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A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is
hard
Laws of Motion
2016
physics
g R ( μ s + tan θ 1 − μ s tan θ ) \sqrt{\frac{g}{R} (\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta})} R g ( 1 − μ s t a n θ μ s + t a n θ )
g R 2 ( μ s + tan θ 1 − μ s tan θ ) \sqrt{\frac{g}{R^2} (\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta})} R 2 g ( 1 − μ s t a n θ μ s + t a n θ )
g R 2 ( μ s + tan θ 1 − μ s tan θ ) \sqrt{gR^2 (\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta})} g R 2 ( 1 − μ s t a n θ μ s + t a n θ )
g R ( μ s + tan θ 1 − μ s tan θ ) \sqrt{gR (\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta})} g R ( 1 − μ s t a n θ μ s + t a n θ )
Explanation To solve this problem, we need to find the maximum safe velocity of a car negotiating a banked curve. \\
Given: \\
• Radius of the curve R R \\
R • Banking angle θ \theta \\
θ • Coefficient of static friction μ s \mu_s \\
μ s
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The forces acting on the car are:
• Gravitational force
acting downwards.
• Normal force
acting perpendicular to the surface.
• Frictional force
acting parallel to the surface.
For a car to move safely on a banked road, the net force towards the center of the curve provides the necessary centripetal force.
The centripetal force
is given by:
F c = m v 2 R \\
F_c = \frac{mv^2}{R} \\
F c = R m v 2 The components of forces along the incline are:
• Normal force component:
N cos θ N \cos\theta \\
N cos θ • Frictional force component:
f s = μ s N f_s = \mu_s N \\
f s = μ s N The components of forces perpendicular to the incline are:
• Normal force component:
N sin θ N \sin\theta \\
N sin θ • Gravitational force component:
m g cos θ mg \cos\theta \\
m g cos θ Balancing the forces perpendicular to the incline:
N cos θ = m g + μ s N sin θ \\
N \cos\theta = mg + \mu_s N \sin\theta \\
N cos θ = m g + μ s N sin θ Solving for
N : N = m g cos θ − μ s sin θ N: \\
N = \frac{mg}{\cos\theta - \mu_s \sin\theta} \\
N : N = c o s θ − μ s s i n θ m g Balancing the forces along the incline for centripetal force:
m v 2 R = N sin θ + μ s N cos θ \\
\frac{mv^2}{R} = N \sin\theta + \mu_s N \cos\theta \\
R m v 2 = N sin θ + μ s N cos θ Substitute
from the previous equation:
m v 2 R = ( m g cos θ − μ s sin θ ) ( sin θ + μ s cos θ ) \\
\frac{mv^2}{R} = \left(\frac{mg}{\cos\theta - \mu_s \sin\theta}\right) (\sin\theta + \mu_s \cos\theta) \\
R m v 2 = ( c o s θ − μ s s i n θ m g ) ( sin θ + μ s cos θ ) Simplify:
v 2 = g R ( μ s + tan θ ) 1 − μ s tan θ \\
v^2 = \frac{gR (\mu_s + \tan\theta)}{1 - \mu_s \tan\theta} \\
v 2 = 1 − μ s t a n θ g R ( μ s + t a n θ ) Therefore, the maximum safe velocity
is:
v = g R ( μ s + tan θ 1 − μ s tan θ ) \\
v = \sqrt{gR \left(\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}\right)} \\
v = g R ( 1 − μ s t a n θ μ s + t a n θ ) This corresponds to Option 4.