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Step-by-Step Solution
Step 1: Identify the Forces Acting on the Car
When a car moves on a banked road of radius $R$ and banking angle $\theta$, the forces acting on it are:
Gravitational force $mg$, acting vertically downward.
Normal reaction $N$ from the road, acting perpendicular to the surface of the bank.
Frictional force $f$, which acts parallel to the surface of the bank and opposes relative motion (or possible slipping).
Step 2: Resolve Forces into Components
To maintain circular motion, the net force towards the center of the curve provides the centripetal force. Break down the forces into two directions:
Along the slope (towards the center of the circular path): Components of $N$ and friction must balance or provide centripetal force.
Perpendicular to the slope: Components of $N$, friction, and $mg$ must balance each other in the vertical/perpendicular direction.
Step 3: Express the Conditions for Maximum Safe Velocity
At maximum safe velocity, friction acts in such a way that it just prevents slipping outward. The expression derived from the force balance and centripetal force requirement for maximum speed on a banked track (with friction) leads to:
$ \frac{V_{\max}^2}{R\,g} \;=\; \frac{\mu_s + \tan\theta}{1 \;-\; \mu_s\,\tan\theta}.$
Step 4: Solve for $V_{\max}$
Rearranging for $V_{\max}$, we get:
$ V_{\max} \;=\; \sqrt{\, R\,g \,\Bigl[\frac{\mu_s + \tan\theta}{1 - \mu_s \,\tan\theta}\Bigr] }\,. $
Final Answer
$ \displaystyle V_{\max} \;=\; \sqrt{\, g\,R \,\frac{\mu_s + \tan\theta}{1 - \mu_s\,\tan\theta}\;}. $
