The Question is from neet-physics laws-of-motion , and it was created by the author VENKATESWARARAO

Question

A car is moving with uniform speed on a banked road inclined at an angle $	heta$. The coefficient of static friction between the tyres and road is $\mu_s$. What is the maximum speed with which the car can negotiate the curve without skidding, if the radius of the curve is $R$?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Circular Motion on a Banked Road with Friction
	**Related Topics**: Centripetal Force, Static Friction, Normal Force, Resolution of Forces
	**Prerequisites**: Understanding of forces, circular motion, trigonometry
## SOLUTION PATHWAY
**Given Information**:
	* Angle of banking: $	heta$
	* Coefficient of static friction: $\mu_s$
	* Radius of the curve: $R$
	* Uniform speed
**Method & Steps**:
	1. Draw a free body diagram of the car on the banked road. The forces acting are:
		* Weight ($mg$) vertically downwards
		* Normal force ($N$) perpendicular to the road surface
		* Frictional force ($f$) along the road surface (upwards to prevent skidding)
		Reasoning: We must analyze the forces acting on the car to determine its maximum speed without skidding.
	2. Resolve the normal force ($N$) and frictional force ($f$) into horizontal and vertical components.
		* Vertical components: $N\cos	heta$ upwards, $f\sin	heta$ downwards
		* Horizontal components: $N\sin	heta$ towards the center, $f\cos	heta$ towards the center
	3.  Apply Newton's second law. In the vertical direction, the net force is zero since the car has no vertical acceleration:
		$N\cos	heta = mg + f\sin	heta$  (Equation 1)
	4. In the horizontal direction, the net force provides the centripetal force:
		$N\sin	heta + f\cos	heta = \frac{mv^2}{R}$  (Equation 2)
	5. The limiting frictional force is $f = \mu_s N$. Substitute this into equations 1 and 2:
		$N\cos	heta = mg + \mu_s N\sin	heta$
		$N\sin	heta + \mu_s N\cos	heta = \frac{mv^2}{R}$
	6. From the first equation, express $N$ in terms of $m$, $g$, $	heta$ and $\mu_s$:
		$N(\cos	heta - \mu_s\sin	heta) = mg$
		$N = \frac{mg}{\cos	heta - \mu_s\sin	heta}$ (Equation 3)
	7. Substitute the expression for $N$ from Equation 3 into the second equation:
		$\frac{mg}{\cos	heta - \mu_s\sin	heta} (\sin	heta + \mu_s\cos	heta) = \frac{mv^2}{R}$
	8. Simplify and solve for $v^2$:
		$v^2 = Rg\frac{\sin	heta + \mu_s\cos	heta}{\cos	heta - \mu_s\sin	heta}$
	9. Take the square root to find the maximum speed $v_{max}$:
		$v_{max} = \sqrt{Rg\frac{\sin	heta + \mu_s\cos	heta}{\cos	heta - \mu_s\sin	heta}}$
		Result: $v_{max} = \sqrt{Rg\frac{\sin	heta + \mu_s\cos	heta}{\cos	heta - \mu_s\sin	heta}}$
## OPTION ANALYSIS
[A] :x: This option is the speed without friction, which is incorrect here.
[B] :x: This option represents the speed on a flat road with friction, which is not the case here.
[C] ✓: This is the correct expression derived above for the maximum speed on a banked road with friction.
[D] :x: This option is incorrect due to wrong signs in the numerator and denominator.
## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly resolving the normal and frictional forces into components. [How to avoid]: Draw a clear free-body diagram and use trigonometry carefully.
		2.[Error 2]: Forgetting that the friction can act either up or down depending on the speed. Here, it's acting upwards to prevent skidding, and we use the limiting static friction. [How to avoid]: Understand the direction of static friction based on the scenario.
	**Quick Tips**:
		1.[Helpful shortcut]: When $\mu_s = 0$, the formula reduces to $v_{max} = \sqrt{Rg	an	heta}$, which is the case of no friction.
		2.[Memory aid]: Remember that the numerator contains sine with + sign and denominator contains cosine with - sign with friction terms.

VENKATESWARARAO · Answer

$v_{max} = \sqrt{Rg	an	heta}$

VENKATESWARARAO · Answer

$v_{max} = \sqrt{\mu_s Rg}$

VENKATESWARARAO · Answer

The correct Option is "$v_{max} = \sqrt{Rg\frac{\sin	heta + \mu_s\cos	heta}{\cos	heta - \mu_s\sin	heta}}$"

VENKATESWARARAO · Answer

$v_{max} = \sqrt{Rg\frac{\cos	heta + \mu_s\sin	heta}{\sin	heta - \mu_s\cos	heta}}$

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