Question

A circular path of radius 75 m is banked at an angle of \(\tan^{-1}(0.2)\). If the coefficient of static friction between the tyres of the car and the circular path is 0.1, then the maximum permissible speed of the car to avoid slipping is

• A. 10 ms\(^{-1}\)
• B. 20 ms\(^{-1}\)
• C. 15 ms\(^{-1}\)
• D. 30 ms\(^{-1}\)

Hint:

The formula for maximum speed on a banked curve with friction is \(v_{max} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}}\). The formula for the minimum speed to prevent sliding down is \(v_{min} = \sqrt{rg \frac{\tan\theta - \mu_s}{1 + \mu_s \tan\theta}}\). Note the sign changes.

Answer 1

The Correct Option is C

The formula for the maximum safe speed on a banked road with friction is:
\( v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta} \right)} \).
We are given the following values:
Radius, \(r = 75\) m.
Acceleration due to gravity, \(g \approx 10\) ms\(^{-2}\).
Coefficient of static friction, \( \mu_s = 0.1 \).
The banking angle \(\theta\) is given by \( \theta = \tan^{-1}(0.2) \), so \( \tan\theta = 0.2 \).
Now, substitute these values into the formula.
\( v_{max} = \sqrt{75 \times 10 \left( \frac{0.1 + 0.2}{1 - (0.1)(0.2)} \right)} \).
\( v_{max} = \sqrt{750 \left( \frac{0.3}{1 - 0.02} \right)} \).
\( v_{max} = \sqrt{750 \left( \frac{0.3}{0.98} \right)} \).
\( v_{max} = \sqrt{\frac{750 \times 30}{98}} = \sqrt{\frac{22500}{98}} = \sqrt{\frac{11250}{49}} \).
\( v_{max} = \frac{\sqrt{11250}}{7} \approx \frac{106}{7} \approx 15.15 \) ms\(^{-1}\).
This is approximately 15 ms\(^{-1}\). Let's recompute with the exact fractions.
\( v_{max} = \sqrt{750 \left( \frac{3/10}{98/100} \right)} = \sqrt{750 \cdot \frac{3}{10} \cdot \frac{100}{98}} = \sqrt{75 \cdot 3 \cdot \frac{100}{98}} = \sqrt{225 \cdot \frac{100}{98}} \).
\( v_{max} = 15 \cdot \frac{10}{\sqrt{98}} = \frac{150}{7\sqrt{2}} \). This calculation seems off. Let's re-check the formula. The formula is correct. Let's approximate \(0.98 \approx 1\). Then \(v_{max} \approx \sqrt{750 \times 0.3} = \sqrt{225} = 15\) ms\(^{-1}\). This approximation works well and gives the correct option.