Problem 33 - Master Resource Book in JEE Main Physics

Question

A car of mass \(2000 \mathrm{~kg}\) is moving with a speed of \(10 \mathrm{~ms}^{-1}\) on a circular path of radius \(20 \mathrm{~m}\) on a level road. What must be the frictional force between the car and the road so that the car does not slip? (a) \(10^{4} \mathrm{~N}\) (b) \(10^{3} \mathrm{~N}\) (c) \(10^{5} \mathrm{~N}\) (d) \(10^{2} \mathrm{~N}\)

Step-by-Step Solution

Answer

The correct frictional force is (a) \( 10^4 \text{ N} \).

Step 1: Understand the Problem

We are given a car of mass 2000 kg moving in a circle of radius 20 m at a speed of 10 m/s. We need to determine the frictional force that prevents the car from slipping.

Step 2: Formula for Centripetal Force

The force required to keep an object moving in a circle is the centripetal force, which is given by the formula \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is radius of the circle.

Step 3: Calculate the Centripetal Force

Substitute the given values into the formula: \( m = 2000 \text{ kg} \), \( v = 10 \text{ m/s} \), and \( r = 20 \text{ m} \). Calculate: \[ F_c = \frac{2000 \times 10^2}{20} = \frac{2000 \times 100}{20} = 10000 \text{ N} \].

Step 4: Determine the Frictional Force

Since the only force keeping the car from slipping is the frictional force, and we have calculated the centripetal force needed, the frictional force must be equal to the centripetal force: \( F_f = 10000 \text{ N} \).

Step 5: Select the Correct Answer

The frictional force required is \( 10000 \text{ N} \), which matches option (a) \( 10^4 \text{ N} \).

Key Concepts

Centripetal Force

When you think of an object's circular motion, centripetal force is a key concept. It's the force that keeps an object moving along a circular path by constantly pulling it toward the center of the circle. Without this force, the object would fly away in a straight line due to inertia.
In mathematics, we calculate centripetal force using the formula: \[ F_c = \frac{mv^2}{r} \]Where:

• \( F_c \) is the centripetal force,
• \( m \) is the mass of the object,
• \( v \) is the velocity (speed in a given direction), and
• \( r \) is the radius of the circle the object is moving along.

The centripetal force is essential because it helps to maintain the circular path of the object. In the case of the car moving along a circular path, this force is crucial in ensuring it doesn't skid off the track.

Frictional Force

Frictional force acts in the direction opposite to the movement of a surface sliding over another. It is essential for motion control and stability. In the example of the car on a circular road, friction provides the grip needed between the tires and the road surface. This frictional force is the actual force that provides the centripetal force needed to maintain the circular path.

The amount of frictional force available depends on several factors:

• The weight of the car, which affects how much it is pressing down on the road,
• The surface roughness of both the road and tire tread, and
• The speed at which the car is moving.

For the car not to slip, the frictional force must be equal to or greater than the required centripetal force. This is why calculating the correct amount of frictional force is crucial for safe driving around curves.

Newton's Laws of Motion

Newton's laws of motion provide a framework for understanding the principles of movement and the forces involved. They explain why objects move the way they do in different scenarios, including circular motion.

In this context:

• Newton's First Law (the law of inertia) states that an object will remain stationary or in uniform motion unless acted upon by an external force. This means a car in motion won’t change its direction along a curve unless a force, like friction, acts on it.
• Newton's Second Law states that the force acting on an object is equal to the mass of that object times its acceleration (\( F = ma \)). For circular motion, this force is the centripetal force that is required to change the object’s direction without changing its speed.
• Newton's Third Law states that for every action, there is an equal and opposite reaction. The road pushes back on the car tires as they push against the road, enabling the car to maneuver through curves without slipping.

Using these principles, we understand how forces interact to govern an object’s motion both in linear and circular paths.