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Q.Derive an expression for maximum velocity of a car while moving on a banked road.

Nagaland NbseNagaland Board of School Education (Class XI) 2021Subjective· 3mImportance★★★★★
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At the maximum safe speed on a banked, rough road, friction acts down the slope (about to make the vehicle skid outward); balancing vertical and horizontal (centripetal) forces gives vmax=rg(tan⁡θ+μs)/(1−μstan⁡θ)v_{max}=\sqrt{rg(\tan\theta+\mu_s)/(1-\mu_s\tan\theta)}.

Consider a vehicle of mass mm going around a curve of radius rr, banked at angle θ\theta, with coefficient of static friction μs\mu_s between the tyres and road. At the maximum possible speed, the vehicle is on the verge of skidding outward (up the slope), so friction acts down the incline, at its limiting value f=μsNf = \mu_s N.

Resolving forces: vertically, Ncos⁡θ−fsin⁡θ=mgN\cos\theta - f\sin\theta = mg; horizontally (providing the centripetal force), Nsin⁡θ+fcos⁡θ=mvmax2rN\sin\theta + f\cos\theta = \dfrac{mv_{max}^2}{r}.

Substituting f=μsNf = \mu_s N:

N(cos⁡θ−μssin⁡θ)=mg⇒N=mgcos⁡θ−μssin⁡θN(\cos\theta - \mu_s\sin\theta) = mg \quad\Rightarrow\quad N = \frac{mg}{\cos\theta - \mu_s\sin\theta}

N(sin⁡θ+μscos⁡θ)=mvmax2rN(\sin\theta + \mu_s\cos\theta) = \frac{mv_{max}^2}{r} …

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