Exercises — Banking of roads — derivation

Kaun sa tool aur KYUN: "Ideal / frictionless" ka matlab hai sirf normal force turning karti hai, isliye $\tan \theta =\frac{v^2}{rg}$ use karo — yeh woh ek formula hai jisme friction ka koi role nahi. $\tan \theta =\frac{20^2}{80\times 9.8}=\frac{400}{784}=0.5102$ $\theta =\arctan (0.5102)=27.0^{\circ }$ Answer: $\theta \approx 27.0^{\circ }$.

KYUN yeh formula: design speed = frictionless speed, isliye $\tan \theta =v^2/(rg)$ ko rearrange karke $v_0=\sqrt{rg\tan \theta }$ banao. $v_0=\sqrt{50\times 9.8\times \tan 30^{\circ }}=\sqrt{50\times 9.8\times 0.5774}=\sqrt{282.9}=16.82\text{m/s}$ Answer: $v_0\approx 16.8\text{m/s}$ (lagbhag $60$ km/h).

KYUN $v_{max}$: tez car outward/up the slope slide karne ki koshish karti hai, isliye friction down the slope point karta hai aur inward pull mein add hota hai — "plus on top, minus on bottom" formula. $\frac{\mu +\tan \theta }{1-\mu \tan \theta }=\frac{0.2+0.5774}{1-0.2\times 0.5774}=\frac{0.7774}{0.8845}=0.8789$ $v_{max}=\sqrt{50\times 9.8\times 0.8789}=\sqrt{430.7}=20.75\text{m/s}$ Answer: $v_{max}\approx 20.8\text{m/s}$. Design speed $16.8$ m/s se zyada — friction aapko aur tez jaane deta hai.

KYUN $v_{min}$: slow car inward/down slip karne ki koshish karti hai, isliye friction up the slope point karta hai aur use bahar rokta hai — signs flip karo: "minus on top, plus on bottom." $\frac{\tan \theta -\mu }{1+\mu \tan \theta }=\frac{0.5774-0.2}{1+0.2\times 0.5774}=\frac{0.3774}{1.1155}=0.3383$ $v_{min}=\sqrt{50\times 9.8\times 0.3383}=\sqrt{165.8}=12.87\text{m/s}$ Answer: $v_{min}\approx 12.9\text{m/s}$. Toh is curve par safe band roughly $12.9$–$20.8$ m/s hai.

Pehle denominator compute karo: $1-\mu \tan \theta =1-0.9\times \tan 40^{\circ }=1-0.9\times 0.8391=1-0.7552=0.2448$. Ab bhi positive hai, isliye: $\frac{\mu +\tan \theta }{1-\mu \tan \theta }=\frac{0.9+0.8391}{0.2448}=\frac{1.7391}{0.2448}=7.105$ $v_{max}=\sqrt{100\times 9.8\times 7.105}=\sqrt{6962.7}=83.4\text{m/s}$ Physical meaning: jaise-jaise $\mu \tan \theta \to 1$, denominator $\to 0$ aur $v_{max}\to \infty$. Jab $\mu \tan \theta \ge 1$ (yaani $\mu \ge \cot \theta$), friction tilt ke relative itna strong hai ki car kabhi bahar nahi phek'i ja sakti — is mechanism se koi upper speed limit nahi (car pehle kisi aur wajah se fail ho jaayegi). Yahan $\mu \tan \theta =0.755<1$, isliye ek finite $v_{max}\approx 83.4$ m/s abhi bhi exist karta hai. Answer: $v_{max}\approx 83.4\text{m/s}$; near-zero denominator runaway limit ki warning sign hai.

$v_{min}$ ke numerator ko dekho: $\tan \theta -\mu$. Agar $\tan \theta \le \mu$ ho, toh yeh $\le 0$ hai. Speed imaginary number nahi ho sakti, isliye physical minimum simply $v_{min}=0$ hai. Matlab: Friction akela (slope par upar ki taraf) car ko gravity component ke against hold karne ke liye kaafi strong hai jo use tilt ke neeche kheench raha hai — rest par bhi. Car banked road par park kar sakti hai bina centre mein slide hue. Tilt "gentle enough" hai ki grip jeet jaata hai. Answer: $v_{min}=0$ jab bhi $\tan \theta \le \mu$; parked car wahan ruki rehti hai.

