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

Nagaland NbseNagaland Board of School Education (Class XI) 2025Subjective· 3mImportance★★★★★
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On a banked road, both the horizontal component of the normal reaction and (up to its limiting value) friction supply the centripetal force; the maximum safe speed corresponds to friction acting at its limiting value, about to make the car skid outward/up the incline.

Consider a car of mass mm going around a curve of radius rr on a road banked at angle θ\theta to the horizontal, with coefficient of static friction μs\mu_s between tyres and road. At maximum speed vmaxv_{max}, the car is on the verge of skidding outward (up the slope), so friction f=μsNf=\mu_sN acts down the incline (towards the centre's opposite side along the slope), helping to prevent outward sliding... more precisely, friction acts down the slope (inward-and-down along the incline) opposing the tendency to slide up/out.

Resolve forces along the vertical and horizontal (or along/perpendicular to incline). Along the vertical:

Ncos⁡θ=mg+fsin⁡θ=mg+μsNsin⁡θN\cos\theta = mg + f\sin\theta = mg + \mu_sN\sin\theta

⇒N(cos⁡θ−μssin⁡θ)=mg\Rightarrow N(\cos\theta - \mu_s\sin\theta) = mg \quad ...(i)

Along the horizontal (net force = centripetal force):

Nsin⁡θ+fcos⁡θ=mvmax2rN\sin\theta + f\cos\theta = \dfrac{mv_{max}^2}{r}

Nsin⁡θ+μsNcos⁡θ=mvmax2rN\sin\theta + \mu_sN\cos\theta = \dfrac{mv_{max}^2}{r}

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