Exercises — Rotating frames — centrifugal force, Coriolis force

Sirf Centrifugal. Coriolis ${\vec{F}}_{\mathrm{Cor}}=-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ hai. Book rotating frame mein rest mein hai, isliye ${\vec{v}}_{\mathrm{rot}}=\vec{0}$, aur $\vec{0}$ ke saath koi bhi cross product $\vec{0}$ hota hai → koi Coriolis nahi. Centrifugal $m{\omega }^2{r}_{\perp }$ hai, jo sirf position par depend karta hai, motion par nahi. Jab tak $r_{\perp }>0$ hai, book ek outward centrifugal force feel karti hai → centrifugal act karta hai.

Sirf centrifugal work kar sakta hai. Coriolis hamesha ${\vec{v}}_{\mathrm{rot}}$ ke perpendicular hota hai (yeh cross product $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ mein baked in hai). Velocity ke perpendicular force se $\vec{F}\cdot \vec{v}=0$ milta hai → zero work; yeh sirf path ko turn karta hai. Centrifugal radially outward point karta hai, isliye agar aap radially move karo toh uski motion ke saath component hoti hai → yeh kar sakta hai work.

Position-only formula: $F_{\mathrm{cf}}=m{\omega }^2{r}_{\perp }=30\times (2.0{)}^2\times 1.8=30\times 4.0\times 1.8=\boxed{216\text{ N}}.$ Yeh real inward (centripetal) grip ki size bhi hai jo seat ko use circling mein rakhne ke liye supply karni padegi.

Radial-outward velocity: ${\vec{v}}_{\mathrm{rot}}=v\hat{r}=4\hat{r}$. $\vec{\omega }=\omega \hat{z}$ ke saath: ${\vec{F}}_{\mathrm{Cor}}=-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}=-2(0.5)(2.0)(\hat{z}\times 4\hat{r}).$ Cylindrical rule $\hat{z}\times \hat{r}=\hat{\theta }$ (counter-clockwise tangential direction) use karte hue: ${\vec{F}}_{\mathrm{Cor}}=-2(0.5)(2.0)(4)\hat{\theta }=-8.0\hat{\theta }\text{ N}.$ Magnitude $8.0\text{ N}$, $-\hat{\theta }$ direction mein point karta hai — yaani spin direction ke opposite. Outward jaata puck spin ke against backward push hota hai, isliye uski path curve ho jaati hai.

Velocity tangential hai: ${\vec{v}}_{\mathrm{rot}}=v\hat{\theta }$. ${\vec{F}}_{\mathrm{Cor}}=-2m\omega \hat{z}\times v\hat{\theta }=-2m\omega v(\hat{z}\times \hat{\theta }).$ Cylindrical coordinates mein $\hat{z}\times \hat{\theta }=-\hat{r}$. Toh ${\vec{F}}_{\mathrm{Cor}}=-2m\omega v(-\hat{r})=+2m\omega v\hat{r}.$ Direction: $+\hat{r}$, radially outward → Coriolis force bug ko axis se door push karta hai jab yeh spin ke saath chalta hai. (Spin ke against chalna $\vec{v}$ flip kar deta aur inward push karta.)

$\frac{F_{\mathrm{cf}}}{F_{\mathrm{Cor}}}=\frac{m{\omega }^2{r}_{\perp }}{2m\omega v}=\frac{\omega r_{\perp }}{2v}=\frac{3\times 1}{2\times 2}=\frac{3}{4}=\mathrm{0.75.}$ Kyunki $0.75<1$, Coriolis (thoda sa) bada hai yahan. Comparison $\omega r_{\perp }$ vs $2v$ tak reduce ho jaata hai: jab bhi aapki speed $v$ rim speed $\omega r_{\perp }$ ke relative badi ho, Coriolis jeetta hai.

