A car is negotiating a curved road of radius R . The road is banked at an angle θ . The coefficient of friction between the tyres of the car and the road is μ _ s . The maximum safe velocity on this road is:

Correct answer: B. gR μ _ s + tan θ 1 - μ _ s tan θ N = mgcos θ + m v ^ 2 R sin θ f _ = ⇒ f _ = μ _ s mgcos θ + μ _ s m v ^ 2 R sin θ mg sinθ + f _ = m v ^ 2 R cosθ mgsin θ + μ _ s mgcos θ + μ _ s m v ^ 2 R sin θ = m v ^ 2 R cos θ gsin θ + μ _ s gcos θ = v ^ 2 R (cos θ - μ _ s sin θ) gR[ tan θ + μ _ s 1 - μ _ s tan θ ] = v ^ 2

NEET 2016

Exam: NEET
Year: 2016
Subject: Physics
Chapter: Laws of Motion

NEET 2016 · Physics · Laws of Motion

A car is negotiating a curved road of radius R . The road is banked...

A car is negotiating a curved road of radius

\text{R}

. The road is banked at an angle

\theta

. The coefficient of friction between the tyres of the car and the road is

{\mu}_{s}

. The maximum safe velocity on this road is: A car is negotiating a curved road of radius R . The road is banked at an angle θ . The coefficient of friction between

A car is negotiating a curved road of radius R . The road is banked at an angle θ . The coefficient of friction between

Options

A $\sqrt{g{R}^{2}\dfrac{{\mu}_{s} + tan \theta}{1 - {\mu}_{s}tan \theta}}$
B $\sqrt{gR\dfrac{{\mu}_{s} + tan \theta}{1 - {\mu}_{s}tan \theta}}$
C $\sqrt{\dfrac{g}{R}\dfrac{{\mu}_{s} + tan \theta}{1 - {\mu}_{s}tan \theta}}$
D $\sqrt{\dfrac{g}{{R}^{2}}\dfrac{{\mu}_{s} + tan\theta}{1 - {\mu}_{s}tan \theta}}$

Correct answer

B. $\sqrt{gR\dfrac{{\mu}_{s} + tan \theta}{1 - {\mu}_{s}tan \theta}}$

Step-by-step solution

N = mgcos \theta + \dfrac{m{v}^{2}}{R}sin \theta

{f}_\max = \muN ⇒

{f}_\max = {\mu}_{s}mgcos \theta + \dfrac{{\mu}_{s}m{v}^{2}}{R}sin \theta

mg sin\theta + {f}_\max = \dfrac{m{v}^{2}}{R} cos\theta

mgsin \theta + {\mu}_{s}mgcos \theta + \dfrac{{\mu}_{s}m{v}^{2}}{R}sin \theta = \dfrac{m{v}^{2}}{R}cos \theta

gsin \theta + {\mu}_{s}gcos \theta = \dfrac{{v}^{2}}{R}\left(cos \theta - {\mu}_{s}sin \theta\right)

gR\left[\dfrac{tan \theta + {\mu}_{s}}{1 - {\mu}_{s}tan \theta}\right] = {v}^{2}

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