/*! 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 69 From a circular dise of radius \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

From a circular dise of radius \(R\) and mass \(9 M\), a small disc or radius \(R / 3\) is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the dise and passing through \(O\) is [IIT JEE 2005] (a) \(4 \mathrm{MR}^{2}\) (b) \(\frac{40}{9} \mathrm{MR}^{2}\) (c) \(10 \mathrm{MR}^{2}\) (d) \(\frac{37}{9} M R^{2}\)

Short Answer

Expert verified
The moment of inertia of the remaining disc is \( \frac{40}{9} MR^2 \).

Step by step solution

01

Moment of Inertia of the Original Disc

To find the moment of inertia of the entire original disc, we use the formula for a solid disc: \( I = \frac{1}{2} M R^2 \), where \( M \) is the mass and \( R \) is the radius. Given the original disc has mass \( 9M \) and radius \( R \), the moment of inertia is \( I = \frac{1}{2} \times 9M \times R^2 = \frac{9}{2} MR^2 \).
02

Determine Mass and Moment of Inertia of Removed Small Disc

The smaller disc that is removed has a radius of \( \frac{R}{3} \). The area ratio of the small disc to the large disc is \( \left(\frac{R/3}{R}\right)^2 = \frac{1}{9} \). Therefore, the mass of the small disc is \( 9M \times \frac{1}{9} = M \). Its moment of inertia is \( I_{small} = \frac{1}{2} \times M \times \left(\frac{R}{3}\right)^2 = \frac{1}{2} \times M \times \frac{R^2}{9} = \frac{MR^2}{18} \).
03

Find the Moment of Inertia of the Remaining Disc

The moment of inertia of the remaining disc is the difference between the moment of inertia of the original disc and the removed small disc: \( I_{remaining} = \frac{9}{2} MR^2 - \frac{MR^2}{18} \). To subtract these, find a common denominator, \( 18 \), resulting in the formula \( I_{remaining} = \frac{81MR^2}{18} - \frac{MR^2}{18} = \frac{80MR^2}{18} = \frac{40MR^2}{9} \).
04

Select the Correct Option

Comparing \( \frac{40}{9} MR^2 \) with the provided options, we see the correct choice is option (b).

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.

Understanding Circular Discs
A circular disc is a fundamental shape you will often encounter in physics. It's defined by its radius and the surface area it encompasses. When dealing with physics problems involving discs, such as calculating moments of inertia, it's important to understand the properties of the disc. A full circular disc is symmetrical around its center.
Its mass is evenly distributed across its surface, making it easier to calculate properties related to rotation, like the moment of inertia.
In this problem, the initial setting is a full disc with a defined total mass and radius, which forms the basis for further calculations.
  • Radius: The distance from the center to any point on the boundary of the disc.
  • Mass: The total weight of the disc, distributed evenly across its plane.
Mass Distribution in a Circular Disc
Mass distribution refers to how mass is spread out over the area of the disc. In a perfectly symmetrical circular disc, mass is uniform, which simplifies calculations for rotational dynamics.
For example, when you're given a problem involving a circular disc, you normally assume the mass is evenly spread unless stated otherwise.
This is important because the uniformity of mass distribution allows you to use standard formulas for various physical calculations.
  • Uniform Distribution: Mass is evenly distributed.
  • Helps in Solver Simplicity: Facilitates straightforward calculations using known formulas.
In the exercise, the mass is entirely distributed throughout the whole disc before the smaller disc is removed, aiding in the computation of the original moment of inertia.
Impact of Removal of Mass
Removing a section from a body affects its physical properties, such as center of mass, symmetry, and moment of inertia. In this problem, we remove a smaller disc from the larger one, altering its moment of inertia.
The smaller disc has its own radius and mass, which must be calculated in relation to the original disc to determine the effect of removal accurately.
Removing mass shifts how the remaining mass is distributed and alters the rotational dynamics:
  • Change in Mass Distribution: Removal leads to uneven mass distribution.
  • Moment of Inertia Affected: Must be recalculated by subtracting the moment of inertia of the removed part from the whole.
Here, by considering the removed smaller disc as an entity, you subtract its moment of inertia from that of the full disc, simplifying the problem to finding the moment of inertia of the leftover portion.
Physics Problem Solving Approach
Tackling physics problems involves a structured approach, starting with identifying known values and the desired outcome. For the disc problem concerning moment of inertia, the process is divided into manageable steps.
First, calculate the moment of inertia of the entire disc, which serves as your base. Next, determine the properties of the mass removed, considering its scale relative to the whole. This involves finding both its mass and moment of inertia.
Finally, adjust the moment of inertia for the larger disc by removing the smaller disc's contribution.
  • Structured Steps: Break down into known values and stepwise calculations.
  • Check Units: Ensure consistent units across calculations.
  • Eliminate options: Compare findings with given choices for accuracy.
Employing this methodology, the solution becomes manageable, and each step checks your understanding of physical concepts being applied.

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 net external torque on a system of particles about an axis is zero. Which of the following are compatible with it? \(\quad\) INCERT Exemplar] (a) The forces may be acting radially from a point on the axis (b) The forces may be acting on the axis of rotation (c) The forces may be acting parallel to the axis of rotation (d) The torque caused by some forces may be equal and opposite to that caused by other forces

A circular platform is mounted on a vertical friction less axle. Its radius is \(r=2 \mathrm{~m}\) and its moment of inertia \(I=200 \mathrm{~kg}-\mathrm{m}^{2} .\) It is initially at rest. A \(70 \mathrm{~kg}\) man stands on the edge of the platform and begins to walk along the edge at speed \(v_{0}=1.0 \mathrm{~ms}^{-1}\) relative to the ground. The angular velocity of the platform is \(0.7 \mathrm{rad} \mathrm{s}^{-1}\). When the man has walked once around the platform, so that he is at his original position on it, his angular displacement relative to ground is (a) \(\frac{6}{5} \pi\) (b) \(\frac{5}{6} \pi\) (c) \(\frac{4}{5} \pi\) (d) \(\frac{5}{4} \pi\)

A solid cylinder of mass \(20 \mathrm{~kg}\) has length \(1 \mathrm{~m}\) and radius \(0.2 \mathrm{~m}\). Then its moment of inertia in \(\mathrm{kg} \mathrm{m}^{2}\) about its geometrical axis is \([\) Kerala CET 2005] (a) \(0.8\) (b) \(0.4\) (c) \(0.2\) (d) \(20.4\)

Two thin uniform circular rings each of radius \(10 \mathrm{~cm}\) and mass \(0.1 \mathrm{~kg}\) are arranged such that they have a common centre and their planes are perpendicular to each other. The moment of inertia of this system about an axis passing through their common centre and perpendicular to the plane of one of the rings in \(\mathrm{kgm}^{2}\) is (a) \(15 \times 10^{-3}\) (b) \(5 \times 10^{-3}\) (c) \(1.5 \times 10^{-3}\) (d) \(18 \times 10^{-4}\)

A ballet dancer spins with \(2.8\) rev \(\mathrm{s}^{-1}\) with her arms out stretched. When the moment of inertia about the same axis becomes \(0.7 I\), the new rate of spin is (a) \(3.2 \mathrm{rev} \mathrm{s}^{-1}\) (b) \(4.0 \mathrm{rev} \mathrm{s}^{-1}\) (c) \(4.8 \mathrm{rev} \mathrm{s}^{-1}\) (d) \(5.6 \mathrm{rev} \mathrm{s}^{-1}\)

See all solutions

Recommended explanations on Physics 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.