Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 608

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. \(\\{\mathrm{A}\\}\left(\pi^{2} / 3\right)\) \(\\{B\\}\left(4 \pi^{2} / 3\right)\) \(\\{\mathrm{C}\\}\left(8 \pi^{2} / 3\right)\) \(\\{\mathrm{D}\\}\left(16 \pi^{2} / 3\right)\)

Short Answer

Expert verified
The short answer for the given problem is: \(\frac{I_1}{I_2} = \frac{4\pi^2}{3}\), which corresponds to option B.

Step by step solution

01

Find the Moment of Inertia of the Rod

For a uniform rod of mass \(m\) and length \(L\), the moment of inertia (\(I_1\)) about a perpendicular axis passing through one end is given by: \(I_1 = \frac{1}{3}mL^2\) #Step 2: Moment of Inertia of Ring#
02

Find the Moment of Inertia of the Ring

To find the moment of inertia of the same rod bent into a ring, first, determine the radius (\(R\)) of the resulting ring by dividing the length of the rod by the circumference of the circle: \(R = \frac{L}{2\pi}\) Now, the moment of inertia (\(I_2\)) for a uniform ring of mass \(m\) and radius \(R\) about a diameter is given by: \(I_2 = mR^2\) #Step 3: Substitute the expression for R into \(I_2\)#
03

Replace R with the expression found in Step 2

Use \(R=\frac{L}{2\pi}\) to replace R with an expression involving L in the formula for \(I_2\): \(I_2 = m\left(\frac{L}{2\pi}\right)^2\) Simplify this expression: \(I_2 = \frac{mL^2}{4\pi^2}\) #Step 4: Find the ratio \(\frac{I_1}{I_2}\)#
04

Calculate the ratio of \(I_1\) and \(I_2\)

Divide \(I_1\) by \(I_2\): \(\frac{I_1}{I_2} = \frac{\frac{1}{3}mL^2}{\frac{mL^2}{4\pi^2}}\) Now, cancel \(mL^2\) from the numerator and the denominator: \(\frac{I_1}{I_2} = \frac{1/3}{1/(4\pi^2)} = \frac{1}{3} \cdot (4\pi^2) = \frac{4\pi^2}{3}\) Comparing this result with the given options, we can see that the correct answer is: \\({\mathrm{B}}\\) \(\left(4 \pi^{2} / 3\right)\)

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

A body of mass \(\mathrm{m}\) is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimetre. If the angular velocity is doubted, the elongation in the spring is \(5 \mathrm{~cm}\). The original length of spring is... \(\\{\mathrm{A}\\} 16 \mathrm{~cm}\) \(\\{B\\} 15 \mathrm{~cm}\) \(\\{\mathrm{C}\\} 14 \mathrm{~cm}\) \(\\{\mathrm{D}\\} 13 \mathrm{~cm}\)

A car is moving at a speed of \(72 \mathrm{~km} / \mathrm{hr}\) the radius of its wheel is \(0.25 \mathrm{~m}\). If the wheels are stopped in 20 rotations after applying breaks then angular retardation produced by the breaks is \(\ldots .\) \(\\{\mathrm{A}\\}-25.5 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{B}\\}-29.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{C}\\}-33.52 \mathrm{rad} / \mathrm{s}^{2}\) \(\\{\mathrm{D}\\}-45.52 \mathrm{rad} / \mathrm{s}^{2}\)

The height of a solid cylinder is four times that of its radius. It is kept vertically at time \(t=0\) on a belt which is moving in the horizontal direction with a velocity \(\mathrm{v}=2.45 \mathrm{t}^{2}\) where \(\mathrm{v}\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) is in second. If the cylinder does not slip, it will topple over a time \(t=\) \(\\{\mathrm{A}\\} 1\) second \(\\{\mathrm{B}\\} 2\) second \\{C \(\\}\) \\} second \(\\{\mathrm{D}\\} 4\) second

A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is \(\\{\mathrm{A}\\} 2 / 5\) \\{B \\} \(3 / 5\) \(\\{\mathrm{C}\\} 2 / 7\) \(\\{\mathrm{D}\\} 3 / 7\)

A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)
See all solutions

Recommended explanations on English Textbooks

Multiple Choice Questions

Read Explanation

Creative Story

Read Explanation

English Language Study

Read Explanation

Semiotics

Read Explanation

Sociolinguistics

Read Explanation

Summary Text

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