Problem 31 - Master Resource Book in JEE Main Physics

Question

A car is moving on a circular level road of radius of curvature \(300 \mathrm{~m}\). If the coefficient of friction is \(0.3\) and acceleration due to gravity \(10 \mathrm{~ms}^{-2}\), the maximum speed of the car can have is (in \(\mathrm{kmh}^{-1}\) ) (a) 30 (b) 81 (c) 108 [d) 162

Step-by-Step Solution

Answer

The maximum speed of the car is 108 km/h (option c).

Step 1: Identify Forces Involved

On a level circular road, the centripetal force needed to keep the car moving in a circle is provided by the frictional force between the tires and the road. The maximum friction force that can be exerted is given by \( f = \mu N \), where \( \mu \) is the coefficient of friction, and \( N \) is the normal force, which is equal to the weight of the car, \( N = mg \).

Step 2: Set Up the Equation for Maximum Speed

The centripetal force formula is \( f = \frac{mv^2}{r} \), where \( v \) is the speed of the car, \( m \) is the mass, and \( r \) is the radius of curvature. For maximum speed, the frictional force equals the centripetal force: \[ \mu mg = \frac{mv^2}{r} \]

Step 3: Simplify and Solve for Speed

Cancel \( m \) from both sides of the equation to isolate \( v^2 \): \[ \mu g = \frac{v^2}{r} \] Multiply both sides by \( r \) to find: \[ v^2 = \mu gr \] Substitute \( \mu = 0.3 \), \( g = 10 \ \mathrm{ms}^{-2} \), and \( r = 300 \ \mathrm{m} \): \[ v^2 = 0.3 \times 10 \times 300 \] \[ v^2 = 900 \] Take the square root of both sides to find \( v \): \[ v = 30 \ \mathrm{ms}^{-1} \]

Step 4: Convert to Kilometers Per Hour

To convert \( 30 \ \mathrm{ms}^{-1} \) to kilometers per hour, use the conversion factor: \[ 1 \ \mathrm{m/s} = 3.6 \ \mathrm{km/h} \] Calculate \( v \) in \( \mathrm{km/h} \): \[ v = 30 \times 3.6 = 108 \ \mathrm{km/h} \]

Step 5: Select the Correct Answer

Compare the calculated speed with the given options. The calculated maximum speed \( 108 \ \mathrm{km/h} \) matches option (c).

Key Concepts

Centripetal Force

Imagine you're driving a car around a circular track. To keep the car moving in a circle, it needs a special force called centripetal force. This force is directed towards the center of the circle, helping the car continue its circular path without veering off tangent. In this scenario, the tires' grip on the road provides the centripetal force. When we talk about scientific formulas, centripetal force can be expressed as \( f = \frac{mv^2}{r} \), where:

• \( f \) is the centripetal force,
• \( m \) is the mass of the car,
• \( v \) is the speed of the car, and
• \( r \) is the radius of the circle (road).

This equation shows that the required centripetal force increases with higher speed or a smaller radius of curvature. Thus, if the road curve is sharper (smaller radius), or if you're driving faster, a greater centripetal force is necessary to keep a car moving in its circular path.

Coefficient of Friction

Friction is your friend when a car travels around a curve as it prevents the car from slipping off the road. The coefficient of friction is a number that tells us how much frictional force the surface can exert. In our exercise, this was given as \( 0.3 \). A higher coefficient means more grip, while a lower coefficient indicates a slippery surface.

• The maximum frictional force is calculated with \( f = \mu N \), where:
• \( \mu \) is the coefficient of friction, and
• \( N \) is the normal force (equal to the weight of the car \( mg \)).

In circular motion, this frictional force is what provides the necessary centripetal force to keep the car on its track. Thus, the greater the friction, the higher the speed at which a car can round the curve without skidding. For our problem, the coefficient of friction directly influences the maximum speed the car can maintain safely on the circular track.

Radius of Curvature

The radius of curvature is essentially the radius of the circular path that the car is moving along. This distance from the center of the circle (underlying path) to the circle itself determines how "tight" the curve is. Think of it like a flexibility factor for the road:

• A large radius means a broad, gentle curve.
• A small radius implies a sharp turn.

In the given exercise, \( 300 \ m \) is our radius of curvature. This affects how easily your car can take the turn, and it takes place in the centripetal force equation \( \frac{mv^2}{r} \). If the radius is small, even a slight speed increase demands a higher centripetal force. Therefore, during fast driving, wider curves (larger radii) are safer because they demand less friction to provide the necessary centripetal force.

Conversion of Units

Converting units is crucial in physics to ensure clarity and consistency in calculations. In our exercise, maximum speed needs to be in kilometers per hour \( \mathrm{km/h} \), but calculations are initially done in meters per second \( \mathrm{ms}^{-1} \). To convert these units, remember:

• 1 \( \mathrm{m/s} \) is equivalent to 3.6 \( \mathrm{km/h} \).

Why does this conversion work? There are 1000 meters in a kilometer and 3600 seconds in an hour (60 seconds per minute times 60 minutes per hour). Thus, multiplying by 3.6 gives you an easy conversion to \( \mathrm{km/h} \) from \( \mathrm{m/s} \). Hence, in our task, the speed of \( 30 \ \mathrm{ms}^{-1} \) becomes \( 108 \ \mathrm{km/h} \) by multiplying by 3.6. Understanding these conversions helps in everyday scenarios, like speed limits or travel times, ensuring all speeds are understandable in local contexts.