Exercises — Gyroscopic effect — precession of spinning top

(a) $\vec{L}$ points along the spin axis (up-and-tilted; sense set by the right-hand rule around the spin). (b) $\vec{\tau }=\vec{r}\times m\vec{g}$ is horizontal, perpendicular to the vertical plane that contains the axis. Gravity is straight down, $\vec{r}$ is up-the-axis, so their cross product is horizontal — it can never point down. (c) $d\vec{L}=\vec{\tau }dt$ points the same way as $\vec{\tau }$ — horizontal. That is why the axis swings sideways (precesses) instead of falling. See the picture of these three vectors below.

Top (faster precession): arm length $r$, mass $m$, gravity $g$ — together they are the torque $rmg$. Bottom (slower precession): moment of inertia $I$ and spin rate $\omega$ — together they are the spin angular momentum $L=I\omega$. Tilt $\theta$: it cancelled. It appears once in $\tau =rmg\sin \theta$ (numerator) and once in the horizontal projection $L\sin \theta$ (denominator), so $\Omega$ is independent of tilt in the fast-spin limit.

Step 1 — torque. $\tau =rmg=0.10\times 2.0\times 9.8=1.96\text{N⋅m}$. Step 2 — angular momentum. $L=I\omega =0.050\times 150=7.5{\text{kg⋅m}}^2/\text{s}$. Step 3 — precession rate. $\Omega =\tau /L=1.96/7.5=0.2613\text{rad/s}$. Step 4 — period. $T=2\pi /\Omega =6.2832/0.2613=24.0\text{s}$. The axis sweeps a full horizontal circle about once every 24 seconds.

Step 1. $I=m{R}^2=3.0\times (0.35{)}^2=3.0\times 0.1225=0.3675{\text{kg⋅m}}^2$. Step 2. $L=I\omega =0.3675\times 25=9.1875{\text{kg⋅m}}^2/\text{s}$. Step 3. $\tau =rmg=0.25\times 3.0\times 9.8=7.35\text{N⋅m}$. Step 4. $\Omega =\tau /L=7.35/9.1875=0.800\text{rad/s}$.

$\Omega =rmg/(I\omega )$, so $\Omega \propto 1/\omega$. Tripling $\omega$ makes $\Omega$ one-third as big. Period is the inverse of rate: $T=2\pi /\Omega$, so $T\propto \omega$. Tripling $\omega$ triples the period. $T_2=3\times 6.0=18.0\text{s}$.

At any $\theta$: $\tau =rmg\sin \theta =0.50\sin \theta$, and the horizontal projection is $L\sin \theta =2.0\sin \theta$. $\theta =30°$: $\Omega =\frac{0.50\sin 30°}{2.0\sin 30°}=\frac{0.50}{2.0}=0.25\text{rad/s}$. $\theta =75°$: $\Omega =\frac{0.50\sin 75°}{2.0\sin 75°}=\frac{0.50}{2.0}=0.25\text{rad/s}$. The $\sin \theta$ divides out of top and bottom — same $\Omega =0.25\text{rad/s}$ for both tilts. The figure below shows why: the horizontal circle traced by the tip of $\vec{L}$ has radius $L\sin \theta$, exactly matching how the torque grows.

Write $\vec{r}=r\hat{x}$ (its horizontal lean; the vertical part is parallel to $\vec{g}$ and contributes nothing to the cross product), and $m\vec{g}=-mg\hat{z}$. (a) $\vec{\tau }=(r\hat{x})\times (-mg\hat{z})=-rmg(\hat{x}\times \hat{z})$. Now $\hat{x}\times \hat{z}=-\hat{y}$, so $\vec{\tau }=-rmg(-\hat{y})=+rmg\hat{y}$. $\vec{\tau }$ points along $+y$. (b) $d\vec{L}=\vec{\tau }dt$ is along $+y$, so the axis (which leaned toward $+x$) swings toward $+y$. (c) Going $+x\to +y\to -x\to -y$ is counter-clockwise seen from above (looking down the $+z$ axis). See the figure.

Step 1 — required $\Omega$. $\Omega =2\pi /T=6.2832/10.0=0.62832\text{rad/s}$. Step 2 — invert the precession formula. $\Omega =rmg/(I\omega )\Rightarrow \omega =\frac{rmg}{I\Omega }$. Step 3 — torque. $rmg=0.12\times 1.5\times 9.8=1.764\text{N⋅m}$. Step 4 — solve. $\omega =\frac{1.764}{0.020\times 0.62832}=\frac{1.764}{0.0125664}=140.4\text{rad/s}$.

(a) $\tau =rmg=0.03\times 0.20\times 9.8=0.0588\text{N⋅m}$. $L=I\omega =0.004\times 8.0=0.032{\text{kg⋅m}}^2/\text{s}$. $\Omega =\tau /L=0.0588/0.032=1.8375\text{rad/s}$. (b) Ratio $\Omega /\omega =1.8375/8.0=0.2297$, i.e. about $23\%$. This is not $\ll 1$. The fast-spin approximation is shaky: the neglected physics is nutation — the axis will visibly bob up and down as it precesses, and the true averaged $\Omega$ will differ slightly from this value. For a clean steady precession you'd want $\omega$ ten-or-more times larger.

(a) As $\omega \to 0$, the denominator $I\omega \to 0$, so $\Omega \to \infty$. The formula predicts an infinitely fast precession. (b) Physically a non-spinning top just falls over — no precession at all. The contradiction is the formula flagging its own breakdown: with $\omega =0$ there is no angular momentum to redirect, so $\vec{\tau }=d\vec{L}/dt$ simply builds up $\vec{L}$ in the direction of the torque from zero — the top topples. The formula $\Omega =rmg/(I\omega )$ is only valid in the fast-spin limit where a large $\vec{L}$ already exists to be steered; "$\Omega \to \infty$" is its way of saying "I do not apply here."

(a) ${\Omega }_A=\tau /L_A=2.0/4.0=0.50\text{rad/s}$. ${\Omega }_B=\tau /L_B=2.0/12.0=0.1\overline{6}=0.1667\text{rad/s}$. (b) ${\Omega }_A/{\Omega }_B=0.50/0.1667=3.0$. Gyro B precesses 3× slower — exactly the ratio $L_B/L_A=3$. Bigger $L$ ⇒ slower response to the same torque. (c) A bullet given a large spin $L$ has a tiny $\Omega$ for any given cross-wind or gravity torque, so its nose barely wanders off-axis during flight — that is spin-stabilisation, the same principle used in a Gyroscope and Navigation system.

• Precession rate formula? ::: $\Omega =\tau /(L\sin \theta )=rmg/(I\omega )$.
• Direction of $d\vec{L}$? ::: Same as $\vec{\tau }=\vec{r}\times m\vec{g}$ (horizontal).
• Triple the spin ⇒ period does what? ::: Triples ($T\propto \omega$).
• Why does $\sin \theta$ cancel? ::: It multiplies both the torque and the horizontal $L$-projection.
• When does the formula fail? ::: When $\omega$ is not $\gg \Omega$ (nutation matters); at $\omega =0$ it wrongly gives $\infty$.