Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 72

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 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 child builds a simple cart consisting of a \(60.0 \mathrm{~cm}\) by \(1.20 \mathrm{~m}\) sheet of plywood of mass \(8.00 \mathrm{~kg}\) and four wheels, each \(20.0 \mathrm{~cm}\) in diameter and with a mass of \(2.00 \mathrm{~kg}\). It is released from the top of a \(15.0^{\circ}\) incline that is \(30.0 \mathrm{~m}\) long. Find the speed at the bottom. Assume that the wheels roll along the incline without slipping and that friction between the wheels and their axles can be neglected.

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

Using the conservation of mechanical energy, calculate the final speed and the acceleration of a cylindrical object of mass \(M\) and radius \(R\) after it rolls a distance \(s\) without slipping along an inclined plane of angle \(\theta\) with respect to the horizontal

A 25.0 -kg boy stands \(2.00 \mathrm{~m}\) from the center of a frictionless playground merry-go-round, which has a moment of inertia of \(200 . \mathrm{kg} \mathrm{m}^{2} .\) The boy begins to run in a circular path with a speed of \(0.600 \mathrm{~m} / \mathrm{s}\) relative to the ground. a) Calculate the angular velocity of the merry-go-round. b) Calculate the speed of the boy relative to the surface of the merry-go- round.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Force

Read Explanation

Dynamics

Read Explanation

Linear Momentum

Read Explanation

Nuclear Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation
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.

Sign-up for free