Question

A car of 800 kg is taking a turn on a banked road of radius 300 m and angle of banking 30°. If the coefficient of static friction is 0.2, then the maximum speed with which the car can negotiate the turn safely: (Given $g = 10 \, \text{m/s}^2$, $\sqrt{3} = 1.73$).

• A. 70.4 m/s
• B. 51.4 m/s
• C. 264 m/s
• D. 102.8 m/s

Answer 1

The Correct Option is B

The problem asks for the maximum safe speed with which a car can navigate a banked circular turn, considering the forces of gravity, normal reaction, and static friction.

Concept Used:

When a car is taking a turn on a banked road, the necessary centripetal force is provided by the horizontal components of the normal force and the frictional force. For the car to move at the maximum safe speed (\(v_{\text{max}}\)), it has a tendency to slide up the incline. Therefore, the force of static friction (\(f_s\)) acts down the incline.

By balancing the forces in the vertical direction and equating the net horizontal force to the required centripetal force (\( \frac{mv^2}{R} \)), we can derive the formula for the maximum safe speed:

\[ v_{\text{max}} = \sqrt{gR \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)} \]

where:

• \(g\) is the acceleration due to gravity.
• \(R\) is the radius of the turn.
• \(\mu_s\) is the coefficient of static friction.
• \(\theta\) is the angle of banking.

Step-by-Step Solution:

Step 1: List the given parameters.

• Mass of the car, \(m = 800 \, \text{kg}\) (Note: the mass is not needed for the final calculation as it cancels out).
• Radius of the turn, \(R = 300 \, \text{m}\).
• Angle of banking, \(\theta = 30^\circ\).
• Coefficient of static friction, \(\mu_s = 0.2\).
• Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\).
• Given value, \(\sqrt{3} = 1.73\).

Step 2: Calculate the value of \(\tan\theta\).

\[ \tan \theta = \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{1}{1.73} \approx 0.578 \]

Step 3: Substitute the given values into the formula for maximum speed.

\[ v_{\text{max}} = \sqrt{10 \times 300 \left( \frac{0.2 + 0.578}{1 - 0.2 \times 0.578} \right)} \]

Step 4: Perform the numerical calculation step-by-step.

First, calculate the value of the expression in the parenthesis:

\[ \text{Numerator} = 0.2 + 0.578 = 0.778 \] \[ \text{Denominator} = 1 - (0.2 \times 0.578) = 1 - 0.1156 = 0.8844 \] \[ \frac{\text{Numerator}}{\text{Denominator}} = \frac{0.778}{0.8844} \approx 0.8797 \]

Now, substitute this back into the main equation:

\[ v_{\text{max}} = \sqrt{3000 \times 0.8797} \] \[ v_{\text{max}} = \sqrt{2639.1} \]

Step 5: Calculate the final value of the maximum speed.

\[ v_{\text{max}} \approx 51.37 \, \text{m/s} \]

Rounding to one decimal place, we get 51.4 m/s.

The maximum speed with which the car can negotiate the turn safely is approximately 51.4 m/s.