-1}},\qquad v_f=\frac{L_f}{mr}=\frac{1.08}{(0.50)(0.40)} \]

Add the angular impulse, then invert \(L=mvr\) for the final speed. A

\[ L_f\approx 1.1\ \mathrm{kg\,m^2\,s^{-1}},\qquad v_f=5.4\ \mathrm{m\,s^{-1}} \]

Reading. Because the force is not central, \(\vec L\) is not conserved — it grows linearly at the rate set by the torque, and the puck speeds up. Contrast with a purely radial guide force, which would leave \(L\) unchanged.

Problems

Show from \(\vec L=\vec r\times\vec p\) that a particle moving in a straight line at constant velocity, passing a perpendicular distance \(b\) from the origin, has constant angular momentum \(L=mvb\).
Solution
At any instant \(\vec r\) makes angle \(\theta\) with \(\vec v\), and \(|\vec r\times\vec p|=mvr\sin\theta\). But \(r\sin\theta\) is exactly the perpendicular distance from \(O\) to the line of motion, the fixed impact parameter \(b\). Hence \(L=mvb=\) const, with no force needed (torque \(=0\)). For \(m=2\,\mathrm{kg},\,v=3\,\mathrm{m/s},\,b=0.5\,\mathrm{m}\): \(L=3\,\mathrm{kg\,m^2/s}\).
A satellite at perigee \(r_1=7.0\times10^{6}\,\mathrm{m}\) moves at \(v_1=9.0\,\mathrm{km/s}\). At apogee \(r_2=4.2\times10^{7}\,\mathrm{m}\), find \(v_2\) using angular-momentum conservation.
Solution
Gravity is central, so \(v_1 r_1=v_2 r_2\) at the apsides (where \(\vec v\perp\vec r\)). \(v_2=v_1 r_1/r_2=(9.0\times10^3)(7.0\times10^6)/(4.2\times10^7)=(9.0\times10^3)(0.1667)=1.5\times10^3\,\mathrm{m/s}=1.5\,\mathrm{km/s}\).
A \(0.20\,\mathrm{kg}\) ball on a string moves in a horizontal circle of radius \(0.60\,\mathrm{m}\) at \(4.0\,\mathrm{m/s}\). The string is pulled through a central hole, shortening the radius to \(0.30\,\mathrm{m}\). Find the new speed and the ratio of kinetic energies.
Solution
The string tension is central, so \(L=mvr\) is conserved: \(v'=v\,r/r'=(4.0)(0.60)/(0.30)=8.0\,\mathrm{m/s}\). With \(K=\tfrac12 mv^2\): \(K'/K=(v'/v)^2=(8/4)^2=4\). The energy quadruples because the puller does work while shortening the radius; \(L\) alone is conserved, not \(K\).
A particle has \(\vec r=(3,0,0)\,\mathrm{m}\) and \(\vec p=(0,4,0)\,\mathrm{kg\,m/s}\). Compute \(\vec L\). Then a force \(\vec F=(0,0,2)\,\mathrm{N}\) acts; compute \(\vec\tau\) and state \(d\vec L/dt\).
Solution
\(\vec L=\vec r\times\vec p=(3,0,0)\times(0,4,0)=(0,0,12)\,\mathrm{kg\,m^2/s}\). \(\vec\tau=\vec r\times\vec F=(3,0,0)\times(0,0,2)=(0\cdot2-0\cdot0,\ 0\cdot0-3\cdot2,\ 0)=(0,-6,0)\,\mathrm{N\,m}\). Thus \(d\vec L/dt=(0,-6,0)\,\mathrm{kg\,m^2/s^2}\): \(L_z\) is momentarily unchanged but \(L_y\) decreases — the force is not central, so \(\vec L\) is not conserved.
For a central force the areal velocity is \(dA/dt=L/(2m)\). A planet has \(L/m=1.2\times10^{15}\,\mathrm{m^2/s}\) and orbital period \(T=3.0\times10^{7}\,\mathrm{s}\). Find the total area of the orbital ellipse and comment on Kepler's second law.
Solution
\(dA/dt=L/(2m)=\tfrac12(1.2\times10^{15})=6.0\times10^{14}\,\mathrm{m^2/s}\), constant (equal areas in equal times = Kepler II, a direct consequence of \(\vec L=\)const). Total area \(=\int_0^T (dA/dt)\,dt=(6.0\times10^{14})(3.0\times10^7)=1.8\times10^{22}\,\mathrm{m^2}\). This equals the whole ellipse area \(\pi ab\); the constancy of \(dA/dt\) means the planet sweeps it uniformly in time even though its speed varies.