Problem 30 - Master Resource Book in JEE Main Physics

Question

A body moves along a circular path of radius \(5 \mathrm{~m}\). The coefficient of friction between the surface of path and the body is \(0.5\). The angular velocity, in \(\mathrm{rad} / \mathrm{s}\), with which the body should move so that it does not leave the path is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) 4 (b) 3 (c) 2 (d) 1

Step-by-Step Solution

Answer

The angular velocity is 1 rad/s (option d).

Step 1: Understand the Problem

A body is moving along a circular path with a given radius, and we need to find the maximum angular velocity it can have without slipping. This is determined by the frictional force between the body and the path.

Step 2: Identify the Forces

The necessary centripetal force to keep the body moving in a circle is provided by the frictional force. Thus, the maximum frictional force is equal to the centripetal force needed to keep the body on its path.

Step 3: Use the Formula for Friction Force

The maximum frictional force is given by\[ F_{ ext{friction}} = ext{Coefficient of friction} \times ext{Normal force} = ext{Coefficient of friction} \times mg \]Here, the mass is not given, so it will cancel out later.

Step 4: Use the Formula for Centripetal Force

The centripetal force required for circular motion is given by the equation\[ F_{ ext{centripetal}} = m \cdot R \cdot \omega^2 \]where \( R \) is the radius of the circle and \( \omega \) is the angular velocity we need to find.

Step 5: Equate Frictional and Centripetal Forces

Set the maximum frictional force equal to the centripetal force:\[ ext{Coefficient of friction} \times mg = m \cdot R \cdot \omega^2 \]The mass \( m \) cancels out from both sides, simplifying to:\[ ext{Coefficient of friction} \times g = R \cdot \omega^2 \]

Step 6: Substitute Known Values and Solve for Angular Velocity

Insert the known values: the coefficient of friction is 0.5, acceleration due to gravity \( g \) is 10 m/s², and radius \( R \) is 5 m:\[ 0.5 \times 10 = 5 \cdot \omega^2 \]This simplifies to:\[ 5 = 5 \cdot \omega^2 \]\[ \omega^2 = 1 \]Solve for \( \omega \):\[ \omega = \sqrt{1} = 1 \]

Step 7: Verify the Solution

The calculation shows that the angular velocity \( \omega \) should be 1 rad/s to prevent slipping.

Key Concepts

Circular Motion

When an object travels in a path that forms a circle, it's known as circular motion. This can be seen in the movement of a car around a roundabout or the Earth's orbit around the sun. Circular motion is characterized by a constant change in direction. This implies an acceleration, even if the speed is constant.

• The motion requires a force directed towards the center of the circle, known as centripetal force.
• Speed around the circle is often referred to in terms of angular speed or angular velocity.
• Understanding circular motion requires analyzing the interplay between different forces, such as friction and centripetal force.

A clear understanding of circular motion is crucial, especially in problems involving angular velocity and the forces that keep an object on its circular path.

Frictional Force

The frictional force acts between two surfaces in contact, resisting their relative motion. In circular motion, friction plays a crucial role in preventing an object from slipping or skidding off its path.

• This force depends on the coefficient of friction and the normal force exerted by an object on the surface.
• In our case, it's vital for providing the necessary centripetal force that keeps the object in circular motion.
• No movement can occur without sufficient frictional force to counteract the other forces, like gravity.

The frictional force is often the unsung hero, quietly providing the grip needed to maintain rotation without making objects slide away.

Centripetal Force

Centripetal force is the force required to make an object follow a curved path. It's the backbone of circular motion, always pointing inward toward the circle's center to maintain rotation.

The Role of Centripetal Force

• Acts perpendicular to the motion of the object along the curve.
• Necessary to counteract the object's natural tendency to travel in a straight line.
• In our example, it's provided by the frictional force.

Without sufficient centripetal force, an object would cease to follow a circular path. It's key to understanding why objects stay on their paths in circular motion scenarios.

Coefficient of Friction

The coefficient of friction is a dimensionless scalar value that reflects how rough or smooth a surface is. It helps determine how much frictional force can be developed between two surfaces.

• It multiplies with the normal force to give the frictional force.
• A higher coefficient indicates more friction; a lower one indicates less.
• In calculations, it allows us to relate the frictional force with forces acting on different axes, such as normal force and gravity.

In our specific exercise, a coefficient of 0.5 gives us a precise measure of how much frictional force is available to keep an object moving without slipping.

Angular Speed Calculation

Angular speed is a measure of how quickly an object rotates or revolves around a point or axis. In simpler terms, it's how fast something spins. It's an essential part of understanding circular motion.

Calculating Angular Speed

• The exercise involves determining the maximum angular speed using known values.
• Equating the expressions for frictional and centripetal forces enables the calculation of angular speed.
• The formula derived, \( \frac{ \text{coefficient of friction} \times g }{ R } = \omega^2\), helps in solving for \( \omega \).

Understanding how to calculate angular speed allows us to find how fast an object needs to go to stay on its circular path without slipping away.