A car is negotiating a curved road of radius $R$. The road is banked at an angle $ heta$. The coefficient of

Question

A car is negotiating a curved road of radius $R$. The road is banked at an angle $	heta$. The coefficient of friction between the tyre of the car and the road is $\mu_s$. The maximum safe velocity on this road is:

Accepted Answer

1. **Forces Analysis:** Consider the car as a particle. The forces acting on it are: weight $mg$ (downwards), normal reaction $N$ (perpendicular to the road), and static friction $f$ (parallel to the road). For maximum speed ($v_{max}$), friction acts down the slope to prevent sliding up.
2. **Equations of Motion:** 
   - Vertical equilibrium: $N \cos 	heta = mg + f \sin 	heta$. Since $f = \mu_s N$ (limiting case), $N \cos 	heta - \mu_s N \sin 	heta = mg$.
   - Horizontal centripetal force: $N \sin 	heta + f \cos 	heta = \frac{mv^2}{R}$. Substituting $f = \mu_s N$, $N \sin 	heta + \mu_s N \cos 	heta = \frac{mv_{max}^2}{R}$.
3. **Derivation:** Divide the horizontal equation by the vertical equation:
   $$ \frac{\frac{mv_{max}^2}{R}}{mg} = \frac{N(\sin 	heta + \mu_s \cos 	heta)}{N(\cos 	heta - \mu_s \sin 	heta)} $$
   $$ \frac{v_{max}^2}{Rg} = \frac{\sin 	heta + \mu_s \cos 	heta}{\cos 	heta - \mu_s \sin 	heta} $$
   Divide the numerator and denominator of the RHS by $\cos 	heta$:
   $$ \frac{v_{max}^2}{Rg} = \frac{	an 	heta + \mu_s}{1 - \mu_s 	an 	heta} $$
   $$ v_{max} = \sqrt{Rg \left( \frac{\mu_s + 	an 	heta}{1 - \mu_s 	an 	heta} ight)} $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.10, Equation 4.21).

Step-by-Step Solution

Practice All PHYSICS Questions