Problem 569 - JEE Physics

Question

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

Step-by-Step Solution

Answer

The center of mass of the rod is located at \(\frac{2}{3}L\).

Step 1: Understanding the center of mass formula for continuous objects

In the case of continuous mass distribution, the center of mass coordinates can be calculated as follows. For the x-coordinate of the center of mass: \[ x_{cm} = \frac{\int xdm}{\int dm} \] Here, \(dm = Axdx\), where Ax is the mass per unit length and dx is an infinitesimal length element on the rod.

Step 2: Calculate the mass of the rod

To find the total mass (M) of the rod, we integrate the dm function over the length of the rod: \[ M = \int_0^L dm = \int_0^L Axdx \]

Step 3: Integrate

Integrate the mass function over the length of the rod: \[ M = A\int_0^L xdx = A\left[\frac{1}{2}x^2\right]_0^L = \frac{1}{2}AL^2\]

Step 4: Calculate the numerator of the center of mass formula

Now we need to calculate the integral in the numerator of the center of mass formula: \[ \int_0^L xdm = \int_0^L x(Axdx) = A\int_0^L x^2dx\]

Step 5: Integrate

Integrate the function over the length of the rod: \[ A\int_0^L x^2dx = A\left[\frac{1}{3}x^3\right]_0^L = \frac{1}{3}AL^3\]

Step 6: Calculate the center of mass

Now we can plug our results into the center of mass formula: \[ x_{cm} = \frac{\frac{1}{3}AL^3}{\frac{1}{2}AL^2} \]

Step 7: Simplify the result

Simplify the expression to find the x-coordinate of the center of mass: \[ x_{cm} = \frac{\frac{1}{3}AL^3}{\frac{1}{2}AL^2} = \frac{2}{3}L \] So, the center of mass of the rod is located at \(\frac{2}{3}L\). The correct answer is option C.

Key Concepts

Mass Distribution

In physics, mass distribution describes how mass is spread across an object. When evaluating mass distribution, it's crucial to know how much mass is present at each point in an object, especially for continuous objects. For a straight rod with varying mass along its length, the mass per unit length is described by a function—in this case, \(Ax\), where \(A\) is a constant and \(x\) is the position along the rod.
This means the rod's mass isn’t uniform, but increases linearly from the origin to \( L \). Understanding mass distribution helps in calculating the center of mass, which is the point where the mass is balanced on either side. It is essential for predicting the rod's behavior when interacting with forces, like gravity.
Knowing the specific mass distribution informs how to integrate, leading to more accurate calculations regarding the rod's properties.

Integration

Integration is a fundamental tool in calculus used to find the total of a continuously varying quantity. In the context of mass distribution, integration allows us to sum up the infinitesimal masses across an object's length. For the rod, the integral of the mass per unit length function \(Ax\) over its length from 0 to \(L\) gives the total mass \(M\). The integration process captures how masses accumulate as you move along the rod.
To solve for total mass, compute the integral \( \int_0^L Axdx \), which results in \( \frac{1}{2}AL^2 \). Many physics problems require such integration to find all kinds of totals, such as area, volume, or in this case, mass.

• Definite integrals, like the one used here, provide real values for the entire length of the object.
• Understanding how to set limits of integration (from 0 to \(L\)) is crucial for accurate results.

Continuous Objects

Continuous objects, unlike discrete ones, have mass distributed over a continuum. This differentiation is critical when calculating certain properties such as center of mass. For the rod, treating it as continuous means considering an infinitesimally small element of mass \(dm\), related by \(dm = Axdx\).
This approach provides a more accurate depiction of physical phenomena because it accounts for changes in mass along the length of the object.
By utilizing continuous concepts, we identify every tiny segment's contribution to the overall properties of the object and integrate over these contributions to get a comprehensive understanding of properties like mass and center of mass. It's an approach essential for real-world applications where objects are not just collections of discrete parts but form a smooth distribution of matter.

Calculus in Physics

Calculus, the mathematical study of change, is integral to solving numerous physics problems. It provides tools to describe dynamically varying systems. In this exercise, calculus helps in finding the center of mass for a rod with its mass varying linearly along its length.
The integral calculus used finds total attributes such as mass (\(M = \int_0^L dm\)) and moment (\( \int_0^L x dm\)), which allow determining the center of mass, a pivotal point that indicates balance in the system.
Physics frequently uses calculus to understand not just static objects but dynamic systems influenced by forces. By leveraging these mathematical techniques, physicists can predict how objects will behave under various conditions, making calculus indispensable for advanced problem-solving and theoretical exploration.

• Derivatives and integrals describe motion and forces.
• Calculus helps quantify continuous distributions essential in real-world physics.