For a curved track of radius R, banked at angle θ (Take v0 = $\sqrt{Rg\tan\theta}$ )

For No friction & safely turn$\frac{mv_{0}^{2}}{R}\cos\theta = mg\sin\theta$$\boxed{v_{0} = \sqrt{Rg\tan\theta}}$Vmin < V0 < VmaxIn Vmin & Vmax must be come into play for safely turn maximum value of angle of banking$\boxed{\theta = \tan^{- 1}\mu_{0}}$

Circular MotionHardBloom L3

Question

For a curved track of radius R, banked at angle θ (Take v₀ = $\sqrt{Rg\tan\theta}$ )

Options

A.a vehicle moving with a speed v₀ is able to negotiate the curve without calling friction into play at all

B.a vehicle moving with any speed V > V₀ is always able to negotiate the curve, with friction called into play

C.a vehicle moving with any speed V < V₀ must have the force of friction into play

D.the minimum value of the angle of banking for a vehicle parked on the banked road can stay there without slipping, is given by θ = tan^–1 µ₀ (µ₀ = coefficient of static friction)

Solution

For No friction & safely turn

$\frac{mv_{0}^{2}}{R}\cos\theta = mg\sin\theta$

$\boxed{v_{0} = \sqrt{Rg\tan\theta}}$

V_min < V₀ < V_max

In V_min & V_max must be come into play for safely turn maximum value of angle of banking

$\boxed{\theta = \tan^{- 1}\mu_{0}}$

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