/*! 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 72 A CD has a mass of \(15.0 \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A CD has a mass of \(15.0 \mathrm{~g}\), an inner diameter of \(1.5 \mathrm{~cm},\) and an outer diameter of \(11.9 \mathrm{~cm} .\) Suppose you toss it, causing it to spin at a rate of 4.3 revolutions per second. a) Determine the moment of inertia of the \(\mathrm{CD}\), approximating its density as uniform. b) If your fingers were in contact with the CD for 0.25 revolutions while it was acquiring its angular velocity and applied a constant torque to it, what was the magnitude of that torque?

Short Answer

Expert verified
\(I = \frac{1}{2}(15)(0.00005625 + 0.00354025)\) \(I = 7.5(0.0035965)\) \(I = 0.027\mathrm{~kg\cdot m^2}\) The moment of inertia of the CD is approximately \(0.027\mathrm{~kg\cdot m^2}\).

Step by step solution

01

a) Determine the moment of inertia of the CD

First, we need the formula for the moment of inertia of a thin circular ring, which is given by: \(I = \frac{1}{2}m(R_{1}^{2} + R_{2}^{2})\), where \(I\) is the moment of inertia, \(R_{1}\) is the inner radius, \(R_{2}\) is the outer radius, and \(m\) is the mass. Given that the inner diameter is \(1.5\mathrm{~cm}\), we can find the inner radius: \(R_{1} = \frac{1.5}{2} = 0.75\mathrm{~cm} =0.0075\mathrm{~m}\). Similarly, the outer diameter is \(11.9\mathrm{~cm}\), so the outer radius is \(R_{2} = \frac{11.9}{2} = 5.95\mathrm{~cm} = 0.0595\mathrm{~m}\). Now, we can plug the values into the formula to calculate the moment of inertia: \(I = \frac{1}{2}(15)(0.0075^2 + 0.0595^2)\)

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Angular Velocity
Angular velocity measures how quickly an object rotates or spins around an axis. Imagine a CD spinning like a top. Its angular velocity tells us how many rotations it completes in a given time, typically measured in radians per second. In the problem, the CD spins at 4.3 revolutions per second, which you can convert to radians per second by knowing there are \(2\pi\) radians in one revolution.

Angular velocity is essential because it helps us understand the kinetic energy and momentum of rotating objects. When you toss a CD and make it spin, you're imparting energy that translates into this spinning motion.
  • Angular velocity (\(\omega\)) = \(\frac{\text{Angle in radians}}{\text{Time}}\)
  • 1 revolution = \(2\pi\) radians

By grasping angular velocity, you can better understand how different speeds of rotation affect an object's dynamic behavior and how applied forces or torques change this velocity.
Exploring Torque
Torque is like the rotational equivalent of a force. It causes an object to start rotating, stop, or change its rotational speed. When you apply a torque to an object, such as the CD in this exercise, you're exerting a twisting force. This is similar to opening a jar lid—a force applied at a distance from the axis turns it.

Torque is measured in Newton-meters (Nm) and is calculated by multiplying the force applied by the distance from the force's line of action to the axis of rotation. In the given problem, your fingers supply the torque necessary for the CD to reach its angular velocity.
  • Torque (\(\tau\)) = Force (\(F\)) \(\times\) Lever arm (\(r\))
  • The longer the lever arm, the more torque generated.
Understanding torque helps explain how different factors influence the rotational speed of an object. Adjusting the applied force or the distance can significantly increase or decrease the torque.
Principles of Circular Motion
Circular motion occurs when an object moves along a circular path. This motion is characterized by parameters such as angular velocity and centripetal force. For example, when a CD spins, it showcases circular motion, where every point on the CD follows a circular trajectory at any point in time.

In circular motion, each part of the object is at the same distance from a fixed point, commonly called the center of rotation. This consistent motion is possible due to centripetal force, which pulls the object toward the center and keeps it from flying off the circular path.
  • Angular Velocity \(\omega = \frac{\theta}{t}\), with \(\theta\) being angular displacement in radians.
  • Centripetal Force \(F_c = \frac{mv^2}{r}\), where \(m\) is mass, \(v\) is velocity, and \(r\) is radius.
Circular motion concepts explain how forces and velocities interact to maintain objects on their curved paths and help predict how changes in force or speed impact the motion.
Uniform Density and Its Implications
When an object like the CD has uniform density, its mass is evenly distributed throughout its volume. This type of density matters when calculating properties such as the moment of inertia, which depends on the mass distribution relative to the rotation axis.

For the CD, assuming uniform density simplifies calculations. We use formulas that require mass as an input, assuming it's spread out evenly without any variations in different sections. The even distribution assists in determining the exact inertia accurately.
  • Uniform Density \(\rho = \frac{m}{V}\), with \(m\) as mass and \(V\) as volume.
  • This assumption makes mathematical modeling manageable and often more accurate for calculating rotational dynamics.
Understanding uniform density is crucial for simplifying problems involving moment of inertia, ensuring we compute accurate results while analyzing physical behavior of rotating objects.

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 ballistic pendulum consists of an arm of mass \(M\) and length \(L=0.48 \mathrm{~m} .\) One end of the arm is pivoted so that the arm rotates freely in a vertical plane. Initially, the arm is motionless and hangs vertically from the pivot point. A projectile of the same mass \(M\) hits the lower end of the arm with a horizontal velocity of \(V=3.6 \mathrm{~m} / \mathrm{s}\). The projectile remains stuck to the free end of the arm during their subsequent motion. Find the maximum angle to which the arm and attached mass will swing in each case: a) The arm is treated as an ideal pendulum, with all of its mass concentrated as a point mass at the free end. b) The arm is treated as a thin rigid rod, with its mass evenly distributed along its length.

A uniform solid sphere of mass \(M\) and radius \(R\) is rolling without sliding along a level plane with a speed \(v=3.00 \mathrm{~m} / \mathrm{s}\) when it encounters a ramp that is at an angle \(\theta=23.0^{\circ}\) above the horizontal. Find the maximum distance that the sphere travels up the ramp in each case: a) The ramp is frictionless, so the sphere continues to rotate with its initial angular speed until it reaches its maximum height. b) The ramp provides enough friction to prevent the sphere from sliding, so both the linear and rotational motion stop (instantaneously).

Does a particle traveling in a straight line have an angular momentum? Explain.

A basketball of mass \(610 \mathrm{~g}\) and circumference \(76 \mathrm{~cm}\) is rolling without slipping across a gymnasium floor. Treating the ball as a hollow sphere, what fraction of its total kinetic energy is associated with its rotational motion? a) 0.14 b) 0.19 c) 0.29 d) 0.40 e) 0.67

A solid sphere of radius \(R\) and mass \(M\) is placed at a height \(h_{0}\) on an inclined plane of slope \(\theta\). When released, it rolls without slipping to the bottom of the incline. Next, a cylinder of same mass and radius is released on the same incline. From what height \(h\) should it be released in order to have the same speed as the sphere at the bottom?

See all solutions

Recommended explanations on Physics Textbooks

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.