/*! 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 48 A cylinder of mass \(m\) has a l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 48

A cylinder of mass \(m\) has a length \(\ell\) that is \(\sqrt{3}\) times its radius \(R\). What is the ratio of its moment of inertia about its own axis and that of an axis passing through its centre and perpendicular to its axis?

Short Answer

Expert verified
The ratio of the moment of inertia about its own axis to that of an axis passing through its center and perpendicular to its axis is \(\frac{I_1}{I_2} = 1\).

Step by step solution

01

Find the moment of inertia for the own axis

In this case, the own axis is the central axis of the cylinder. To find the moment of inertia, we can use the formula for a solid cylinder rotating about its central axis: \[I_1 = \frac{1}{2}mR^2\] It's given that the mass of the cylinder is `m` and the length `l` is \(\sqrt{3}\) times the radius `R` which can be written as: \[ \ell = \sqrt{3}R \]
02

Find the moment of inertia for the perpendicular axis

We now need to find the moment of inertia of the cylinder about the axis passing through its center and perpendicular to its axis. To do this, we can use the perpendicular axis theorem, which states that the moment of inertia about a perpendicular axis passing through the center of mass is equal to the sum of the moment of inertia around two other perpendicular axes that also pass through the center of mass. In this case, since the cylinder is symmetrical, we can use the following equation: \[I_2 = \frac{1}{4}mR^2 + \frac{1}{12}m(\sqrt{3}R)^2\]
03

Find the ratio of the moments of inertia

To find the ratio of the moment of inertia about its own axis to that of an axis passing through its center and perpendicular to its axis, we can divide the first moment by the second: \[\frac{I_1}{I_2} = \frac{\frac{1}{2}mR^2}{\frac{1}{4}mR^2 + \frac{1}{12}m(\sqrt{3}R)^2}\]
04

Simplify the ratio

Now we simplify the expression for the ratio: \[\frac{I_1}{I_2} = \frac{\frac{1}{2}mR^2}{\frac{mR^2}{4} + \frac{m(3R^2)}{12}} = \frac{\frac{1}{2}mR^2}{\frac{mR^2}{4} + \frac{mR^2}{4}} = \frac{\frac{1}{2}mR^2}{\frac{mR^2}{2}}\] We can now cancel out the mass and radius terms from the equation: \[\frac{I_1}{I_2} = \frac{\frac{1}{2}}{\frac{1}{2}}\] The final answer to the ratio of the moments of inertia is: \[\frac{I_1}{I_2} = 1\]

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physics for JEE MAIN

Physics is a crucial part of the JEE MAIN syllabus, which tests students on the fundamental concepts of mechanics, electromagnetism, thermodynamics, and more. Understanding the principles of rotational motion and the moment of inertia is vital for successfully navigating the physics section of this competitive exam. The problem involving the cylinder's moment of inertia is an exemplary illustration of how rotational dynamics are applied in JEE MAIN. It embodies the critical concepts of rotational inertia and the application of theorems such as the perpendicular axis theorem, which is often addressed in questions within the exam.


When studying for JEE MAIN, it's essential to focus on the conceptual clarity and problem-solving skills. Grasping the core ideas in rotational dynamics can significantly improve students' problem-solving speed and accuracy, which is pivotal in an exam where every second counts.

Cylinder Moment of Inertia

The moment of inertia is a physical quantity that measures the rotational inertia of a body when torque is applied. For a solid cylinder rotating about its central axis, the formula to calculate the moment of inertia is:

\[I = \frac{1}{2}mR^2\]


Here, \(m\) stands for the mass of the cylinder, and \(R\) signifies the radius. The moment of inertia is an intrinsic property that depends on how the mass is distributed with respect to the axis of rotation. For JEE MAIN aspirants, understanding how to derive and apply these formulas in various scenarios is critical, as it will be required to solve questions related to rotating bodies.

Perpendicular Axis Theorem

The perpendicular axis theorem is a pivotal concept often used in the context of planar bodies that are symmetrical about an axis. It asserts that the moment of inertia of a body about an axis perpendicular to its plane is equal to the sum of the moments of inertia about two orthogonal axes lying in the plane of the body. Mathematically, for axes \(x\), \(y\), and \(z\) that are mutually orthogonal and meet at a point in the plane of the body, the theorem can be expressed as:

\[I_z = I_x + I_y\]


For a cylinder, the application of this theorem simplifies finding the moment of inertia for axes that are not the central axis. It is a handy tool for JEE MAIN students, especially in problems where they need to consider multiple axes of rotation and combine rotational effects accordingly.

Rotation Dynamics

Rotation dynamics is the study of motion of bodies when they rotate around a central axis. This area of physics encompasses not only the calculation of the moment of inertia but also the study of torque, angular velocity, angular acceleration, and kinetic energy in rotational motion. In the context of the given problem, rotation dynamics principles help us understand how the distribution of mass affects the tendency of a cylinder to resist changes to its state of spin.


For JEE MAIN, mastering rotation dynamics is essential as it allows students to relate linear motion principles with their rotational counterparts and solve complex problems effectively. Whether it's a simple spinning wheel or a more complex system with multiple rotating parts, a clear understanding of rotation dynamics equips students with the necessary tools to tackle a wide range of questions in the physics section of the exam.

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 particle of mass \(m\) is moving along the line \(y=3 x+5\) with speed \(v .\) The magnitude of angular momentum about origin is (A) \(\sqrt{\frac{5}{2}} m v\) (B) \(\frac{5}{2} m v\) (C) \(\frac{1}{2} m v\) (D) \(\frac{1}{\sqrt{3}} m v\)

A string is wrapped several times round a solid cylinder of mass \(m\) and then the end of the string is held stationary while the cylinder is released from rest with no initial motion. The acceleration of the cylinder will be \(\frac{n g}{3}\), then the value of \(n\) is.

A hollow cylinder, a spherical shell, a solid cylinder and a solid sphere are allowed to roll on an inclined rough surface of coefficient of friction \(\mu\) and inclination \(\theta\). The correct statements are (A) if cylindrical shell can roll on inclined plane, all other objects will also roll. (B) if all the objects are rolling and have same mass, the \(\mathrm{KE}\) of all the objects will be same at the bottom of inclined plane. (C) work done by the frictional force will be zero, if objects are rolling. (D) frictional force will be equal for all the objects, if mass is same.

A point \(P\) moves in counter-clockwise direction on a circular path as shown in Fig. 6.44. The movement of \(P\) is such that it sweeps out a length \(s=t^{3}+5\), where s is in metres and \(t\) is in seconds. The radius of the path is \(20 \mathrm{~m}\). The acceleration of \(P\) when \(t=2\) is nearly (A) \(13 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(12 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(7.2 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(14 \mathrm{~m} / \mathrm{s}^{2}\)

A man of \(50 \mathrm{~kg}\) is riding a bicycle. Man changes angular speed of wheel of radius \(0.5 \mathrm{~m}\) by \(4 \mathrm{rad} / \mathrm{s}\) in \(1 \mathrm{~s}\). (Assume pure rolling) (A) The angular acceleration of wheel is \(4 \mathrm{rad} / \mathrm{s}^{2}\) (B) Net horizontal force on the man is \(100 \mathrm{~N}\) (C) Net force on the man is \(100 \mathrm{~N}\) (D) Net force on the man is \(\sqrt{26} \times 100 \mathrm{~N}\)
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Atoms and Radioactivity

Read Explanation

Waves Physics

Read Explanation

Thermodynamics

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.