Step 1 — units convert karo. $90\text{km/h}=90\times \frac{1000}{3600}=25\text{m/s}$. KYUN: hamare formulas SI metres aur seconds use karte hain. Step 2 — design angle (ideal → $\tan \theta =v^2/rg$): $\tan \theta =\frac{25^2}{200\times 9.8}=\frac{625}{1960}=0.3189⟹\theta =\arctan (0.3189)=17.7^{\circ }$ Step 3 — $\mu =0.15$ ke saath real max speed: $\frac{\mu +\tan \theta }{1-\mu \tan \theta }=\frac{0.15+0.3189}{1-0.15\times 0.3189}=\frac{0.4689}{0.9522}=0.4925$ $v_{max}=\sqrt{200\times 9.8\times 0.4925}=\sqrt{965.3}=31.07\text{m/s}$ Answer: $\theta \approx 17.7^{\circ }$; real $v_{max}\approx 31.1\text{m/s}$ ($\approx 112$ km/h), $25$ m/s design speed se safely upar.

KYUN same formula: dono problems mein ek single force (yahan string tension, wahan normal force) $\theta$ se tilt hoti hai; uska vertical part $mg$ balance karta hai aur horizontal part $m{v}^2/r$ supply karta hai. Identical geometry ⇒ identical $\tan \theta =v^2/(rg)$. Step 1 — circle ka radius: $r=L\sin \theta =1.2\times \sin 30^{\circ }=1.2\times 0.5=0.6\text{m}$. Step 2 — $v$ ke liye solve karo: $v=\sqrt{rg\tan \theta }=\sqrt{0.6\times 9.8\times \tan 30^{\circ }}=\sqrt{0.6\times 9.8\times 0.5774}=\sqrt{3.395}=1.843\text{m/s}$. Answer: $v\approx 1.84\text{m/s}$.

Strategy: Hamare paas do unknowns ($\theta ,\mu$) aur do conditions ($v_{max},v_{min}$) hain. Dono likho, phir combine karo. Maano $t=\tan \theta$. Dono formulas se: $\frac{v_{max}^2}{rg}=\frac{t+\mu }{1-\mu t},\frac{v_{min}^2}{rg}=\frac{t-\mu }{1+\mu t}$ Numbers compute karo: $rg=120\times 9.8=1176$. $M\equiv \frac{v_{max}^2}{rg}=\frac{900}{1176}=0.7653,m\equiv \frac{v_{min}^2}{rg}=\frac{225}{1176}=0.1913$ Fractions clear karo. Har equation ko uske denominator se multiply karo taaki do clean linear-in-products relations milein: M(1-\mu t) = t+\mu \implies M - M\mu t = t + \mu \tag{A} m(1+\mu t) = t-\mu \implies m + m\mu t = t - \mu \tag{B} Step 1 — (B) ko (A) se subtract karo taaki lone $t$ terms cancel ho jaayein aur $\mu$ isolate ho: (M - m) - \mu t(M + m) = 2\mu \implies M - m = \mu\big(2 + t(M+m)\big) \tag{i} Step 2 — ab (A) aur (B) add karo ek doosri, independent relation paane ke liye: (M + m) + \mu t(m - M) = 2t \implies M+m = 2t + \mu t (M-m) \tag{ii} Equations (i) aur (ii) $t$ aur $\mu$ mein do equations hain. Numbers substitute karo $M=0.7653$, $m=0.1913$ (toh $M+m=0.9566$, $M-m=0.5740$):

