Question bank — Gyroscopic effect — precession of spinning top

False. Gravity's torque is perpendicular to $\vec{L}$ so it can't pull the top down, but friction at the pivot and air drag slowly bleed $\omega$ away — as $\omega$ drops the top eventually wobbles and falls.

True. With $\vec{\tau }=0$ we get $d\vec{L}/dt=0$, so $\vec{L}$ is frozen in direction — the axis just sits there. No torque, no precession.

False. Tilt raises the torque ($\tau =rmg\sin \theta$) but it raises the horizontal part of $\vec{L}$ that must rotate ($L\sin \theta$) by the same factor; the $\sin \theta$'s cancel, leaving $\Omega =rmg/(I\omega )$, independent of $\theta$.

False. $\omega$ sits in the denominator of $\Omega =rmg/(I\omega )$, so doubling $\omega$ halves $\Omega$ — the top precesses more lazily, not faster.

False. Precession is driven by the gravity torque about the pivot. Kill the torque and $\vec{L}$ holds its direction — the axle simply floats pointing the same way (this is exactly why gyroscopes are used for navigation).

False. $d\vec{L}=\vec{\tau }dt$ points along $\vec{\tau }$, which is horizontal — 90° away from the downward pull. That 90° offset is the entire gyroscopic surprise.

False in general. $\Omega =rmg/(I\omega )$ has $m$ on top and hidden inside $I$ (bigger mass usually means bigger $I=I\omega$). If you scale mass uniformly, $m$ and $I$ both grow and the effect largely cancels; only the distribution matters.

True. The steady $\Omega =rmg/(I\omega )$ is precisely the fast-spin, no-wobble result; nutation is the extra bobbing that averages out on top of it.

The cross product of the (near-vertical) $\vec{r}$ with the vertical $\vec{g}$ is horizontal, not down. A cross product is perpendicular to both inputs, so $\vec{\tau }$ can never lie along the vertical $\vec{g}$.

The $\sin \theta$ in the torque is cancelled by the $\sin \theta$ in the horizontal projection $L\sin \theta$ during Step 5. The correct steady result is $\Omega =rmg/(I\omega )$, tilt-free.

A perpendicular torque does plenty — it can't change $|\vec{L}|$, but it continuously rotates $\vec{L}$'s direction, which is precession. "Does nothing to the magnitude" is not "does nothing."

You must use $I$ about the spin axis. A point mass spinning about an axis through itself has tiny $I$, so tiny $L$ — the formula only behaves once $I$ is the correct moment of inertia (see Moment of Inertia).

It is a vector equation. The whole phenomenon lives in the direction of $d\vec{L}$; treating it as a scalar erases precession entirely (see Vector nature of dL/dt).

Flip $\omega$ and $\vec{L}$ reverses, so with the same torque $d\vec{L}=\vec{\tau }dt$ now sweeps $\vec{L}$ the opposite way — the precession sense reverses.

For a fast top the torque immediately redirects $\vec{L}$ sideways; there is no falling phase, only a tiny initial dip (nutation) before steady precession sets in.

Large $\vec{L}$ makes $\Omega =\tau /(I\omega )$ tiny, so any disturbing torque only turns the axis very slowly — it "resists" tipping, holding its orientation across flight.

It appears twice: once in the driving torque $\tau =rmg\sin \theta$, once in the radius $L\sin \theta$ of the horizontal circle $\vec{L}$'s tip traces. Dividing one by the other removes it exactly.

Only the component of the force perpendicular to $\vec{r}$ twists the top, and we also need the direction of the twist. The cross product $\vec{r}\times m\vec{g}$ delivers both the $\sin \theta$ factor and the perpendicular direction in one operation.

The formula assumes constant $\omega$. Friction slowly reduces $\omega$, which raises $\Omega =rmg/(I\omega )$ and eventually breaks the fast-spin approximation; the top then nutates violently and topples.

It is the rotational twin of $\vec{F}=d\vec{p}/dt$. Torque sets the rate of change of the momentum vector's direction, and following that vector geometry is what predicts sideways precession rather than falling.

Because $d\vec{L}=\vec{\tau }dt$ is horizontal, the tip of $\vec{L}$ is nudged sideways at every instant. A continuous sideways nudge on a vector of fixed tilt sweeps out a horizontal circle.

$\Omega =rmg/(I\omega )\to \infty$, meaning the formula blows up — physically the fast-spin approximation dies and the "top" simply topples over like an ordinary falling stick.

The gravity torque $\tau =rmg\sin \theta =0$, so there is no torque and (ideally) no precession — a perfectly upright spinning top ("sleeping top") stands still without walking around.

The torque is maximal ($\sin 90^{\circ }=1$) and so is the horizontal projection of $\vec{L}$; they still cancel, giving the same $\Omega =rmg/(I\omega )$ — a horizontally-held gyroscope precesses at the identical rate.

Then $\tau =rmg=0$ regardless of tilt, so $\Omega =0$: a body pivoted through its centre of mass feels no gravity torque and does not precess — it just spins in place.

A parallel torque changes $|\vec{L}|$ (speeds up or slows the spin) rather than its direction, so it produces angular acceleration of the spin, not precession.

$\Omega \to 0$: the axis is essentially frozen. The top becomes an ideal direction-holder — the ideal exploited in inertial navigation.