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

This problem requires determining the maximum safe velocity for a car traversing a banked circular curve, taking into account gravitational force, the normal reaction force, and static friction.

Concept Utilized:

During a car's movement around a banked turn, the required centripetal force is supplied by the horizontal components of both the normal force and the frictional force. For the car to maintain the maximum safe speed (\(v_{\text{max}}\)), it tends to slide upwards along the incline. Consequently, the static friction force (\(f_s\)) acts downwards along the incline.

By achieving equilibrium in the vertical forces and equating the net horizontal force to the centripetal force required (\( \frac{mv^2}{R} \)), the formula for the maximum safe speed is derived:

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

where:

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

Step-by-Step Solution:

Step 1: Enumerate the provided parameters.

• Car's mass, \(m = 800 \, \text{kg}\) (Note: mass is irrelevant to the final calculation as it cancels out).
• Curve radius, \(R = 300 \, \text{m}\).
• Banking angle, \(\theta = 30^\circ\).
• Static friction coefficient, \(\mu_s = 0.2\).
• Gravitational acceleration, \(g = 10 \, \text{m/s}^2\).
• Given value, \(\sqrt{3} = 1.73\).

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

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

Step 3: Insert the given values into the maximum speed formula.

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

Step 4: Execute the numerical computation sequentially.

First, calculate the value of the expression within the parentheses:

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

Now, substitute this result back into the primary equation:

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

Step 5: Calculate the final maximum speed value.

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

Rounding to one decimal place yields 51.4 m/s.

The maximum velocity at which the car can safely navigate the turn is approximately 51.4 m/s.