Question bank — Angular momentum L = Iω (fixed axis), L = r × p (general)

False — it has zero $L$ only about points on its line of motion; about an off-line origin the lever arm $r_{\perp }\ne 0$, so $L=p{r}_{\perp }$ is nonzero and stays constant.

False — a body sliding (translating) with $\omega =0$ still carries orbital angular momentum $\vec{L}=\vec{r}\times \vec{p}$ about any origin off its path; "not spinning" and "zero $L$" are different things.

False — $\vec{L}=\vec{r}\times \vec{p}$ depends on the chosen origin; move the origin and both $\vec{r}$ and $\vec{L}$ change. There is no such thing as "the" angular momentum without naming a point.

False — $L$ also depends on the perpendicular distance $r_{\perp }$ (the lever arm) and the direction of motion; same $m$ and $v$ can give wildly different $L$.

False — that holds only about a principal/symmetry axis. In general $I$ is a tensor and $\vec{L}$ can tilt off the spin axis, which is exactly why an unbalanced wheel wobbles (see the tilt figure below).

False — the skater pulling arms in keeps $L=I\omega$ fixed but $KE=\frac{1}{2}I{\omega }^2=L^2/(2I)$ rises as $I$ shrinks; her muscles supply the extra energy. $L$ conserved does not imply $KE$ conserved.

False — $L$ has units $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}=\mathrm{J}\mathrm{s}$ (energy × time), not joules. The extra factor of time is the giveaway.

False — only the torque $\vec{\tau }=\vec{r}\times \vec{F}$ changes $\vec{L}$. A force pointing straight toward or away from the origin (central force) has $\vec{r}\parallel \vec{F}$, so $\vec{\tau }=0$ and $L$ is conserved — this is exactly why planets obey Kepler's Second Law.

False — $I=\sum m_i{r}_i^2$ depends on the axis chosen; the same disc has $I=\frac{1}{2}M{R}^2$ about its centre but a larger value about its edge (parallel-axis theorem).

The correct magnitude is $L=rp\sin \theta =p{r}_{\perp }$. Dropping $\sin \theta$ ignores that only the component of motion going around the origin counts; the radial part contributes nothing.

Wrong scope — $\vec{r}\times \vec{p}$ is per particle. For the whole rigid body you must sum $m_i{r}_i^2$ over all particles, giving $L=I\omega$; the centre's own $\vec{r}\times \vec{p}$ misses the internal spin entirely.

The correct condition is zero external torque, not zero force. A force can act yet produce no torque (central force), or a torque can exist with balanced forces (a couple). Conservation of $L$ is tied to ${\vec{\tau }}_{ext}=0$.

No external torque acts; her own muscles are internal forces that reduce $I$. To keep $L=I\omega$ fixed, $\omega$ must rise — the speed-up is a bookkeeping consequence, not an external push.

For circular motion this is actually fine — the velocity is perpendicular to the radius so $\sin \theta =1$. The error is generalising that shortcut to non-circular motion, where $\sin \theta \ne 1$ and $r_{\perp }$ must be used.

They are different quantities: $\vec{\tau }=d\vec{L}/dt$ is the rate of change of $\vec{L}$, like force is the rate of change of linear momentum. A constant $\vec{L}$ can coexist with zero $\vec{\tau }$; a nonzero $\vec{\tau }$ can point sideways to $\vec{L}$ and merely turn it.

The cross product $\vec{r}\times \vec{p}$ automatically discards motion pointing straight toward/away from the origin (that produces no turning) and keeps only the "swirl" component — exactly the part that measures rotation about the origin. See Cross Product and figure s01.

Because $I$ is the rotational analogue of mass — it measures resistance to changing spin. But unlike mass, $I$ depends on how far the mass sits from the axis ($I=\sum m_i{r}_i^2$), which is why moving mass inward changes it. Warning: $I$ is a scalar only about a symmetry axis; off-axis (figure s02) each particle's ${\vec{r}}_i\times {\vec{p}}_i$ tilts a little, and the tilts add up so that $\vec{L}$ points off $\vec{\omega }$.

With no force there is no torque, so $\vec{L}$ cannot change. Geometrically (figure s03): the ball's straight path is a fixed line, and the perpendicular distance from a fixed origin to a fixed line is a single unchanging number — that is $r_{\perp }$. So even as $\vec{r}$ lengthens and rotates, its perpendicular drop onto the path stays put, and $p$ is constant (no force), leaving $L=p{r}_{\perp }$ fixed.

Because $\vec{L}=\vec{r}\times \vec{p}$ requires a position vector $\vec{r}$, and "position" is meaningless until you fix the point it is measured from. Different origins give different lever arms and hence different $L$.

Gravity is a central force pointing along $\vec{r}$ toward the Sun, so $\vec{\tau }=\vec{r}\times \vec{F}=0$ and $\vec{L}$ is conserved. Geometrically (figure s03): in a tiny time $dt$ the planet moves $\vec{v}dt$, sweeping a thin triangle of area $dA=\frac{1}{2}|\vec{r}\times \vec{v}dt|=\frac{1}{2m}|\vec{r}\times \vec{p}|dt=\frac{L}{2m}dt$. So $dA/dt=L/(2m)$, a constant — equal areas in equal times.

$I$ is a property of the whole body about one axis; a single particle just contributes $m_i{r}_i^2$. The moment of inertia only becomes meaningful once every particle's contribution is added up. See Moment of Inertia.

Exactly zero — $\vec{r}$ and $\vec{p}$ are antiparallel, so $\sin \theta =0$ and $\vec{r}\times \vec{p}=\vec{0}$. All the motion is radial; none of it "goes around."

Zero, because $\vec{r}=\vec{0}$ gives $\vec{r}\times \vec{p}=\vec{0}$ regardless of how fast it moves. A body has no lever arm about a point it occupies.

The net force is zero but the net torque is not, so $d\vec{L}/dt\ne 0$ and $\vec{L}$ changes. This is the clean case proving conservation depends on torque, not force.

Yes, and it gives $L=0$ about that axis — a degenerate but consistent case. The formula holds; it simply reports that a non-spinning rigid body carries no spin angular momentum.

Split it: ${\vec{L}}_O={\vec{r}}_{cm}\times {\vec{p}}_{cm}+{\vec{L}}_{spin}$. The first term is the orbital part (the centre of mass moving past the origin, as if all mass sat there), the second is the spin part ($I_{cm}\omega$) about the centre. Figure s04 shows the two pieces stacking to give the total.

$\vec{L}=I\vec{\omega }$ flips sign too, pointing the opposite way along the axis, since $I>0$ is a positive scalar. Magnitude is unchanged; only the right-hand-rule direction reverses.

No — $\vec{L}$ tilts off the axis (figure s02), and its component perpendicular to $\vec{\omega }$ sweeps around, forcing the bearings to supply a torque each turn. Only the axial component behaves like the simple $I\omega$.