Problem 9 - Master Resource Book in JEE Main Physics

Question

A particle moves along a circle of radius \(\left(\frac{20}{\pi}\right) \mathrm{m}\) with constant tangential acceleration. If the velocity of the particle is \(80 \mathrm{~ms}^{-1}\), at the end of seconds revolution after motion has begun, the tangential acceleration is (a) \(40 \mathrm{~ms}^{-2}\) (b) \(640 \pi \mathrm{ms}^{-2}\) (c) \(1609 \pi \mathrm{ms}^{-2}\) (d) \(40 \pi \mathrm{ms}^{-2}\)

Step-by-Step Solution

Answer

The tangential acceleration is \( 40 \text{ m/s}^2 \), so option (a) is correct.

Step 1: Understand the Problem

The problem asks for the tangential acceleration of a particle moving in a circle. We know the particle completes its second revolution and reaches a velocity of 80 m/s. We need to calculate the tangential acceleration that allows the particle to achieve this velocity after two complete revolutions.

Step 2: Determine the Distance Covered

Determine the distance traveled by the particle during the two revolutions. The circumference of the circle is \( C = 2\pi r \), where \( r = \frac{20}{ ext{π}} \) meters. So, \( C = 2 \pi \cdot \frac{20}{ ext{π}} = 40 \text{ meters} \). For two revolutions, the total distance traveled is \( 2 \times 40 = 80 \text{ meters} \).

Step 3: Use the Kinematic Equation

The kinematic equation relating initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), and distance (\(s\)) is: \[ v^2 = u^2 + 2as \]Given \( v = 80 \text{ m/s} \), \( u = 0 \text{ m/s} \) (starting from rest), and \( s = 80 \text{ meters} \), substitute the values into the equation to solve for \( a \).

Step 4: Substitute and Solve for Acceleration

Substitute the values into the kinematic equation:\[ 80^2 = 0^2 + 2a \times 80 \]This simplifies to:\[ 6400 = 160a \]Solve for \( a \):\[ a = \frac{6400}{160} = 40 \text{ m/s}^2 \]

Step 5: Validate and Choose the Correct Option

From the given options, (a) is the correct answer since we calculated the tangential acceleration as \( 40 \text{ m/s}^2 \), which matches option (a).

Key Concepts

Circular Motion

Circular motion refers to the movement of a particle or object along the circumference of a circle. In circular motion, the object is constantly changing direction, making velocity a vector since it has both magnitude and direction. The principles of circular motion are crucial in understanding the behavior of objects in orbits, wheels, and other circular paths.

In this problem, a particle is moving with a constant tangential acceleration along a circle with a specific radius. The concept of tangential acceleration is vital here, as it refers to the change in the speed of the particle along the tangent to the circle. Unlike radial or centripetal acceleration, which deals with changes in direction, tangential acceleration affects the speed. This makes it essential for calculating how fast an object can go in circular paths.

• Circular Path: Defined by the circle's circumference, calculated as \(C = 2\pi r\).
• Radius: The distance from the center of the circle to any point on its circumference, here given as \(\frac{20}{\pi}\).
• Revolutions: The number of times the particle completes a full circle; in this case, two revolutions.

Kinematic Equations

Kinematic equations are tools that help us solve problems related to motion. They link various physical quantities such as velocity, acceleration, distance, and time. These equations are crucial in understanding how an object accelerates over time.

In our exercise, the kinematic equation \(v^2 = u^2 + 2as\) is used. This equation shows the relationship between the final velocity \(v\), initial velocity \(u\), acceleration \(a\), and the distance traveled \(s\). The given problem provides these variables: the particle starts from rest (* \(u = 0\)*), covers a distance of 80 meters after two revolutions, and achieves a final velocity of 80 m/s.

• Initial Velocity \(u\): The velocity of the particle at the start; here it is zero, indicating motion begins from rest.
• Final Velocity \(v\): The velocity after two revolutions, given as 80 m/s.
• Acceleration \(a\): What we solve for using the kinematic equation.
• Distance \(s\): Total path length traveled during the two revolutions, calculated as 80 meters.

Velocity in Circular Motion

Velocity in circular motion is intriguing because it involves both speed and direction changes. While speed might remain constant, the direction of motion constantly changes as the object moves along the circular path. This creates a unique aspect where velocity in circular motion is not just a linear quantity.

In the given problem, the velocity of 80 m/s is attained at the end of two revolutions. This indicates how quickly the particle can travel around the circle. Achieving a final velocity requires calculating the tangential acceleration that facilitated this motion.

• Constant Speed: In cases without tangential acceleration, speed might remain constant; however, the given problem involves a change in speed due to acceleration.
• Vector Quantity: As a vector, velocity has both magnitude (speed) and direction, essential for understanding motion in a circular path.
• Tangential Acceleration: Directly impacts the change in speed as observed with the particle reaching 80 m/s after two complete rounds.

Understanding velocity in circular motion helps us grasp how particles and objects behave when constrained by paths that require continuous direction change, such as planets orbiting stars or electrons moving in magnetic fields.