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Chapter 8: Problem 54

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 each object. (a) a hoop (b) a solid cylinder (c) a solid sphere (d) a hollow spherical shell

Short Answer

Expert verified
The magnitudes of the angular momenta of the hoop, solid cylinder, solid sphere and hollow spherical shell are \(2.7216 kg m^2/s\), \(1.3608 kg m^2/s\), \(1.08864 kg m^2/s\), and \(1.8144 kg m^2/s\) respectively.

Step by step solution

01

Find Moments of Inertia

Use the given formulas for moments of inertia for each type of object with radius \(r=0.180\) m and mass \(m = 2.40\) kg. \n For a hoop \(I = m r^2 = 2.40 kg * (0.180 m)^2 = 0.07776 kg m^2\), \n for a solid cylinder \(I = 0.5 m r^2 = 0.5 * 2.40 kg * (0.180 m)^2 = 0.03888 kg m^2\), \n for a solid sphere \(I = 0.4 m r^2 = 0.4 * 2.40 kg * (0.180 m)^2 = 0.031104 kg m^2\), \n and for a hollow spherical shell \(I = 2/3 m r^2 = 2/3 * 2.40 kg * (0.180 m)^2 = 0.05184 kg m^2\). \n This gives the moments of inertia for each type of object.
02

Find Angular Momentum

Find the magnitude of the angular momentum \(L\) for each type of object using the formula \(L=I\omega\), with the given angular speed \(\omega=35.0 rad/s\) and the moments of inertia found in Step 1. For a hoop, \(L = 0.07776 kg m^2 * 35.0 rad/s = 2.7216 kg m^2/s\). \n For a solid cylinder, \(L = 0.03888 kg m^2 * 35.0 rad/s = 1.3608 kg m^2/s\). \n For a solid sphere, \(L = 0.031104 kg m^2 * 35.0 rad/s = 1.08864 kg m^2/s\). \n And for a hollow spherical shell, \(L = 0.05184 kg m^2 * 35.0 rad/s = 1.8144 kg m^2/s\). \n This gives the magnitudes of the angular momenta of each type of object.

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Key Concepts

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

Moments of Inertia
The moment of inertia, often represented by the symbol \(I\), is a measure of an object's resistance to changes in its rotation. It is comparable to mass in linear motion, explaining how difficult it is to alter an object's rotational speed. The moment of inertia depends on both the mass of the object and the distance the mass is from the axis of rotation. For different shapes, there are specific formulas to calculate it:
  • For a hoop (or thin ring) rotating about its center, the moment of inertia is calculated as \(I = mr^2\).
  • A solid cylinder has its moment of inertia as \(I = 0.5mr^2\).
  • For a solid sphere, \(I = 0.4mr^2\).
  • The hollow spherical shell requires the formula \(I = \frac{2}{3}mr^2\).
Knowing the moment of inertia helps to predict how these objects will behave under various rotational forces.
Angular Speed
Angular speed, denoted as \(\omega\), measures how fast an object rotates or revolves relative to another point, typically in radians per second. It can help quantify the rotational motion of an object. Imagine it as how many 'turns' the object makes in a given period.For all objects discussed in our problem, their angular speed is given as \(35.0 \, \text{rad/s}\). This means these objects are spinning around their respective axes quite quickly. Angular speed is a crucial part of calculating other properties of rotation like angular momentum, as seen in our problem.In contrast to linear speed, which measures how fast something moves along a path, angular speed deals purely with rotational movement. This is important in many applications, from the wheels of a car to the planets orbiting around stars.
Rotational Dynamics
Rotational dynamics explores how objects behave as they rotate. It's the study of the forces and torques that influence rotational motion. Just like linear dynamics focuses on objects moving in straight lines, rotational dynamics examines objects turning around a pivot or axis.Some key components of rotational dynamics include:
  • Torque: It is the rotational equivalent of force. Torque causes an object to start rotating and is calculated as the product of force and the radius (distance from pivot point).
  • Angular momentum: This concept combines both angular speed and moment of inertia to describe the rotational state of an object. It is given by \(L = I \omega\). This formula was applied in the textbook exercise to understand how various object shapes maintain their rotation.
  • Conservation of Angular Momentum: If no external torques act on the system, the total angular momentum remains constant. This principle explains why an ice skater can spin faster by pulling in their arms.
Understanding rotational dynamics helps predict, manipulate, and optimize how objects rotate, essential for engineering tasks, sports analysis, and even astrophysics.

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Most popular questions from this chapter

QIC A \(0.00500-\mathrm{kg}\) bullet traveling horizontally with a speed of \(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) enters an \(18.0-\mathrm{kg}\) door, embedding itself \(10.0 \mathrm{~cm}\) from the side opposite the hinges as in Figure P8.56. The \(1.00-\mathrm{m}\)-wide door is free to swing on its hinges. (a) Before it hits the door, does the bullet have angular momentum relative the door's axis of rotation? Explain. (b) Is mechanical energy con-served in this collision? Answer without doing a calculation. (c) At what angular speed does the door swing open immediately after the collision? (The door has the same moment of inertia as a rod with axis at one end.) (d) Calculate the energy of the door-bullet system and determine whether it is less than or equal to the kinetic energy of the bullet before the collision.

An object of mass \(M=\) \(12.0 \mathrm{~kg}\) is attached to a cond that is wrapped around a wheel of radius \(r=10.0 \mathrm{~cm}\) (Fig. P8.70). The acceleration of the object down the frictionless incline is measured to be \(a=2.00 \mathrm{~m} / \mathrm{s}^{2}\) andthe incline makes an angle \(\theta=37.0^{\circ}\) with the horizontal. Assuming the axle of the wheel to be frictionless, determine (a) the tension in the rope, (b) the moment of inertia of the wheel, and (c) the angular speed of the wheel \(2.00 \mathrm{~s}\) after it begins rotating, starting from rest.

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.

An airliner lands with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). Each wheel of the plane has a radius of \(1.25 \mathrm{~m}\) and a moment of inertia of \(110 \mathrm{~kg} \cdot \mathrm{m}^{2}\). At touchdown, the wheels begin to spin under the action of friction. Each wheel supports a weight of \(1.40 \times 10^{4} \mathrm{~N}\), and the wheels attain their angular speed in \(0.480 \mathrm{~s}\) while rolling without slipping. What is the coefficient of kinetic friction between the wheels and the runway? Assume that the speed of the plane is constant.

Two astronauts (Fig. P8.72), each having a mass of \(75.0 \mathrm{~kg}\), are connected by a \(10.0-\mathrm{m}\) rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum and (b) the rotational energy of the system. By pulling on the rope, the astronauts shorten the distance between them to \(5.00 \mathrm{~m}\). (c) What is the new angular momentum of the system? (d) What are their new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronauts in shortening the rope?
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