A car of mass 'm' moves on a banked road having radius 'r' and banking angle . To avoid slipping from banked road, the maximum permissible speed of the car is . The coefficient of friction between the wheels of the car and the banked road is [20

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Solution

Q.
A car of mass 'm' moves on a banked road having radius 'r' and banking angle $θ$ . To avoid slipping from banked road, the maximum permissible speed of the car is $v_{0}$ . The coefficient of friction $μ$ between the wheels of the car and the banked road is [2025]

1 $μ = \frac{v_{0}^{2} + r g \tan θ}{r g - v_{0}^{2} \tan θ}$

2 $μ = \frac{v_{0}^{2} + r g \tan θ}{r g + v_{0}^{2} \tan θ}$

3 $μ = \frac{v_{0}^{2} - r g \tan θ}{r g + v_{0}^{2} \tan θ}$

4 $μ = \frac{v_{0}^{2} - r g \tan θ}{r g - v_{0}^{2} \tan θ}$

Ans.
(3)

Given maximum possible speed $= v_{0}$

For this

So, $N = m g \cos θ + \frac{m v_{0}^{2}}{r} \sin θ$

$f_{r} = μ N = μ (m g \cos θ + \frac{m v_{0}^{2}}{r} \sin θ)$

And $\frac{m v_{0}^{2}}{r} \cos θ = m g \sin θ + f_{r}$

$\Rightarrow \frac{m v_{0}^{2}}{r} \cos θ - m g \sin θ = μ (m g \cos θ + \frac{m v_{0}^{2}}{r} \sin θ)$

$\Rightarrow (v_{0}^{2} - g r \tan θ) = μ (v_{0}^{2} \tan θ + g r)$

$\Rightarrow μ = \frac{v_{0}^{2} - g r \tan θ}{g r + v_{0}^{2} \tan θ}$

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