Question:
A circular race track of radius 240 m is banked at an angle of 45°. If the coefficient of friction between the wheels of a race car and the road is 0.2, the maximum permissible speed to avoid slipping is:
[Acceleration due to gravity = 10 m/s\(^2\)]
A circular race track of radius 240 m is banked at an angle of 45°. If the coefficient of friction between the wheels of a race car and the road is 0.2, the maximum permissible speed to avoid slipping is:
[Acceleration due to gravity = 10 m/s\(^2\)]
[Acceleration due to gravity = 10 m/s\(^2\)]
Show Hint
For a car on a banked track with friction, resolve forces along and perpendicular to the slope and use centripetal force to find maximum/minimum speed. The formula:
\[
v_\text{max} = \sqrt{ r g \frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta} }
\]
is useful for any similar problems.
Updated On: Jun 19, 2026
- 40 m/s
- 60 m/s
- 72 m/s
- 50 m/s
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the problem.
We have a car moving on a banked circular track of radius \(r = 240~\text{m}\), with a banking angle \(\theta = 45^\circ\) and coefficient of friction \(\mu = 0.2\). We are asked to find the maximum speed \(v_\text{max}\) so that the car does not slip.
Step 2: Forces on the car.
Consider the forces acting on the car:
- Normal force \(N\) perpendicular to the surface.
- Frictional force \(f = \mu N\) acting parallel to the surface to prevent slipping.
- Component of gravity along the slope \(mg \sin \theta\).
- Centripetal force required for circular motion \(m v^2 / r\).
Step 3: Resolving forces.
- Horizontal component contributing to centripetal force: \[ F_\text{centripetal} = N \sin\theta + f \cos\theta \] - Maximum friction acts up the slope to help circular motion.
- Vertical component balancing weight: \[ N \cos\theta - f \sin\theta = mg \]
Step 4: Expressing friction in terms of normal force.
\[ f = \mu N \Rightarrow N \cos\theta - \mu N \sin\theta = mg \] \[ N(\cos\theta - \mu \sin\theta) = mg \Rightarrow N = \frac{mg}{\cos\theta - \mu \sin\theta} \]
Step 5: Maximum speed formula.
Centripetal force equation: \[ \frac{m v_\text{max}^2}{r} = N \sin\theta + f \cos\theta = N (\sin\theta + \mu \cos\theta) \] Substitute \(N\): \[ \frac{m v_\text{max}^2}{r} = \frac{mg (\sin\theta + \mu \cos\theta)}{\cos\theta - \mu \sin\theta} \] \[ v_\text{max} = \sqrt{ r g \frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta} } \]
Step 6: Substituting numerical values.
\(\theta = 45^\circ, \mu = 0.2, r = 240~\text{m}, g = 10~\text{m/s}^2\)
\(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707\)
\[ v_\text{max} = \sqrt{ 240 \cdot 10 \cdot \frac{0.707 + 0.2 \cdot 0.707}{0.707 - 0.2 \cdot 0.707} } = \sqrt{ 2400 \cdot \frac{0.707(1 + 0.2)}{0.707(1 - 0.2)} } = \sqrt{ 2400 \cdot \frac{0.8484}{0.5656} } = \sqrt{ 2400 \cdot 1.5 } \] \[ v_\text{max} = \sqrt{3600} = 60~\text{m/s} \]
Step 7: Conclusion.
The maximum permissible speed of the car to avoid slipping is 60 m/s.
We have a car moving on a banked circular track of radius \(r = 240~\text{m}\), with a banking angle \(\theta = 45^\circ\) and coefficient of friction \(\mu = 0.2\). We are asked to find the maximum speed \(v_\text{max}\) so that the car does not slip.
Step 2: Forces on the car.
Consider the forces acting on the car:
- Normal force \(N\) perpendicular to the surface.
- Frictional force \(f = \mu N\) acting parallel to the surface to prevent slipping.
- Component of gravity along the slope \(mg \sin \theta\).
- Centripetal force required for circular motion \(m v^2 / r\).
Step 3: Resolving forces.
- Horizontal component contributing to centripetal force: \[ F_\text{centripetal} = N \sin\theta + f \cos\theta \] - Maximum friction acts up the slope to help circular motion.
- Vertical component balancing weight: \[ N \cos\theta - f \sin\theta = mg \]
Step 4: Expressing friction in terms of normal force.
\[ f = \mu N \Rightarrow N \cos\theta - \mu N \sin\theta = mg \] \[ N(\cos\theta - \mu \sin\theta) = mg \Rightarrow N = \frac{mg}{\cos\theta - \mu \sin\theta} \]
Step 5: Maximum speed formula.
Centripetal force equation: \[ \frac{m v_\text{max}^2}{r} = N \sin\theta + f \cos\theta = N (\sin\theta + \mu \cos\theta) \] Substitute \(N\): \[ \frac{m v_\text{max}^2}{r} = \frac{mg (\sin\theta + \mu \cos\theta)}{\cos\theta - \mu \sin\theta} \] \[ v_\text{max} = \sqrt{ r g \frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta} } \]
Step 6: Substituting numerical values.
\(\theta = 45^\circ, \mu = 0.2, r = 240~\text{m}, g = 10~\text{m/s}^2\)
\(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707\)
\[ v_\text{max} = \sqrt{ 240 \cdot 10 \cdot \frac{0.707 + 0.2 \cdot 0.707}{0.707 - 0.2 \cdot 0.707} } = \sqrt{ 2400 \cdot \frac{0.707(1 + 0.2)}{0.707(1 - 0.2)} } = \sqrt{ 2400 \cdot \frac{0.8484}{0.5656} } = \sqrt{ 2400 \cdot 1.5 } \] \[ v_\text{max} = \sqrt{3600} = 60~\text{m/s} \]
Step 7: Conclusion.
The maximum permissible speed of the car to avoid slipping is 60 m/s.
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