Exercises — Rotating frames — centrifugal force, Coriolis force

Centrifugal only. Coriolis is ${\vec{F}}_{\mathrm{Cor}}=-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$. The book is at rest in the rotating frame, so ${\vec{v}}_{\mathrm{rot}}=\vec{0}$, and any cross product with $\vec{0}$ is $\vec{0}$ → no Coriolis. Centrifugal is $m{\omega }^2{r}_{\perp }$, which depends only on position, not motion. As long as $r_{\perp }>0$ the book feels an outward centrifugal force → centrifugal acts.

Only centrifugal can do work. Coriolis is always perpendicular to ${\vec{v}}_{\mathrm{rot}}$ (that is baked into the cross product $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$). A force perpendicular to velocity gives $\vec{F}\cdot \vec{v}=0$ → zero work; it only turns the path. Centrifugal points radially outward, so if you move radially it has a component along your motion → it can do work.

Position-only formula: $F_{\mathrm{cf}}=m{\omega }^2{r}_{\perp }=30\times (2.0{)}^2\times 1.8=30\times 4.0\times 1.8=\boxed{216\text{ N}}.$ This is also the size of the real inward (centripetal) grip the seat must supply to keep her circling.

Radial-outward velocity: ${\vec{v}}_{\mathrm{rot}}=v\hat{r}=4\hat{r}$. With $\vec{\omega }=\omega \hat{z}$: ${\vec{F}}_{\mathrm{Cor}}=-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}=-2(0.5)(2.0)(\hat{z}\times 4\hat{r}).$ Using the cylindrical rule $\hat{z}\times \hat{r}=\hat{\theta }$ (the counter-clockwise tangential direction): ${\vec{F}}_{\mathrm{Cor}}=-2(0.5)(2.0)(4)\hat{\theta }=-8.0\hat{\theta }\text{ N}.$ Magnitude $8.0\text{ N}$, pointing $-\hat{\theta }$ — i.e. opposite to the spin direction. The outward-moving puck gets shoved backward against the rotation, so its path curves.

Velocity is tangential: ${\vec{v}}_{\mathrm{rot}}=v\hat{\theta }$. ${\vec{F}}_{\mathrm{Cor}}=-2m\omega \hat{z}\times v\hat{\theta }=-2m\omega v(\hat{z}\times \hat{\theta }).$ In cylindrical coordinates $\hat{z}\times \hat{\theta }=-\hat{r}$. So ${\vec{F}}_{\mathrm{Cor}}=-2m\omega v(-\hat{r})=+2m\omega v\hat{r}.$ Direction: $+\hat{r}$, radially outward → the Coriolis force pushes the bug away from the axis when it walks with the spin. (Walking against the spin would flip $\vec{v}$ and push it inward.)

$\frac{F_{\mathrm{cf}}}{F_{\mathrm{Cor}}}=\frac{m{\omega }^2{r}_{\perp }}{2m\omega v}=\frac{\omega r_{\perp }}{2v}=\frac{3\times 1}{2\times 2}=\frac{3}{4}=\mathrm{0.75.}$ Since $0.75<1$, Coriolis is (slightly) larger here. The comparison reduces to $\omega r_{\perp }$ vs $2v$: whenever your speed $v$ is large relative to the rim speed $\omega r_{\perp }$, Coriolis wins.

Step 1 — velocity in the rotating frame. The puck is fixed in the ground frame, but the frame spins under it at $\omega$. Relative to the frame it therefore travels backward along the rim: ${\vec{v}}_{\mathrm{rot}}=-\omega r_{\perp }\hat{\theta }=-(2.5)(1.5)\hat{\theta }=-3.75\hat{\theta }$ m/s. Speed $=3.75$ m/s.

Step 2 — centrifugal force (outward). $F_{\mathrm{cf}}=m{\omega }^2{r}_{\perp }=0.4\times (2.5{)}^2\times 1.5=0.4\times 6.25\times 1.5=3.75\text{ N},\text{along }+\hat{r}.$

Step 3 — Coriolis force. With ${\vec{v}}_{\mathrm{rot}}=-\omega r_{\perp }\hat{\theta }$: ${\vec{F}}_{\mathrm{Cor}}=-2m\omega \hat{z}\times (-\omega r_{\perp }\hat{\theta })=2m{\omega }^2{r}_{\perp }(\hat{z}\times \hat{\theta })=2m{\omega }^2{r}_{\perp }(-\hat{r}).$ $|{\vec{F}}_{\mathrm{Cor}}|=2\times 0.4\times 6.25\times 1.5=7.5\text{ N},\text{along }-\hat{r}(\text{inward}).$

