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Chapter 9: Problem 52

A string is wrapped several times around the rim of a small hoop with a radius of 0.0800 \(\mathrm{m}\) and a mass of 0.180 \(\mathrm{kg}\) . If the free end of the string is held in place and the hoop is released from rest (see Figure 9.30), calculate the angular speed of the rotating hoop after it has descended 0.750 \(\mathrm{m} .\)

Short Answer

Expert verified
The angular speed of the hoop is approximately 47.90 rad/s.

Step by step solution

01

Understanding the Problem

The problem involves a hoop with a string wrapped around it, which descends a distance of 0.750 m as it unwinds. We need to find the angular speed of the hoop at this point.
02

Establishing the Energy Balance

Since the hoop is released from rest, its initial kinetic energy is zero. As it unwinds, gravitational potential energy is converted into rotational kinetic energy. We will use the principle of energy conservation to solve the problem.
03

Applying the Energy Conservation Principle

The initial potential energy is given by \( U_i = mgh \), where \( m = 0.180 \ \mathrm{kg} \) is the mass, \( g = 9.81 \ \mathrm{m/s^2} \) is the acceleration due to gravity, and \( h = 0.750 \ \mathrm{m} \) is the descent height. This energy converts into rotational kinetic energy \( K = \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular speed.
04

Calculating the Moment of Inertia

For a hoop, the moment of inertia \( I \) about its center is \( I = mr^2 \), where \( r = 0.0800 \ \mathrm{m} \) is the radius. Substituting the values, we find \( I = 0.180 \ \mathrm{kg} \times (0.0800 \ \mathrm{m})^2 = 0.001152 \ \mathrm{kg \, m^2} \).
05

Solving for Angular Speed

Using the conservation equation: \( mgh = \frac{1}{2}I\omega^2 \), replace \( mh = 0.180 \times 9.81 \times 0.750 \) and \( I \) from the previous step. This gives us \( 1.32375 = \frac{1}{2} \times 0.001152 \times \omega^2 \). Solving for \( \omega \), we get \( \omega = \sqrt{\frac{2 \times 1.32375}{0.001152}} \approx 47.90 \ \mathrm{rad/s} \).

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Key Concepts

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

Energy Conservation
Energy conservation is a fundamental principle stating that energy cannot be created or destroyed, only transformed from one form to another. In this problem, as the hoop descends, its gravitational potential energy is converted into rotational kinetic energy. This transformation allows us to calculate the hoop's angular speed using only its physical characteristics and the distance it descends.
  • Gravitational potential energy is initially stored in the system when the hoop is at rest.
  • As the hoop unwinds and falls, this potential energy converts into rotational kinetic energy.
  • The total energy in the system remains constant, allowing energy equations to be used for solving problems.

By rigorously applying energy conservation principles, we can confidently determine dynamic values, like the hoop's angular speed, when it reaches a certain depth.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It plays a role similar to mass in linear motion. In this scenario, the moment of inertia determines how the hoop responds as it begins rotating.
  • The formula for the moment of inertia of a hoop around its central axis is given by \( I = mr^2 \).
  • Here, \( m \) represents the mass of the hoop, and \( r \) represents its radius.
Recognizing the role of moment of inertia is crucial for solving problems involving rotational dynamics, as it provides a way to relate mass distribution to angular response.
Gravitational Potential Energy
Gravitational potential energy is energy an object possesses due to its position in a gravitational field. The higher the object, the more gravitational potential energy it has.
To calculate this energy for the hoop, we use the equation:
\( U_i = mgh \), where:
  • \( m \) is the mass of the hoop (0.180 kg).
  • \( g \) is the acceleration due to gravity (approximately 9.81 \( \mathrm{m/s^2} \)).
  • \( h \) is the descent height (0.750 m).
This calculated energy is initially stored in the hoop before it descends. As it unwinds, gravitational potential energy diminishes, transforming into kinetic forms.
Rotational Kinetic Energy
Rotational kinetic energy is the energy due to an object's rotation, comparable to linear kinetic energy for moving objects. For rotating bodies like the hoop, this kinetic energy measures how much effort it takes to set the body into rotation.
  • The formula for rotational kinetic energy is \( K = \frac{1}{2}I\omega^2 \).
  • Here, \( I \) is the moment of inertia, and \( \omega \) is the angular speed.
In this exercise, the hoop's descent transforms its potential energy into this form of kinetic energy, enabling us to calculate the angular speed. This conversion underscores the interconnected nature of different energy forms and the power of energy conservation components to unravel dynamic motion.

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Most popular questions from this chapter

\(\bullet\) The spin cycles of a washing machine have two angular speeds, 423 \(\mathrm{rev} / \mathrm{min}\) and 640 \(\mathrm{rev} / \mathrm{min.}\) The internal diameter of the drum is 0.470 \(\mathrm{m} .\) (a) What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed? (b) What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed? (c) Find the laundry's maximum tangential speed and the maximum radial acceleration, in terms of \(g\) .

\(\bullet\) A twirler's baton is made of a slender metal cylinder of mass \(M\) and length \(L .\) Each end has a rubber cap of mass \(m,\) and you can accurately treat each cap as a particle in this problem. Find the total moment of inertia of the baton about the usual twirling axis (perpendicular to the baton through its center).

A flywheel in a motor is spinning at 500.0 \(\mathrm{rpm}\) when a power failure suddenly occurs. The flywheel has mass 40.0 \(\mathrm{kg}\) and diameter 75.0 \(\mathrm{cm} .\) The power is off for 30.0 \(\mathrm{s}\) , and during this time the flywheel slown down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 200.0 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Compound objects. Moment of inertia is a scalar. There- fore, if several objects are connected together, the moment of inertia of this compound object is simply the scalar (algebraic) sum of the moments, of inertia of each of the component. objects. Use this principle to answer each of the following questions about the moment of inertia of compound objects: (a) A thin uniform 2.50 \(\mathrm{kg}\) bar 1.50 \(\mathrm{m}\) long has a small 1.25 \(\mathrm{kg}\) mass glued to each end. What is the moment of inertia of this object about an axis perpendicular to the bar through its center? (b) What is the moment of inertia of the object in part (a) about an axis perpendicular to the bar at one end? (c) \(A 725\) g metal wire is bent into the shape of a hoop 60.0 \(\mathrm{cm}\) in diameter. Six wire spokes, each of mass 112 g, are added from the center of the hoop to the rim. What is the moment of inertia of this object about an axis perpendicular to it through its center?

\(\bullet\) A size-5 soccer ball of diameter 22.6 \(\mathrm{cm}\) and mass 426 \(\mathrm{g}\) rolls up a hill without slipping, reaching a maximum height of 5.00 \(\mathrm{m}\) above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it then have?
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