/*! 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 28 Four spheres of diameter \(2 a\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 28

Four spheres of diameter \(2 a\) and mass \(M\) are placed with their centres on the four corners of a square of side \(b\). Then the moment of inertia of the system about an axis along one of the sides of the square is (a) \(\frac{4}{5} M a^{2}+2 M b^{2}\) (b) \(\frac{8}{5} M a^{2}+2 M b^{2}\) (c) \(\frac{8}{5} M a^{2}\) (d) \(\frac{4}{5} M a^{2}+4 M b^{2}\)

Short Answer

Expert verified
The answer is (d) \(\frac{4}{5} M a^2 + 4 M b^2\).

Step by step solution

01

Understanding the Concept

The moment of inertia ( I ) of a system is calculated with respect to a particular axis. It measures the resistance to angular acceleration. For this exercise, we have four spheres placed at the corners of a square, and we need to calculate the moment of inertia about an axis along one side.
02

Calculating Moment of Inertia of a Sphere

The moment of inertia of each sphere about its own center is given by the formula: \( I = \frac{2}{5} M R^2 \), where \( R \) is the radius of the sphere. Since the diameter is \( 2a \), the radius \( R = a \). Thus, each sphere's moment of inertia about its center is \( \frac{2}{5} M a^2 \).
03

Using Parallel Axis Theorem

The Parallel Axis Theorem states that if you know the moment of inertia of a body about an axis through its center of mass (\( I_{CM} \)), you can find the moment of inertia about any parallel axis by adding \( M d^2 \), where \( d \) is the distance between the two axes. For spheres at a distance \( b \) from the axis along the side of the square, this additional term is \( M b^2 \).
04

Calculating Moment of Inertia for Each Sphere

Applying the Parallel Axis Theorem for each sphere:- Moment of inertia for each sphere with respect to the axis: \( I = \frac{2}{5} M a^2 + M b^2 \).
05

Total Moment of Inertia for the System

Since there are four spheres, the total moment of inertia \( I_{ ext{total}} \) is the sum of the individual moments of inertia:\[ I_{ ext{total}} = 4 \left( \frac{2}{5} M a^2 + M b^2 \right) \].
06

Simplify the Total Moment of Inertia Expression

Simplifying the expression:\[ I_{ ext{total}} = 4 \times \left( \frac{2}{5} M a^2 \right) + 4 \times \left( M b^2 \right) \= \frac{8}{5} M a^2 + 4 M b^2 \].

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.

Parallel Axis Theorem
The Parallel Axis Theorem is a useful tool for calculating the moment of inertia of an object about any axis that is parallel to the axis through its center of mass. It's very handy when dealing with complex systems like the one in our exercise. Initially, you calculate the moment of inertia about the center of mass (commonly known as the centroid) using a straightforward formula.
  • The formula for the moment of inertia about the center of mass axis is typically given or derived, like \(I = \frac{2}{5} M R^2\) for spheres.
  • The theorem tells us that to find the moment of inertia about a parallel axis, you add \ M d^2 \, where \(d\) is the perpendicular distance between the centers of the two axes.
This method simplifies the calculations, as it allows us to recalculate moment of inertia without changing our point of view entirely, simply by adding a straightforward factor.
Sphere Moment of Inertia
For spheres, the moment of inertia is calculated about an axis through the center of the sphere. When calculating, one must know the formula specific to spheres due to their symmetrical shape. It's expressed as: \( I = \frac{2}{5} M R^2 \). Here, \( M\) is the mass, and \( R\) is the radius, which you can find if you know the diameter of the sphere.
  • For our exercise, the diameter is given as \2a\, so the radius \(R\) is simply \a\.
  • Plugging in the radius gives us the basic moment of inertia for each sphere: \( \frac{2}{5} M a^2 \).
This calculation is a foundational step before applying the Parallel Axis Theorem to account for displacement when spheres are part of a system like in our exercise.
Angular Acceleration
Angular acceleration describes the rate of change of angular velocity. In terms of inertia, it is a measure of how a shape responds to the forces that cause rotation. Think of it as the rotational equivalent of linear acceleration. It highlights how the distribution of mass in a body can affect its rotational motion.
  • A system with high moment of inertia resists changes in rotational speed more than one with lower inertia.
  • In the context of our spheres placed at the corners of a square, understanding these distributions helps explain why the system鈥檚 resistance to rolling or spinning varies depending on the axis.
Grasping the relationship between inertia and angular acceleration is essential, especially in physics problems involving rotating bodies.
Resistance to Rotation
Resistance to rotation, also known as rotational inertia, dictates how much torque is needed to achieve a desired angular acceleration. It's closely related to the moment of inertia, as the distribution of mass around an axis directly influences how easily the object can be rotated.
  • The higher the moment of inertia, the greater the resistance to change in its motion.
  • In our exercise, all four spheres have the same moment of inertia, influencing how they resist rolling around the axis along one side of the square.
When studying rotational dynamics, it鈥檚 crucial to remember that both the distance of mass from the axis and the mass's distribution affect how easily an object moves.

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 cubical block of mass \(M\) and edge a slides down a rough inclined plane of inclination \(\theta\) with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude (a) zero (b) \(\mathrm{Mga}\) (c) \(M g a \sin \theta\) (d) \(\frac{M g a \sin \theta}{2}\)

A particle moves in a force field given by: \(\vec{F}=\hat{r} F(r)\), where \(\hat{r}\) is a unit vector along the position vector \(\vec{r}\) then which is true? (a) The torque acting on the particle is not zero. (b) The torque acting on the particle produces an angular acceleration in it. (c) The angular momentum of the particle is conserved. (d) The angular momentum of the particle increases.

A uniform rod of mass \(M\) and length \(L\) is pivoted at one end such that it can rotate in a vertical plane. There is negligible friction at the pivot. The free end of the rod is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle \(\theta\) with the vertical is (a) \(g \sin \theta\) (b) \(\frac{g}{L} \sin \theta\) (c) \(\frac{3 g}{2 L} \sin \theta\) (d) \(6 g L \sin \theta\)

Two wheels are mounted side by side and each is marked with a dot on its rim. The two dots are aligned with the wheels at rest, then one wheel is given a constant angular acceleration of \(\pi / 2 \mathrm{rad} / \mathrm{s}^{2}\) and the other \(\pi / 4 \mathrm{rad} / \mathrm{s}^{2}\). Then the two dots become aligned again for the first time after (a) 2 seconds (b) 4 seconds (c) 1 second (d) 8 seconds

A cricket mat of mass \(50 \mathrm{~kg}\) is rolled loosely in the form of a cylinder of radius \(2 \mathrm{~m}\). Now again it is rolled tightly so that the radius becomes \(\frac{3}{4}\) of original value; then the ratio of moment of inertia of mat in the two cases is (a) \(1: 3\) (b) \(4: 3\) (c) \(16: 9\) (d) \(3: 5\)
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Scientific Method Physics

Read Explanation

Linear Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation

Electromagnetism

Read Explanation

Quantum Physics

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.