Step 4 — net fictitious force. $F_{\mathrm{net},\mathrm{fict}}=F_{\mathrm{cf}}-F_{\mathrm{Cor}}=3.75-7.5=-3.75\text{ N}=3.75\text{ N inward}.$ Interpretation: in the rotating frame the puck moves in a circle of radius $r_{\perp }$ at speed $\omega r_{\perp }$, so it needs a centripetal (inward) force $m{v}_{\mathrm{rot}}^2/r_{\perp }=m{\omega }^2{r}_{\perp }=3.75$ N. The fictitious forces supply exactly that $3.75$ N inward. Consistent — the bookkeeping works. ✓

For horizontal motion at latitude $\lambda$, only the vertical component of Earth's spin, ${\omega }_{\oplus }\sin \lambda$, produces a horizontal Coriolis deflection. The horizontal Coriolis magnitude is: $F_{\mathrm{Cor}}=2m{\omega }_{\oplus }\sin \lambda v.$ $=2\times (2.0\times 10^5)\times (7.29\times 10^{-5})\times \sin 45^{\circ }\times 40.$ $=2\times 2.0\times 10^5\times 7.29\times 10^{-5}\times 0.7071\times 40\approx 824.7\text{ N}.$ Direction: in the Northern Hemisphere, horizontal motion is deflected to the right of travel. Facing north, right is east → the train presses on its east (right) rail. (Consistent with Coriolis Effect in Weather and the Foucault Pendulum logic: everything veers right in the N. hemisphere.)

(a) Centrifugal (outward, along the rod, $+\hat{r}$): $F_{\mathrm{cf}}=m{\omega }^2{r}_{\perp }=0.1\times 16\times 0.25=0.4\text{ N}.$ This is what accelerates the bead outward along the frictionless rod in the rotating frame.

(b) Coriolis. Velocity is radial outward: ${\vec{v}}_{\mathrm{rot}}=v\hat{r}$. ${\vec{F}}_{\mathrm{Cor}}=-2m\omega \hat{z}\times v\hat{r}=-2m\omega v(\hat{z}\times \hat{r})=-2m\omega v\hat{\theta }.$ $|{\vec{F}}_{\mathrm{Cor}}|=2\times 0.1\times 4\times 1.2=0.96\text{ N},\text{direction }-\hat{\theta }(\text{against the spin}).$

(c) The real force from the rod. The rod is rigid and frictionless: it cannot push along its own length ($\hat{r}$), only sideways ($\hat{\theta }$). To keep the bead on the straight rod it must cancel the sideways Coriolis push. So the rod supplies a normal contact force of $0.96\text{ N}$ in the $+\hat{\theta }$ direction (a real force in both frames). The outward centrifugal $0.4$ N is left unopposed → the bead accelerates outward, exactly as observed.

Step 1 — precession rate. ${\Omega }_{\mathrm{prec}}={\omega }_{\oplus }\sin \lambda =7.29\times 10^{-5}\times \sin 45^{\circ }=7.29\times 10^{-5}\times 0.7071=5.155\times 10^{-5}\text{ rad/s}.$

Step 2 — convert to degrees per hour. Multiply by $3600$ s/hr and by $180/\pi$ deg/rad: ${\Omega }_{\mathrm{prec}}=5.155\times 10^{-5}\times 3600\times \frac{180}{\pi }\approx 10.6{}^{\circ }/\text{hr}.$

Step 3 — period of one full plane rotation ($360^{\circ }$): $T=\frac{2\pi }{{\Omega }_{\mathrm{prec}}}=\frac{2\pi }{5.155\times 10^{-5}}\approx 1.219\times 10^5\text{ s}\approx 33.9\text{ hr}.$ So at $45^{\circ }$ latitude the Foucault pendulum's plane takes about 34 hours to complete one turn (it is exactly 24 h only at the poles, where $\sin \lambda =1$). This is precisely the mechanism in Foucault Pendulum.

Which force is position-only? ::: Centrifugal, $m{\omega }^2{r}_{\perp }$ — outward, present whenever $r_{\perp }>0$. Which force is velocity-only and does no work? ::: Coriolis, $-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ — perpendicular to motion. What makes horizontal Coriolis vanish at the equator? ::: Only ${\omega }_{\oplus }\sin \lambda$ deflects horizontal motion; at $\lambda =0$, $\sin \lambda =0$. In the N. hemisphere, moving objects veer to their…? ::: Right.