• (i): $0.5740=\mu (2+0.9566t)$.
• (ii): $0.9566=2t+0.5740\mu t$. Step 3 — refinement se solve karo. (i) se, $\mu =\frac{0.5740}{2+0.9566t}$; ise (ii) mein daalo. Try karo $\theta =22^{\circ }\Rightarrow t=0.4040$: (i) deta hai $\mu =0.5740/2.3865=0.2405$; check karo (ii): $2(0.4040)+0.5740(0.2405)(0.4040)=0.8081+0.0558=0.8639$ — $0.9566$ se neeche hai, isliye $\theta$ badhao. Try karo $\theta =25^{\circ }\Rightarrow t=0.4663$: (i) deta hai $\mu =0.5740/2.4461=0.2347$; check karo (ii): $0.9326+0.5740(0.2347)(0.4663)=0.9326+0.0628=0.9954$ — thoda upar hai. $22^{\circ }$ aur $25^{\circ }$ ke beech interpolate karne par $\theta \approx 24.3^{\circ }$ milta hai, $\mu \approx 0.236$ ke saath. Step 4 — directly verify karo $\theta =24.3^{\circ }$ ($t=0.4515$), $\mu =0.236$ ke saath:
  • $v_{max}=\sqrt{1176\times \frac{0.4515+0.236}{1-0.236\times 0.4515}}=\sqrt{1176\times \frac{0.6875}{0.8934}}=\sqrt{1176\times 0.7695}=\sqrt{905}=30.1$ m/s ✓
  • $v_{min}=\sqrt{1176\times \frac{0.4515-0.236}{1+0.236\times 0.4515}}=\sqrt{1176\times \frac{0.2155}{1.1066}}=\sqrt{1176\times 0.1947}=\sqrt{229}=15.1$ m/s ✓ Answer: $\theta \approx 24.3^{\circ }$, $\mu \approx 0.236$.

Bilkul exactly se shuru karo: $v_{max}^2=rg\frac{\tan \theta +\mu }{1-\mu \tan \theta }$, $v_{min}^2=rg\frac{\tan \theta -\mu }{1+\mu \tan \theta }$. KYUN expand karte hain: hum "small $\mu$" ke liye behaviour chahte hain, isliye sirf $\mu$ ki pehli power tak ke terms rakho; higher powers negligible hain. KYUN $(1-\mu a{)}^{-1}\approx 1+\mu a$: yeh standard geometric / binomial series hai $\frac{1}{1-x}=1+x+x^2+\dots$ (equivalently $(1-x{)}^{-1}$ ka first-order Taylor expansion $x=0$ ke aas-paas). $x=\mu a$ small hone par, $x^2$ aur aage ke har term negligible hain, isliye $\frac{1}{1-\mu a}\approx 1+\mu a$. Yahi woh tool hai jo ek awkward fraction ko simple sum mein badal deta hai jise hum term by term subtract kar sakte hain. $\frac{a+\mu }{1-\mu a}\approx (a+\mu )(1+\mu a)\approx a+\mu +\mu a^2=a+\mu (1+a^2)$ use karte hue $a=\tan \theta$ ke saath (${\mu }^2$ drop karke): $v_{max}^2\approx rg[\tan \theta +\mu (1+{\tan }^2\theta )]=rg\tan \theta +\mu rg{\sec }^2\theta$ Similarly $\frac{a-\mu }{1+\mu a}\approx (a-\mu )(1-\mu a)\approx a-\mu (1+a^2)$ ($(1+\mu a{)}^{-1}\approx 1-\mu a$ use karte hue, same series $x=-\mu a$ ke saath): $v_{min}^2\approx rg\tan \theta -\mu rg{\sec }^2\theta$ Subtract karo: $v_{max}^2-v_{min}^2\approx 2\mu rg{\sec }^2\theta ■$ Sanity check Problem 2 ke numbers se ($r=50,\theta =30^{\circ },\mu =0.2$): exact $v_{max}^2-v_{min}^2=430.7-165.8=264.9$; approx $2\times 0.2\times 50\times 9.8\times {\sec }^{2}30^{\circ }=196\times 1.3333=261.3$. $1.4\%$ ke andar — small-$\mu$ formula $\mu =0.2$ par bhi kaam karta hai. (Yahan ${\sec }^{2}30^{\circ }=1/{\cos }^{2}30^{\circ }=1.3333$.) Answer: $v_{max}^2-v_{min}^2\approx 2\mu rg{\sec }^2\theta$.