Problem 19 - Master Resource Book in JEE Main Physics

Question

A straight rod of length \(L\) extends from \(x=a\) to \(x=L+a\). Find the gravitational force it, exerts on a point mass \(m\) at \(x=0\) if the linear density of rod \(\mu=A+B x^{2}\) (a) \(\mathrm{Gm}\left[\frac{A}{\mathrm{a}}+\mathrm{BL}\right]\) (b) \(G m\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right]\) (c) \(G m\left[B L+\frac{A}{a+L}\right]\) (d) \(G m\left[B L-\frac{A}{a}\right]\)

Step-by-Step Solution

Answer

(b) \( G m\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right] \)

Step 1: Identify the Problem

We need to find the gravitational force exerted by a rod with variable linear density on a point mass. The rod has a length \( L \) and extends from \( x = a \) to \( x = L + a \). The linear density of the rod is \( \mu = A + Bx^2 \) where \( A \) and \( B \) are constants. The point mass \( m \) is located at \( x = 0 \).

Step 2: Understand Gravitational Differential Force

The gravitational force between a differential mass element \( dm \) of the rod and the point mass \( m \) is determined by Newton's law of universal gravitation: \( dF = \frac{G m \, dm}{r^2} \) where \( G \) is the gravitational constant, and \( r \) is the distance between the point mass at \( x = 0 \) and the differential element at position \( x \). Since \( r = x \) in this case, \( dF = \frac{G m \, dm}{x^2} \).

Step 3: Express Differential Mass Element

The differential mass element \( dm \) can be expressed in terms of the linear density: \( dm = \mu \, dx = (A + Bx^2) \, dx \).

Step 4: Set Up the Integral for Total Force

The total gravitational force \( F \) can be obtained by integrating the differential force over the length of the rod from \( x = a \) to \( x = L + a \):\[F = \int_{a}^{L+a} \frac{G m (A + Bx^2)}{x^2} \, dx\]

Step 5: Evaluate the Integral

Split the integral into two parts:\[F = G m \left( \int_{a}^{L+a} \frac{A}{x^2} \, dx + \int_{a}^{L+a} Bx^0 \, dx \right)\]1. For the first integral: \[ \int_{a}^{L+a} \frac{A}{x^2} \, dx = -\frac{A}{x} \Bigg|_a^{L+a} = \frac{A}{a} - \frac{A}{L+a} \]2. For the second integral: \[ B \times \int_{a}^{L+a} 1 \, dx = B \left[ x \right]_a^{L+a} = B (L) \]

Step 6: Combine Terms to Find Total Gravitational Force

Adding the results of the two integrals, we have:\[F = G m \left( \frac{A}{a} - \frac{A}{L+a} + BL \right)\]This simplifies to:\[F = G m \left[ A \left(\frac{1}{a} - \frac{1}{L+a}\right) + BL \right]\]Which matches option \( (b) \).

Key Concepts

Linear Density

Linear density is a measure of how mass is distributed along a one-dimensional object, like a rod. It tells us how much mass is contained in each unit of length. In equations, linear density is often represented by the Greek letter \( \mu \), and it can vary along the length of the rod.
In this exercise, the linear density of the rod is given as a function \( \mu = A + Bx^2 \). This expression means that:

• \( A \) represents the constant part of the linear density.
• \( Bx^2 \) is the part that varies with the position \( x \) along the rod.

This kind of expression indicates that as you move along the rod, the mass per unit length changes depending on \( x \). Such variability in mass distribution requires us to consider each point or small segment of the rod separately if we want to determine forces, masses, or other related physical quantities.

Variable Mass Distribution

In physics, variable mass distribution refers to a situation where the density (or mass per unit length, area, or volume) changes depending on the position within an object. This contrasts with constant mass distribution, where the density is the same throughout.
In our scenario, the rod has a variable linear density, which means the mass changes along its length due to the term \( Bx^2 \).This complex relationship means that when calculating the gravitational force, or any other similar physical interaction, we have to account for how each small portion of the rod contributes separately.
To efficiently handle this changing distribution, we often divide the object into infinitesimally small parts, calculate the desired quantity for each, and then sum all together using calculus. This process allows for accurate calculations even when dealing with intricate changes in mass distribution.

Definite Integral in Physics

The definite integral is a powerful tool in calculus, especially valuable in physics for calculating the total effect from continuous distributions. It allows us to sum up infinitely many small contributions from a range of values to find a total quantity.
In the context of this gravitational force problem, the integral \[F = \int_{a}^{L+a} \frac{G m (A + Bx^2)}{x^2} \, dx\]represents the total gravitational force on a point mass caused by a rod with a given linear density.
The integral is computed over the length of the rod, from \( x = a \) to \( x = L + a \), encompassing all differential segments of the rod.

• The first part of the integral calculates the contribution due to the constant density \( A \).
• The second part deals with the variable density \( Bx^2 \).

This approach provides an exact solution by effectively summing all the differential gravitational forces exerted by each part of the rod on the point mass. Integrals are indispensable in physics whenever variable distribution of properties like mass, charge, or force appear. By mastering their use, you can tackle a wide variety of scenarios with ease and precision.