Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 71

The mass of a flywheel is \(5.6 \times 10^{4} \mathrm{kg} .\) This particular flywheel has its mass concentrated at the rim of the wheel. If the radius of the wheel is \(2.6 \mathrm{m}\) and it is rotating at 350 rpm, what is the magnitude of its angular momentum?

Short Answer

Expert verified
Answer: The magnitude of the flywheel's angular momentum is approximately 1.3837 * 10^7 kg*m^2/s.

Step by step solution

01

Determine the moment of inertia of the flywheel

Using the formula for a hoop's moment of inertia, calculate the moment of inertia (I) of the flywheel: I = MR^2 M = 5.6 * 10^4 kg (mass of the flywheel) R = 2.6 m (radius of the flywheel) I = (5.6 * 10^4 kg) * (2.6 m)^2 I = 3.7776 * 10^5 kg * m^2 (moment of inertia)
02

Convert rotational speed to angular velocity

The given rotational speed of the flywheel is 350 rpm. To convert rpm to radians per second, we can use the formula: ω = (2 * π radians * rotational_speed) / 60 seconds rotational_speed = 350 rpm ω = (2 * π * 350) / 60 ω ≈ 36.65 rad/s (angular velocity)
03

Calculate the angular momentum

Now that we have the moment of inertia and angular velocity, we can calculate the angular momentum (L) of the flywheel using the formula: L = Iω L = (3.7776 * 10^5 kg*m^2) * (36.65 rad/s) L ≈ 1.3837 * 10^7 kg*m^2/s The magnitude of the flywheel's angular momentum is approximately 1.3837 * 10^7 kg*m^2/s.

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

The Moon's distance from Earth varies between \(3.56 \times 10^{5} \mathrm{km}\) at perigee and \(4.07 \times 10^{5} \mathrm{km}\) at apogee. What is the ratio of its orbital speed around Earth at perigee to that at apogee?
Two children standing on opposite sides of a merry-goround (see Fig. 8.5 ) are trying to rotate it. They each push in opposite directions with forces of magnitude \(10.0 \mathrm{N} .\) (a) If the merry-go-round has a mass of $180 \mathrm{kg}\( and a radius of \)2.0 \mathrm{m},$ what is the angular acceleration of the merry-go-round? (Assume the merry-go-round is a uniform disk.) (b) How fast is the merry-go-round rotating after \(4.0 \mathrm{s} ?\)
A uniform diving board, of length \(5.0 \mathrm{m}\) and mass \(55 \mathrm{kg}\) is supported at two points; one support is located \(3.4 \mathrm{m}\) from the end of the board and the second is at \(4.6 \mathrm{m}\) from the end (see Fig. 8.19 ). What are the forces acting on the board due to the two supports when a diver of mass \(65 \mathrm{kg}\) stands at the end of the board over the water? Assume that these forces are vertical. (W) tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]
The radius of a wheel is \(0.500 \mathrm{m} .\) A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude $5.00 \mathrm{N},$ unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. (a) How much rope unwinds while the wheel makes 1.00 revolution? (b) How much work is done by the rope on the wheel during this time? (c) What is the torque on the wheel due to the rope? (d) What is the angular displacement \(\Delta \theta\), in radians, of the wheel during 1.00 revolution? (e) Show that the numerical value of the work done is equal to the product \(\tau \Delta \theta\)
A spinning flywheel has rotational inertia $I=400.0 \mathrm{kg} \cdot \mathrm{m}^{2} .\( Its angular velocity decreases from \)20.0 \mathrm{rad} / \mathrm{s}\( to zero in \)300.0 \mathrm{s}$ due to friction. What is the frictional torque acting?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Waves Physics

Read Explanation

Physics of Motion

Read Explanation

Particle Model of Matter

Read Explanation

Wave Optics

Read Explanation

Classical Mechanics

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