Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 624

According to the theorem of parallel axis \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^{2}\) the graph between \(\mathrm{I} \rightarrow \mathrm{d}\) will be

Short Answer

Expert verified
The graph between the moment of inertia (I) and distance (d) based on the parallel axis theorem will be a parabolic curve that opens upwards, showing a positive and quadratic relation between I and d. This is due to the equation \(I = I_{cm} + md^2\), where Icm and m are fixed values for a specific object.

Step by step solution

01

Understanding The Variables and Formula

The equation for parallel axis theorem is given by: \(I = I_{cm} + md^{2}\) where: - I: Moment of inertia about an axis parallel to the axis passing through the center of mass - Icm: Moment of inertia about the axis passing through the center of mass - m: Mass of the object - d: Distance between the two parallel axes The equation represents the relationship between the moment of inertia around an axis parallel to the one passing through the center of mass and distance between both axes, mass and moment of inertia around the center of mass.
02

Fixing the Known Values

In order to create a graph between I and d, we need to fix the values of mass (m) and moment of inertia about the center of mass (Icm). These values will remain constant so that we can understand how the moment of inertia (I) varies with distance (d). For example, let's choose m = 5 kg and Icm = 10 kg m^2. (Feel free to choose any other values for clarification)
03

Writing the Equation with Fixed Values

Now that we have our fixed values for mass and moment of inertia about the center of mass, let's substitute them into the formula: \(I = 10 + 5 * d^{2}\)
04

Creating a Table of Values

To create a graph between I and d, we'll need to create a table with a variety of values for distance (d) and use the equation to calculate the corresponding moment of inertia (I). The table may look like this: d | I --------|------- 0 | 10 1 | 15 2 | 30 3 | 55 4 | 90
05

Plotting the Graph

With the table of values for I and d, we can now plot the graph. On the horizontal x-axis, represent the distance (d), and on the vertical y-axis, represent the moment of inertia (I). Plot the points from the table and connect them with a smooth curve. Considering the equation has a quadratic form (\(I = 10 + 5d^2\)), the graph will be a parabolic curve that opens upwards. #Conclusion# The graph between the moment of inertia (I) and distance (d) will be a parabolic curve that opens upwards, based on the theorem of the parallel axis, with the fixed values for mass (m) and moment of inertia about the center of mass (Icm). The graph will show a positive and quadratic relation between I and d.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The centre of mass of a systems of two particles is (A) on the line joining them and midway between them (B) on the line joining them at a point whose distance from each particle is proportional to the square of the mass of that particle. (C) on the line joining them at a point whose distance from each particle inversely proportional to the mass of that particle. (D) On the line joining them at a point whose distance from each particle is proportional to the mass of that particle.

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The acceleration of the centre of mass of the disc is \(\\{\mathrm{A}\\} \mathrm{g} \sin \theta\) \(\\{B\\}[(2 g \sin \theta) / 3]\) \(\\{\mathrm{C}\\}[(\mathrm{g} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(2 \mathrm{~g} \cos \theta) / 3]\)

The moment of inertia of a uniform rod about a perpendicular axis passing through one of its ends is \(\mathrm{I}_{1}\). The same rod is bent in to a ring and its moment of inertia about a diameter is \(\mathrm{I}_{2}\), Then \(\left[\mathrm{I}_{1} / \mathrm{I}_{2}\right]\) is. \(\\{\mathrm{A}\\}\left(\pi^{2} / 3\right)\) \(\\{B\\}\left(4 \pi^{2} / 3\right)\) \(\\{\mathrm{C}\\}\left(8 \pi^{2} / 3\right)\) \(\\{\mathrm{D}\\}\left(16 \pi^{2} / 3\right)\)

A cylinder of mass \(\mathrm{M}\) has length \(\mathrm{L}\) that is 3 times its radius what is the ratio of its moment of inertia about its own axis and that about an axis passing through its centre and perpendicular to its axis? \(\\{\mathrm{A}\\} 1\) \(\\{\mathrm{B}\\}(1 / \sqrt{3})\) \(\\{\mathrm{C}\\} \sqrt{3}\) \(\\{\mathrm{D}\\}(\sqrt{3} / 2)\)

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

Recommended explanations on English Textbooks

Lexis and Semantics Summary

Read Explanation

Pragmatics

Read Explanation

Essay Writing Skills

Read Explanation

Language and Social Groups

Read Explanation

Language Analysis

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation
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.

Sign-up for free