Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 560

One circular rig and one circular disc both are having the same mass and radius. The ratio of their moment of inertia about the axes passing through their centers and perpendicular to their planes, will be \(\\{\mathrm{A}\\} 1: 1\) \(\\{\mathrm{B}\\} 2: 1\) \(\\{C\\} 1: 2\) \(\\{\mathrm{D}\\} 4: 1\)

Short Answer

Expert verified
The ratio of the moment of inertia for the circular rig to the circular disc is \(\{B\} 2:1\), as obtained by using the respective moment of inertia formulas \(I_{rig} = M R^2\) and \(I_{disc} = \frac{1}{2} M R^2\), and simplifying the equation for the ratio \(\frac{I_{rig}}{I_{disc}}\).

Step by step solution

01

Determine the moment of inertia formulas for the circular rig and the circular disc

We need to know the formulas for the moment of inertia for both objects. For a circular rig (hollow circle) with mass M and radius R, the moment of inertia is given by: \(I_{rig} = M R^2\). For a circular disc (solid circle) with mass M and radius R, the moment of inertia is given by: \(I_{disc} = \frac{1}{2} M R^2\).
02

Set up the equation for the ratio of moments of inertia

Now that we have the formulas for the moment of inertia, we need to set up the equation for the ratio of their moments of inertia: \(\frac{I_{rig}}{I_{disc}} = \frac{M R^2}{\frac{1}{2} M R^2}\).
03

Simplify the equation

Notice that the mass (M) and radius squared (R^2) appear in both the numerator and the denominator. We can simplify the equation by canceling them out: \(\frac{I_{rig}}{I_{disc}} = \frac{M R^2}{\frac{1}{2} M R^2} = \frac{2}{1}\). The ratio of the moment of inertia is 2:1.
04

Choose the correct answer

Now that we have simplified the equation and found the ratio of the moment of inertia for the circular rig to the circular disc, we can choose the correct answer choice: The correct answer is \(\{B\} 2:1\).

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 moment of inertia of a uniform circular disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about any of its diameter is \((1 / 4) \mathrm{MR}^{2}\), what is the moment of inertia of the disc about an axis passing through its centre and normal to the disc? \(\\{\mathrm{A}\\} \mathrm{MR}^{2}\) \\{B \(\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(3 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\} 2 \mathrm{MR}^{2}\)

Two spheres each of mass \(\mathrm{M}\) and radius \(\mathrm{R} / 2\) are connected with a mass less rod of length \(2 \mathrm{R}\) as shown in figure. What will be moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod? \(\\{\mathrm{A}\\}(21 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(2 / 5) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(5 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(5 / 21) \mathrm{MR}^{2}\)

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

A circular disc of radius \(R\) is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through an angle of \(60 ?\) and released. Its angular velocity when it reaches the equilibrium position will be \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / 3 \mathrm{R})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / 3 \mathrm{R})}\) \(\\{\mathrm{C}\\} \sqrt{(2 \mathrm{~g} / \mathrm{R})}\) \(\\{\mathrm{D}\\} 2 \sqrt{(2 \mathrm{~g} / \mathrm{R})}\)

A player caught a cricket ball of mass \(150 \mathrm{gm}\) moving at a rate of \(20 \mathrm{~m} / \mathrm{s}\) If the catching process is Completed in \(0.1\) sec the force of the flow exerted by the ball on the hand of the player ..... N \(\\{\mathrm{A}\\} 3\) \(\\{B\\} 30\) \(\\{\mathrm{C}\\} 150\) \(\\{\mathrm{D}\\} 300\)
See all solutions

Recommended explanations on English Textbooks

Language and Social Groups

Read Explanation

Essay Prompts

Read Explanation

Language Acquisition

Read Explanation

Textual Analysis

Read Explanation

English Language Study

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