Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 594

A circular disc \(\mathrm{x}\) of radius \(\mathrm{R}\) is made from an iron plate of thickness \(t\). and another disc \(Y\) of radius \(4 R\) is made from an iron plate of thickness \(t / 4\) then the rotation between the moment of inertia \(\mathrm{I}_{\mathrm{x}}\) and \(\mathrm{I}_{\mathrm{y}}\) is \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{y}}=64 \mathrm{I}_{\mathrm{x}}\) \(\\{B\\} I_{y}=32 I_{x}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{y}}=16 \mathrm{I}_{\mathrm{x}}\) \(\\{\mathrm{D}\\} \mathrm{I}_{\mathrm{y}}=\mathrm{I}_{\mathrm{x}}\)

Short Answer

Expert verified
The relationship between the moments of inertia of the discs is \(I_y = 64 I_x\), which corresponds to option A.

Step by step solution

01

Moment of Inertia for a Uniform Circular Disc

The moment of inertia for a thin uniform circular disc rotating about an axis perpendicular to the plane of the disc and passing through its center is given by: \[ I = \frac{1}{2} M R^{2} \] Where I is the moment of inertia, M is the mass of the disc, and R is the radius of the disc.
02

Express Mass in terms of Density, Thickness and Area

We can write the mass of the disc in terms of its density ρ, thickness t, and area A, where A is the area of the disc given by πR². So, \(M = \rho A t = \rho(\pi R^{2})t \). Now, let's substitute the mass into the formula for the moment of inertia of our discs.
03

Moment of Inertia In terms of Density, Thickness and Radius

Substitute the mass expression into the moment of inertia formula: \[ I = \frac{1}{2}(\rho \pi R^{2} t) R^{2} \]
04

Calculate the Moment of Inertia for Disc X and Disc Y

Now, we'll plug in the values for disc X and disc Y and find their moments of inertia. For disc X: Radius RX = R Thickness tX = t \[I_{x} = \frac{1}{2}(\rho \pi R^{2} t) R^{2} \] For disc Y: Radius RY = 4R Thickness tY = t/4 \[I_{y} = \frac{1}{2}(\rho \pi (4R)^{2} \frac{t}{4}) (4R)^{2} \]
05

Find the Relationship between Ix and Iy

Now, let's divide Iy by Ix to find the relationship between them. \[\frac{I_{y}}{I_{x}} = \frac{\frac{1}{2}(\rho \pi (4R)^{2} \frac{t}{4}) (4R)^{2}}{\frac{1}{2}(\rho \pi R^{2} t) R^{2}} \] Simplify the equation: \[\frac{I_{y}}{I_{x}} = \frac{(4R)^{2} (4R)^{2} \frac{t}{4}}{R^{2} R^{2} t} \] \[\frac{I_{y}}{I_{x}} = \frac{256 R^{4} t}{4 R^{4} t} \] \[\frac{I_{y}}{I_{x}} = 64 \] From the given options, the relationship between the moments of inertia is Iy = 64 Ix, which corresponds to option A.

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

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 \(\\} \sqrt{2} \mathrm{MR}^{2}\)

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 circular disc of mass \(\mathrm{m}\) and radius \(\mathrm{r}\) is rolling on a smooth horizontal surface with a constant speed v. Its kinetic energy is \(\\{\mathrm{A}\\}(1 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{B}\\}(1 / 2) \mathrm{mv}^{2}\) \(\\{\mathrm{C}\\}(3 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{D}\\} \mathrm{mv}^{2}\)

A particle performing uniform circular motion has angular momentum \(L\)., its angular frequency is doubled and its \(K . E\). halved, then the new angular momentum is \(\\{\mathrm{A}\\} 1 / 2\) \\{B \(\\} 1 / 4\) \(\\{\mathrm{C}\\} 2 \mathrm{~L}\) \(\\{\mathrm{D}\\} 4 \mathrm{~L}\)

A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first two second it rotate through an angle \(\theta_{1}\), in the next \(2 \mathrm{sec}\). it rotates through an angle \(\theta_{2}\), find the ratio \(\left(\theta_{2} / \theta_{1}\right)=\) \(\\{\mathrm{A}\\} 1\) \(\\{B\\} 2\) \(\\{\mathrm{C}\\} 3\) \(\\{\mathrm{D}\\} 4\)
See all solutions

Recommended explanations on English Textbooks

English Grammar Summary

Read Explanation

Rhetoric

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Free Response Essay

Read Explanation

English Grammar

Read Explanation

Linguistic Terms

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