Exercises — Banking of roads — derivation

WHAT tool & WHY: "Ideal / frictionless" means only the normal force does the turning, so we use $\tan \theta =\frac{v^2}{rg}$ — the one formula where friction plays no part. $\tan \theta =\frac{20^2}{80\times 9.8}=\frac{400}{784}=0.5102$ $\theta =\arctan (0.5102)=27.0^{\circ }$ Answer: $\theta \approx 27.0^{\circ }$.

WHY this formula: design speed = frictionless speed, so rearrange $\tan \theta =v^2/(rg)$ into $v_0=\sqrt{rg\tan \theta }$. $v_0=\sqrt{50\times 9.8\times \tan 30^{\circ }}=\sqrt{50\times 9.8\times 0.5774}=\sqrt{282.9}=16.82\text{m/s}$ Answer: $v_0\approx 16.8\text{m/s}$ (about $60$ km/h).

WHY $v_{max}$: fast car tends to slide outward/up the slope, so friction points down the slope and adds to the inward pull — the "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: $v_{max}\approx 20.8\text{m/s}$. Higher than the design speed $16.8$ m/s — friction lets you push faster.

WHY $v_{min}$: slow car tends to slip inward/down, so friction points up the slope and holds it out — flip the signs: "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: $v_{min}\approx 12.9\text{m/s}$. So the safe band on this curve is roughly $12.9$–$20.8$ m/s.

Compute the denominator first: $1-\mu \tan \theta =1-0.9\times \tan 40^{\circ }=1-0.9\times 0.8391=1-0.7552=0.2448$. Still positive, so: $\frac{\mu +\tan \theta }{1-\mu \tan \theta }=\frac{0.9+0.8391}{0.2448}=\frac{1.7391}{0.2448}=7.105$ $v_{max}=\sqrt{100\times 9.8\times 7.105}=\sqrt{6962.7}=83.4\text{m/s}$ The physical meaning: as $\mu \tan \theta \to 1$ the denominator $\to 0$ and $v_{max}\to \infty$. When $\mu \tan \theta \ge 1$ (i.e. $\mu \ge \cot \theta$), friction is so strong relative to the tilt that the car can never be flung outward — no upper speed limit from this mechanism (the car would fail some other way first). Here $\mu \tan \theta =0.755<1$, so a finite $v_{max}\approx 83.4$ m/s still exists. Answer: $v_{max}\approx 83.4\text{m/s}$; the near-zero denominator is the warning sign of the runaway limit.

Look at the numerator of $v_{min}$: $\tan \theta -\mu$. If $\tan \theta \le \mu$, this is $\le 0$. A speed cannot be an imaginary number, so the physical minimum is simply $v_{min}=0$. Meaning: friction alone (up the slope) is strong enough to hold the car against the gravity component pulling it down the tilt — even at rest. The car can park on the banked road without sliding into the centre. The tilt is "gentle enough" that grip wins. Answer: $v_{min}=0$ whenever $\tan \theta \le \mu$; a parked car stays put.

Step 1 — convert units. $90\text{km/h}=90\times \frac{1000}{3600}=25\text{m/s}$. Why: our formulas use SI metres and seconds. 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 — real max speed with $\mu =0.15$: $\frac{\mu +\tan \theta }{1-\mu \tan \theta }=\frac{0.15+0.3189}{1-0.15\times 0.3189}=\frac{0.4689}{0.9522}=0.4925$ $v_{max}=\sqrt{200\times 9.8\times 0.4925}=\sqrt{965.3}=31.07\text{m/s}$ Answer: $\theta \approx 17.7^{\circ }$; real $v_{max}\approx 31.1\text{m/s}$ ($\approx 112$ km/h), safely above the $25$ m/s design speed.

WHY same formula: in both problems a single force (here string tension, there normal force) tilts by $\theta$; its vertical part balances $mg$ and its horizontal part supplies $m{v}^2/r$. Identical geometry ⇒ identical $\tan \theta =v^2/(rg)$. Step 1 — radius of the circle: $r=L\sin \theta =1.2\times \sin 30^{\circ }=1.2\times 0.5=0.6\text{m}$. Step 2 — solve for $v$: $v=\sqrt{rg\tan \theta }=\sqrt{0.6\times 9.8\times \tan 30^{\circ }}=\sqrt{0.6\times 9.8\times 0.5774}=\sqrt{3.395}=1.843\text{m/s}$. Answer: $v\approx 1.84\text{m/s}$.

