Problem 45 - Master Resource Book in JEE Main Physics

Question

A particle is moving in a circle of radius \(R\) in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at \(t=0\) is \(v_{0}\), the time taken to complete the first revolution is [Kerala CET 2005] (a) \(\frac{R}{v_{0}}\) (b) \(\frac{R}{v_{0}}\left(1-e^{-2 \pi}\right)\) (c) \(\frac{R}{v_{0}} e^{-2 \pi}\) (d) \(\frac{2 \pi R}{v_{0}}\)

Step-by-Step Solution

Answer

Option (b) \(\frac{R}{v_0}(1-e^{-2\pi})\) is correct.

Step 1: Understand the Problem

We have a particle moving in a circle with radius \( R \). The problem states that the normal (centripetal) and tangential components of the acceleration are equal, and we need to find the time taken for the particle to complete the first revolution with an initial speed \( v_0 \).

Step 2: Express Components of Acceleration

The normal (centripetal) acceleration is given by \( a_n = \frac{v^2}{R} \), where \( v \) is the speed of the particle. The tangential acceleration \( a_t \) is given by \( \frac{dv}{dt} \). We know \( a_n = a_t \), so we have \( \frac{v^2}{R} = \frac{dv}{dt} \).

Step 3: Set Up the Differential Equation

From the equation \( \frac{v^2}{R} = \frac{dv}{dt} \), we can rearrange to get \( R dv = v^2 dt \), which is a separable differential equation. We need to integrate both sides to solve it.

Step 4: Integrate Both Sides

Integrate \( R dv = v^2 dt \) from initial time \( t=0 \) to time \( t \), and initial speed \( v = v_0 \) to speed \( v \). The setup becomes \( R \int_{v_0}^{v} dv = \int_{0}^{t} v^2 dt \).

Step 5: Solve the Integrals

The integration of \( R dv \) is straightforward and yields \( R(v - v_0) \). On the right, integrating \( \int v^2 dt \) is tricky because \( v \) is a function of \( t \). Using \( v^2 = R \frac{dv}{dt} \), we get \( \int \, R^2 \frac{1}{v^2} dv = \int dt \). This simplifies as \( -R \left( \frac{1}{v} - \frac{1}{v_0} \right) = t \).

Step 6: Evaluate at One Complete Revolution

One revolution means the particle covers a distance of \( 2\pi R \) which implies the change in angle \( \theta \) is \( 2\pi \). Since \( v = v_0 e^{\theta} \), and \( \theta = 2\pi \), we have \( v = v_0 e^{2\pi} \) after one revolution.

Step 7: Find Time to Complete Revolution

Use the result from our integral \( t = -R \left( \frac{1}{v} - \frac{1}{v_0} \right) \), where \( v = v_0 e^{2\pi} \) for one complete revolution:\[ t = -R \left( \frac{1}{v_0 e^{2 \pi}} - \frac{1}{v_0} \right) \]This simplifies to \[ t = \frac{R}{v_0}(1 - e^{-2\pi}) \].

Step 8: Select Correct Answer from Options

The time taken to complete one revolution, derived as \( \frac{R}{v_0}(1 - e^{-2\pi}) \), matches option (b).

Key Concepts

Centripetal Acceleration

When dealing with circular motion, it's important to understand the concept of centripetal acceleration. This is the acceleration directed towards the center of the circle along which a particle is moving. It allows the particle to follow a curved path rather than a straight line.

In this exercise, the centripetal acceleration is derived from the equation:

• \( a_n = \frac{v^2}{R} \)

Here, \( R \) represents the radius of the circle, and \( v \) is the velocity of the particle. This formula shows that the centripetal acceleration increases with the square of the speed and decreases if the radius of curvature increases.

Practical understanding of centripetal acceleration helps to analyze systems ranging from planets in orbit to cars taking a bend. Without this acceleration, objects would not be able to maintain a circular path.

Tangential Acceleration

Tangential acceleration is what changes the speed of an object along its path in circular motion. It's essential for understanding how a particle speeds up or slows down as it moves along a curve.

The tangential component of acceleration in this problem is:

• \( a_t = \frac{dv}{dt} \)

This measures the rate of change of the velocity of the particle. If we have tangential and centripetal accelerations equal, as in this exercise, it establishes a special relationship between motion parameters. This rarely means that both accelerations explicitly dictate the manner in which the object's speed must change for it to maintain its motion upon a circular path.

Understanding tangential acceleration is critical in fields like automotive engineering, where controlling the speed of a car on a curve is just as crucial as steering. It focuses not just on maintaining a path but on understanding how the speed integrates into the overall motion.

Differential Equation Solving

To solve complex motion problems, like the one provided in this exercise, differential equations become quite handy. They allow us to model and find solutions to situations where variables like velocity change over time.

In the exercise, beginning with the relationship:

• \( \frac{v^2}{R} = \frac{dv}{dt} \)

We arrive at a differential equation \( R dv = v^2 dt \), which is separable.

To solve it, both sides need to be integrated appropriately:

• Integrating the left, \( \int R dv = R(v - v_0) \)
• On the right, due to \( v^2 = R \frac{dv}{dt} \), we focus on the substitution, leading to an equivalence \( \int dt \).

Differential equation solutions here are ingeniously applied to find the time of a complete revolution. This in calculus was shown through setting up the function according to the condition that relates tangential and centripetal accelerations. It underscores why mastering differential equations is necessary for physics problems - they model how systems evolve with time.