/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 569 A straight rod of length \(L\) h... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

Short Answer

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

Step by step solution

01

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.
02

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: \[ M = \int_0^L dm = \int_0^L Axdx \]
03

Integrate

Integrate the mass function over the length of the rod: \[ M = A\int_0^L xdx = A\left[\frac{1}{2}x^2\right]_0^L = \frac{1}{2}AL^2\]
04

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: \[ \int_0^L xdm = \int_0^L x(Axdx) = A\int_0^L x^2dx\]
05

Integrate

Integrate the function over the length of the rod: \[ A\int_0^L x^2dx = A\left[\frac{1}{3}x^3\right]_0^L = \frac{1}{3}AL^3\]
06

Calculate the center of mass

Now we can plug our results into the center of mass formula: \[ x_{cm} = \frac{\frac{1}{3}AL^3}{\frac{1}{2}AL^2} \]
07

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One quarter sector is cut from a uniform circular disc of radius \(\mathrm{R}\). This sector has mass \(\mathrm{M}\). It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is... \(\\{\mathrm{A}\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(1 / 4) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(1 / 8) \mathrm{MR}^{2}\) \\{D \(\\} \sqrt{2} \mathrm{MR}^{2}\)

Two discs of the same material and thickness have radii \(0.2 \mathrm{~m}\) and \(0.6 \mathrm{~m}\) their moment of inertia about their axes will be in the ratio \(\\{\mathrm{A}\\} 1: 81\) \(\\{\mathrm{B}\\} 1: 27\) \(\\{C\\} 1: 9\) \(\\{\mathrm{D}\\} 1: 3\)

A small object of uniform density rolls up a curved surface with initial velocity 'u'. It reaches up to maximum height of \(3 \mathrm{v}^{2} / 4 \mathrm{~g}\) with respect to initial position then the object is \(\\{\mathrm{A}\\}\) ring \(\\{B\\}\) solid sphere \(\\{\mathrm{C}\\}\) disc \\{D\\} hollow sphere

A uniform rod of length \(\mathrm{L}\) is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length, what maximum speed must be imparted to the lower end so that the rod completes one full revolution? \(\\{\mathrm{A}\\} \sqrt{(2 \mathrm{~g} \mathrm{~L})}\) \(\\{\mathrm{B}\\} 2 \sqrt{(\mathrm{gL})}\) \(\\{C\\} \sqrt{(6 g L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 g \mathrm{~L})}\)

A uniform disc of mass \(500 \mathrm{~kg}\) and radius \(2 \mathrm{~m}\) is rotating at the rate of \(600 \mathrm{r}\).p.m. what is the torque required to rotate the disc in the opposite direction with the same angular speed in a time of \(100 \mathrm{sec}\) ? \(\\{\mathrm{A}\\} 600 \pi \mathrm{Nm}\) \\{B \(\\} 500 \pi \mathrm{Nm}\) \\{C \(\\} .400 \pi \mathrm{Nm}\) \(\\{\mathrm{D}\\} 300 \pi \mathrm{Nm}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.