A rod of uniform cross-scction of mass M and lenghth L is...

A rod of uniform cross-scction of mass $M$ and lenghth $L$ is hinged about an end to swing frecly in a vertical planc. However, its density is non uniform and varies lincarly from hinged end to the frec cnd doubling its value. Find the moment of incrtia of the rod, about the rotation axis passing through the hinge point.

Key Concepts

Calculus Integration in Physics

Calculus integration is a fundamental tool in physics used to compute quantities like moment of inertia for objects with non-uniform properties. By integrating over a differential element (for example, incorporating a variable density function), one can accurately sum contributions across the entire object, accounting for variations in shape and density.

Linear Density Variation

In cases where the mass density varies along the length of an object, such as a rod with a density that changes linearly, it is important to express the mass element as the product of the local density and an infinitesimal length. This allows for an accurate calculation of the moment of inertia by integrating over the variable density distribution.

Moment of Inertia

The moment of inertia quantifies an object’s resistance to angular acceleration about a particular axis. It is calculated by integrating the product of each mass element and the square of its distance from the axis. This concept is particularly crucial when determining rotational behavior in physics.

Continuous Mass Distribution

For objects with a continuous mass spread rather than concentrated masses, the total moment of inertia is obtained by integrating over differential mass elements. Each element contributes based on its distance from the rotation axis, necessitating the use of calculus to sum these infinitesimal contributions.

Transcript

00:01 Hello everyone we are going to understand this question here given in the question given there is a uniform rod having cross section mass of cross section that is m and length is given l length is l and it swings about the point o so we have to find the moment of inertia about the point here, mass density of rod is non -uniform and it varies linearly from hinged end.

00:44 So if lambda is the linear mass density, so lambda x is equal to lambda plus lambda upon l into x.

00:57 Now taking lambda common, so we will get 1 plus x upon l.

01:04 This is the linear mass density.

01:08 So required movement of inertia, so required moment of inertia, required moment of inertia, that is, i is equal to, i is equal to integration of x square into dm.

01:40 Here, dm is the mass of partial length, that is, that's length is d x.

01:48 So here we can write integration limit is from 0 to l into x square and value of dm is lambda linear mass density that is lambda into 1 plus x upon l into d x this is the this is the mass of partial length now doing further calculation so we will get we will get integration and here limit is from 0 to l.

02:38 Now we can write lambda into x square plus x cube upon l into d x.

02:52 So here we can write now taking the integration that will keep x cube upon 3 plus x to the power 4 upon 4 into l.

03:07 Now substituting the value here integration limit is from 0 to l.

03:13 So we can write lambda into l cube upon 3 plus l to the power 4 upon 4l.

03:28 Now we can write lambda into here taking lcm we will get 2l.

03:36 So 4l cube plus 3l cube.

03:42 Now after doing further calculation, we will get 7 lq upon 12.

03:49 We will get 7 lq upon 12.

03:54 This is the required moment of inertia.

03:57 Now calculating the mass of the rod...

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