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Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass \(M\) and radius \(R\) about an axis perpendicular to the hoop's plane at an edge.

Short Answer

Expert verified
The moment of inertia of the hoop about the given axis is \( MR^{2} \).

Step by step solution

01

Recognize the Problem

Our aim is to find the moment of inertia \(I\) of a hoop with mass \(M\) and radius \(R\). The axis is perpendicular to the hoop's plane at its edge.
02

Understanding the relevant physics

The formula we need to use is: \(dI = r^{2} dm\), where \(dI\) is the contribution to the moment of inertia from a mass element, \(dm\) is the mass of that element and \(r\) is the perpendicular distance of the mass element from the axis of rotation.
03

Setting up the integral

We can imagine the ring as made up of a large number of small mass elements. The distance \(r\) for every mass element from the axis of rotation is \(R\). So, the moment of inertia of each element is \(R^{2}dm\). For the entire hoop, we get the total moment of inertia by integrating over the entire mass: \(I= \int R^{2} dm\). Because R is a constant, we can bring it outside of the integral.
04

Evaluating the Integral

We need to evaluate the integral over the entire mass. We therefore integrate from 0 to M: \(I= R^{2} \int_0^{M} dm\). The integral of dm from 0 to M is just M. So, we obtain: \(I= MR^{2}\).
05

Interpret the result

This is the moment of inertia of a thin-walled hoop of mass \(M\) and radius \(R\) about an axis passing through its edge and perpendicular to its plane. The moment of inertia is proportional to the mass and the square of the radius, consistent with more general results for wider objects.

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Key Concepts

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

Hoop
A hoop is a simple yet fascinating object in physics. It's essentially a thin-walled, hollow ring. Think of a hula hoop or a bicycle tire. These objects may look simple, but they have interesting properties, especially in rotation. When an object like a hoop rotates, each part of it tries to continue in a straight line, which results in special physical behaviors.
Hoops are unique because of their uniform shape, which makes them perfect for studying fundamental physical properties like the moment of inertia. Every point on a hoop is the same distance from the center, which simplifies many calculations. This symmetry plays a vital role in understanding the hoop's dynamics when it rotates around different axes.
  • Thin-walled rings, making them ideal for specific calculations.
  • Uniform distribution of mass around a central axis.
  • Essential for exploring concepts like rotational motion and moment of inertia.
The hoop's characteristics make it an excellent subject for exercises in physics, especially when discussing rotations.
Axis of Rotation
The axis of rotation is the imaginary line around which an object like a hoop rotates. In this problem, the axis is perpendicular to the hoop's plane and passes through its edge. Imagine holding a bicycle tire and spinning it like a pinwheel. The rod or line you imagine holding it by is the axis of rotation.
The placement of this axis dramatically affects how easy or difficult it is to spin an object. In general, the further a mass is from the axis, the more effort is needed to rotate it. This relationship is crucial in understanding the moment of inertia. For the hoop, being a ring with all its mass concentrated at the edge, the choice of axis makes a significant impact on the calculation.
  • An imaginary line defining the rotation direction.
  • Greatly influences the rotational dynamics.
  • Determines how mass distribution affects the ease of rotation.
Understanding the axis of rotation helps in visualizing how objects rotate and predicts the energy needed to change these rotations.
Integral Calculus
Integral calculus is the branch of mathematics that helps us calculate quantities like area, volume, and, in the case of the hoop, the moment of inertia. It involves summing up an infinite number of infinitesimally small quantities, which sounds complex but is a powerful tool in physics.
To find the moment of inertia, we used an integral to sum up contributions from each tiny mass element of the hoop. Integral calculus allows us to handle continuous quantities with ease, like how the entire mass of the hoop contributes to its rotation.
  • Deals with the concept of integration and summing small contributions.
  • Essential for calculations where continuous change is involved.
  • Makes it possible to compute properties like the moment of inertia from mass distribution.
In physics, integral calculus helps to tie together the concepts of geometry, algebra, and dynamical systems, making it an indispensable tool.
Mass Distribution
Mass distribution refers to how mass is spread out in an object. For the hoop, all the mass is evenly distributed along its circular path. This even distribution makes it much simpler to apply physics concepts since each part of the hoop is equidistant from the chosen axis of rotation.
This uniform mass distribution allows us to express the moment of inertia in a simple form. Each part of the hoop contributes equally to its rotational behavior about the edge axis. Understanding the distribution is key to determining how an object's shape and mass affect its motion.
  • Crucial for calculating rotational properties.
  • Simplifies the calculations for uniform objects like hoops.
  • Determines the moment of inertia based on distance from the axis.
In this problem, knowing that mass is evenly distributed enabled us to confirm that the moment of inertia is simply proportional to the mass and the radius squared.

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Most popular questions from this chapter

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. \(\mathbf{P 9 . 6 4}\) ). There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is \(3.71 \mathrm{~m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to \(15.0 \mathrm{~kg}\) and allowed to fall through \(5.00 \mathrm{~m},\) it gives \(250.0 \mathrm{~J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0 \mathrm{~kg}\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the \(15.0 \mathrm{~kg}\) mass be moving on Mars just as the drum gained \(250.0 \mathrm{~J}\) of kinetic energy?

A new species of eel is found to have the same mass but onequarter the length and twice the diameter of the American eel. How does its moment of inertia for spinning around its long axis compare to that of the American eel? The new species has (a) half the moment of inertia as the American eel; (b) the same moment of inertia as the American eel; (c) twice the moment of inertia as the American eel; (d) four times the moment of inertia as the American eel.

A uniform sphere made of modeling clay has radius \(R\) and moment of inertia \(I_{1}\) for rotation about a diameter. It is flattened to a disk with the same radius \(R .\) In terms of \(I_{1},\) what is the moment of inertia of the disk for rotation about an axis that is at the center of the disk and perpendicular to its flat surface?

\(9.88^{\circ}\) DATA You are analyzing the motion of a large flywheel that has radius \(0.800 \mathrm{~m}\). In one test run, the wheel starts from rest and turns in a horizontal plane with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration \(a\) of a point on the rim of the flywheel as a function of the angle \(\theta-\theta_{0}\) through which the wheel has turned. You collect these results: $$ \begin{array}{l|cccccccc} \boldsymbol{\theta}-\boldsymbol{\theta}_{\mathbf{0}}(\mathbf{r a d}) & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 \\ \hline \boldsymbol{a}\left(\mathbf{m} / \mathbf{s}^{\mathbf{2}}\right) & 0.678 & 1.07 & 1.52 & 1.98 & 2.45 & 2.92 & 3.39 & 3.87 \end{array} $$ Construct a graph of \(a^{2}\left(\right.\) in \(\left.\mathrm{m}^{2} / \mathrm{s}^{4}\right)\) versus \(\left(\theta-\theta_{0}\right)^{2}\) (in rad \(^{2}\) ). (a) What are the slope and \(y\) -intercept of the straight line that gives the best fit to the data? (b) Use the slope from part (a) to find the angular acceleration of the flywheel. (c) What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of \(135^{\circ} ?\) (d) When the flywheel has turned through an angle of \(90.0^{\circ},\) what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius \(0.180 \mathrm{~m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A small block with mass \(0.340 \mathrm{~kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-9.00 \mathrm{~J}\) of work as the block descends \(3.00 \mathrm{~m}\). What is the magnitude of the angular velocity of the wheel after the block has descended \(3.00 \mathrm{~m} ?\)

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