Strategy: We have two unknowns ($\theta ,\mu$) and two conditions ($v_{max},v_{min}$). Write both, then combine. Let $t=\tan \theta$. From the two formulas: $\frac{v_{max}^2}{rg}=\frac{t+\mu }{1-\mu t},\frac{v_{min}^2}{rg}=\frac{t-\mu }{1+\mu t}$ Compute the numbers: $rg=120\times 9.8=1176$. $M\equiv \frac{v_{max}^2}{rg}=\frac{900}{1176}=0.7653,m\equiv \frac{v_{min}^2}{rg}=\frac{225}{1176}=0.1913$ Clear the fractions. Multiply each equation by its denominator to get two clean linear-in-products relations: 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 — subtract (B) from (A) to make the lone $t$ terms cancel and isolate $\mu$: (M - m) - \mu t(M + m) = 2\mu \implies M - m = \mu\big(2 + t(M+m)\big) \tag{i} Step 2 — now add (A) and (B) to get a second, independent relation: (M + m) + \mu t(m - M) = 2t \implies M+m = 2t + \mu t (M-m) \tag{ii} Equations (i) and (ii) are two equations in $t$ and $\mu$. Substitute the numbers $M=0.7653$, $m=0.1913$ (so $M+m=0.9566$, $M-m=0.5740$):

• (i): $0.5740=\mu (2+0.9566t)$.
• (ii): $0.9566=2t+0.5740\mu t$. Step 3 — solve by refinement. From (i), $\mu =\frac{0.5740}{2+0.9566t}$; feed this into (ii). Try $\theta =22^{\circ }\Rightarrow t=0.4040$: (i) gives $\mu =0.5740/2.3865=0.2405$; check (ii): $2(0.4040)+0.5740(0.2405)(0.4040)=0.8081+0.0558=0.8639$ — below $0.9566$, so raise $\theta$. Try $\theta =25^{\circ }\Rightarrow t=0.4663$: (i) gives $\mu =0.5740/2.4461=0.2347$; check (ii): $0.9326+0.5740(0.2347)(0.4663)=0.9326+0.0628=0.9954$ — slightly above. Interpolating between $22^{\circ }$ and $25^{\circ }$ lands at $\theta \approx 24.3^{\circ }$, giving $\mu \approx 0.236$. Step 4 — verify directly with $\theta =24.3^{\circ }$ ($t=0.4515$), $\mu =0.236$:
  • $v_{max}=\sqrt{1176\times \frac{0.4515+0.236}{1-0.236\times 0.4515}}=\sqrt{1176\times \frac{0.6875}{0.8934}}=\sqrt{1176\times 0.7695}=\sqrt{905}=30.1$ 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: $\theta \approx 24.3^{\circ }$, $\mu \approx 0.236$.

Start exactly: $v_{max}^2=rg\frac{\tan \theta +\mu }{1-\mu \tan \theta }$, $v_{min}^2=rg\frac{\tan \theta -\mu }{1+\mu \tan \theta }$. WHY expand: we want behaviour "for small $\mu$", so keep only terms up to the first power of $\mu$; higher powers are negligible. WHY $(1-\mu a{)}^{-1}\approx 1+\mu a$: this is the standard geometric / binomial series $\frac{1}{1-x}=1+x+x^2+\dots$ (equivalently the first-order Taylor expansion of $(1-x{)}^{-1}$ about $x=0$). With $x=\mu a$ small, every term from $x^2$ onward is negligible, so $\frac{1}{1-\mu a}\approx 1+\mu a$. This is exactly the tool that turns an awkward fraction into a simple sum we can subtract term by term. Using $\frac{a+\mu }{1-\mu a}\approx (a+\mu )(1+\mu a)\approx a+\mu +\mu a^2=a+\mu (1+a^2)$ with $a=\tan \theta$ (dropping ${\mu }^2$): $v_{max}^2\approx rg[\tan \theta +\mu (1+{\tan }^2\theta )]=rg\tan \theta +\mu rg{\sec }^2\theta$ Similarly $\frac{a-\mu }{1+\mu a}\approx (a-\mu )(1-\mu a)\approx a-\mu (1+a^2)$ (using $(1+\mu a{)}^{-1}\approx 1-\mu a$, the same series with $x=-\mu a$): $v_{min}^2\approx rg\tan \theta -\mu rg{\sec }^2\theta$ Subtract: $v_{max}^2-v_{min}^2\approx 2\mu rg{\sec }^2\theta ■$ Sanity check with Problem 2 numbers ($r=50,\theta =30^{\circ },\mu =0.2$): exact $v_{max}^2-v_{min}^2=430.7-165.8=264.9$; approx $2\times 0.2\times 50\times 9.8\times {\sec }^{2}30^{\circ }=196\times 1.3333=261.3$. Within $1.4\%$ — the small-$\mu$ formula works even at $\mu =0.2$. (Here ${\sec }^{2}30^{\circ }=1/{\cos }^{2}30^{\circ }=1.3333$.) Answer: $v_{max}^2-v_{min}^2\approx 2\mu rg{\sec }^2\theta$.