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Each of the following objects has a radius of \(0.180 \mathrm{~m}\) and a mass of \(2.40 \mathrm{~kg}\), and each rotates about an axis through its center (as in Table 8.1) with an angular speed of \(35.0 \mathrm{rad} / \mathrm{s}\). Find the magnitude of the angular momentum of cach object. (a) a hoop (b) a solid cylinder (c) a solid sphere (d) a hollow spherical shell

Short Answer

Expert verified
The magnitude of the angular momentum for each rotating object is as follows: (a) hoop: \(10.4 \mathrm{~kg \cdot m^2/s}\), (b) solid cylinder: \(5.20 \mathrm{~kg \cdot m^2/s}\), (c) solid sphere: \(3.47 \mathrm{~kg \cdot m^2/s}\), (d) hollow spherical shell: \(4.63 \mathrm{~kg \cdot m^2/s}\).

Step by step solution

01

Identify the Key Constants

The mass \(m\) for each object is \(2.40 \mathrm{~kg}\), the radius \(r\) is \(0.180 \mathrm{~m}\), and the angular speed \(ω\) is \(35.0 \mathrm{~rad/s}\).
02

Identify the Moment of Inertia for Each Object

For each of these objects, the moment of inertia \(I\) is defined as follows: (a) For a hoop, \(I = m r^2\). (b) For a solid cylinder, \(I = 1/2 m r^2\). (c) For a solid sphere, \(I = 2/5 m r^2\). (d) For a hollow spherical shell, \(I = 2/3 m r^2\).
03

Calculate the Angular Momentum for Each Object

Angular momentum \(L\) can now be calculated for each object as follows, where angular momentum \(L = Iω\). (a) For a hoop, \(L = m r^2 ω\). (b) For a solid cylinder, \(L = 1/2 m r^2 ω\). (c) For a solid sphere, \(L = 2/5 m r^2 ω\). (d) For a hollow spherical shell, \(L = 2/3 m r^2 ω\). Calculate these for the given values of m, r, and ω (2.40 kg, 0.180 m, and 35.0 rad/s respectively) to find the angular momentum.

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Key Concepts

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

Understanding Moment of Inertia
The moment of inertia is a fundamental concept in physics that describes how mass is distributed in an object about an axis of rotation. It can be likened to the role mass plays in linear motion, where mass resists changes in linear velocity. The moment of inertia does the same for rotational motion, resisting changes in angular velocity. Key points to consider about moment of inertia:
  • The value of the moment of inertia, often denoted as \(I\), depends on the shape and distribution of mass within the object.
  • For different objects, this distribution differs, which is why each shape has a unique formula for calculating \(I\).
For example, in this exercise:
  • A hoop, with its mass concentrated at the rim, has an inertia \(I = m r^2\).
  • A solid cylinder's mass is more evenly distributed from center to edge, leading to \(I = \frac{1}{2} m r^2\).
  • For a solid sphere, with mass distributed throughout its volume, inertia is \(I = \frac{2}{5} m r^2\).
  • A hollow spherical shell, where mass is distributed on the surface, has \(I = \frac{2}{3} m r^2\).
Understanding these calculations is crucial in physics education, as it helps in comprehending how different objects behave when they rotate.
The Role of Angular Speed
Angular speed is the rate at which an object rotates around an axis. It is crucial in the study of rotational dynamics as it defines how quickly an object is spinning and is a key component in calculating angular momentum.Let's explore this further:
  • It's measured in radians per second (rad/s), where one complete rotation equals \(2\pi\) radians.
  • Angular speed, denoted as \(\omega\), indicates the rotational angle covered per unit of time.
  • This parameter is similar to linear speed but applies to circular motion.
In the context of the exercise, the angular speed is given as \(35.0 \, \text{rad/s}\) for all the objects. This constant rate allows us to directly calculate angular momentum from the moment of inertia using the formula \(L = I\omega\).Understanding angular speed is vital not only in physics education but also in real-world applications like engineering, where equipment and devices often rely on precise rotational speeds to function correctly.
Enhancing Physics Education through Core Concepts
Physics education involves delving into core concepts such as moment of inertia and angular speed, which serve as building blocks for understanding complex systems in rotation. A clear grasp of these ideas can greatly enhance a student’s ability to tackle physics problems effectively. Consider these learning strategies:
  • Start with fundamental definitions and properties, using simple examples like hoops and spheres.
  • Visual aids can help - diagrams showing how mass distribution affects moment of inertia are particularly helpful.
  • Practical experiments, like measuring rotational speeds and calculating corresponding moments of inertia, solidify theoretical knowledge.
  • Regularly engaging with exercises that challenge students to apply these concepts to different scenarios is essential.
The integration of these strategies in physics education not only aids in understanding the current exercise but lays a strong foundation for more advanced topics. Effective education in physics ensures students are well-prepared to understand and innovate in fields where these principles are applied.

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Most popular questions from this chapter

A large grinding wheel in the shape of a solid cylinder of radius \(0.330 \mathrm{~m}\) is free to rotate on a frictionless, vertical axle. A constant tangential force of \(250 \mathrm{~N}\) applied to its edge causes the wheel to have an angular acceleration of \(0.940 \mathrm{rad} / \mathrm{s}^{2}\). (a) What is the moment of inertia of the wheel? (b) What is the mass of the wheel? (c) If the wheel starts from rest, what is its angular velocity after \(5.00\) s have elapsed, assuming the force is acting during that time?

A solid, horizontal cylinder of mass \(10.0 \mathrm{~kg}\) and radius \(1.00 \mathrm{~m}\) rotates with an angular speed of \(7.00 \mathrm{rad} / \mathrm{s}\) about a fixed vertical axis through its center. A \(0.250-\mathrm{kg}\) piece of putty is dropped vertically onto the cylinder at a point \(0.900 \mathrm{~m}\) from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.

The large quadriceps muscle in the upper leg terminates at its lower end in a tendon attached to the upper end of the tibia (Fig. \(\mathrm{P} 8.29 \mathrm{a}\) ). The forces on the lower leg when the leg is extended are modeled as in Figure P8.29b, where \(\vec{T}\) is the force of tension in the tendon, \(\vec{w}\) is the force of gravity acting on the lower leg, and \(\overrightarrow{\mathbf{F}}\) is the force of gravity acting on the foot. Find \(\vec{T}\) when the tendon is at an angle of \(25.0^{\circ}\) with the tibia, assuming that \(w=30.0 \mathrm{~N}\), \(F=12.5 \mathrm{~N}\), and the leg is extended at an angle \(\theta\) of \(40.0^{\circ}\) with the vertical. Assume that the center of gravity of the lower leg is at its center and that the tendon attaches to the lower leg at a point one-fifth of the way down the leg

A simple pendulum consists of a small object of mass 3.0 kg hanging at the end of a \(2.0\) -m-long light string that is connected to a pivot point. (a) Calculate the magnitude of the torque (due to the force of gravity) about this pivot point when the string makes a \(5.0^{\circ}\) angle with the vertical. (b) Does the torque increase or decrease as the angle increases? Explain.

A person bending forward to lift a load "with his back" (Fig. P8.17a) rather than "with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure P8.17b of a person bending forward to lift a \(200-\mathrm{N}\) object. The spine and upper body are represented as a uniform horizontal rod of weight \(350 \mathrm{~N}\), pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is \(12.0^{\circ}\). Find the tension in the back muscle and the compressional force in the spine.

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