Question

drive an expression for velocity of car on a banked circular road having coefficient of friction u has write the for optimum velocity​

Answer 1

Banking OF Roads:

In Banking of roads the outer edge is raised above the inner edge of the curved road. Angle to which the outer edge is raised above the inner edge of the curved path is called banking of the curved path.

Due to large amount of friction when the vehicles move on the road, it leads to wear and tear of tyres. To avoid this, the curved road is given an inclination slopping upwards towards the outer circumference.

FREE BODY DIAGRAM OF A MOVING CAR ON A BANKED ROAD (in attachment)

Forces acting on the vehicle:

1) Weight mg acting vertically downwards.

2) Normal reaction R of the road acting at an angle $\theta$ with the vertical.

3) Force of friction f acting downwards along the inclined plane.

Balancing the forces in horizontal and vertical direction.

$\sf R \sin \theta + f \cos \theta = \frac{m {v}^{2} }{r} .....(1)$

$\sf R \cos \theta - f \sin \theta = mg ......(2)$

✒ Dividing equation (1) by equation (2)

$\to\sf \frac{R \sin \theta + f \cos \theta}{R \cos \theta - f \sin \theta} = \frac{m {v}^{2} }{rmg}$

$\to\sf \frac{R \sin \theta + f \cos \theta}{R \cos \theta - f \sin \theta} = \frac{{v}^{2} }{rg}$

✒ Dividing the numerator and denominator with of L.H.S R $\cos \theta$

$\to\sf \frac{\tan \theta + \frac{f}{R} }{1 - \frac{f}{R} \tan \theta} = \frac{{v}^{2} }{rg}$

$\to\sf \frac{\tan \theta + \mu }{1 - \mu \tan \theta} = \frac{{v}^{2} }{rg}$

$\to\sf {v}^{2} = rg[ \frac{ \mu + \tan \theta }{1 - \mu \tan \theta } ]$

$\to\sf v= \sqrt{rg[ \frac{ \mu + \tan \theta }{1 - \mu \tan \theta } ]}$

This expression for the maximum speed with which a vehicle can safely negotiate a curved road banked at an angle $\theta$. The coefficient of friction between the wheels and road is μ.