Step 1 — rotating frame mein velocity. Puck ground frame mein fixed hai, lekin frame uske neeche $\omega$ se spin karta hai. Frame ke relative yeh rim ke saath backward travel karta hai: ${\vec{v}}_{\mathrm{rot}}=-\omega r_{\perp }\hat{\theta }=-(2.5)(1.5)\hat{\theta }=-3.75\hat{\theta }$ 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. ${\vec{v}}_{\mathrm{rot}}=-\omega r_{\perp }\hat{\theta }$ ke saath: ${\vec{F}}_{\mathrm{Cor}}=-2m\omega \hat{z}\times (-\omega r_{\perp }\hat{\theta })=2m{\omega }^2{r}_{\perp }(\hat{z}\times \hat{\theta })=2m{\omega }^2{r}_{\perp }(-\hat{r}).$ $|{\vec{F}}_{\mathrm{Cor}}|=2\times 0.4\times 6.25\times 1.5=7.5\text{ N},\text{along }-\hat{r}(\text{inward}).$

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: rotating frame mein puck $r_{\perp }$ radius ke circle mein $\omega r_{\perp }$ speed se ghoomta hai, isliye use centripetal (inward) force $m{v}_{\mathrm{rot}}^2/r_{\perp }=m{\omega }^2{r}_{\perp }=3.75$ N chahiye. Fictitious forces exactly woh $3.75$ N inward supply karte hain. Consistent hai — bookkeeping kaam karti hai. ✓

Latitude $\lambda$ par horizontal motion ke liye, Earth ke spin ka sirf vertical component ${\omega }_{\oplus }\sin \lambda$, horizontal Coriolis deflection produce karta hai. Horizontal Coriolis magnitude hai: $F_{\mathrm{Cor}}=2m{\omega }_{\oplus }\sin \lambda v.$ $=2\times (2.0\times 10^5)\times (7.29\times 10^{-5})\times \sin 45^{\circ }\times 40.$ $=2\times 2.0\times 10^5\times 7.29\times 10^{-5}\times 0.7071\times 40\approx 824.7\text{ N}.$ Direction: Northern Hemisphere mein, horizontal motion travel ke right mein deflect hoti hai. North face karte hue, right east hai → train apni east (right) rail press karti hai. (Coriolis Effect in Weather aur Foucault Pendulum logic ke saath consistent: N. hemisphere mein sab kuch right veer karta hai.)

(a) Centrifugal (outward, rod ke saath, $+\hat{r}$): $F_{\mathrm{cf}}=m{\omega }^2{r}_{\perp }=0.1\times 16\times 0.25=0.4\text{ N}.$ Yahi hai jo bead ko rotating frame mein frictionless rod ke saath outward accelerate karta hai.

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

(c) Rod se real force. Rod rigid aur frictionless hai: yeh apni length ke saath ($\hat{r}$) push nahi kar sakta, sirf sideways ($\hat{\theta }$). Bead ko straight rod par rakhne ke liye yeh sideways Coriolis push cancel karna padega. Isliye rod $+\hat{\theta }$ direction mein $0.96\text{ N}$ ka normal contact force supply karta hai (dono frames mein real force). Outward centrifugal $0.4$ N unopposed reh jaata hai → bead outward accelerate karta hai, exactly jaisa observe hota hai.

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 — degrees per hour mein convert karo. $3600$ s/hr aur $180/\pi$ deg/rad se multiply karo: ${\Omega }_{\mathrm{prec}}=5.155\times 10^{-5}\times 3600\times \frac{180}{\pi }\approx 10.6{}^{\circ }/\text{hr}.$

Step 3 — ek full plane rotation ka period ($360^{\circ }$): $T=\frac{2\pi }{{\Omega }_{\mathrm{prec}}}=\frac{2\pi }{5.155\times 10^{-5}}\approx 1.219\times 10^5\text{ s}\approx 33.9\text{ hr}.$ Toh $45^{\circ }$ latitude par Foucault pendulum ka plane ek turn complete karne mein lagbhag 34 ghante leta hai (yeh exactly 24 h sirf poles par hota hai, jahan $\sin \lambda =1$). Yeh exactly wahi mechanism hai jo Foucault Pendulum mein hai.

Kaun sa force position-only hai? ::: Centrifugal, $m{\omega }^2{r}_{\perp }$ — outward, jab bhi $r_{\perp }>0$ ho present rehta hai. Kaun sa force velocity-only hai aur koi work nahi karta? ::: Coriolis, $-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ — motion ke perpendicular. Equator par horizontal Coriolis kyon vanish ho jaata hai? ::: Sirf ${\omega }_{\oplus }\sin \lambda$ horizontal motion ko deflect karta hai; $\lambda =0$ par, $\sin \lambda =0$. N. hemisphere mein, moving objects kis taraf veer karte hain? ::: Right.