/*! 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 27 A cylinder with mass \(M\) and r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 27

A cylinder with mass \(M\) and radius \(R\) is rolling without slipping through a distance \(s\) along an inclined plane that makes an angle \(\theta\) with respect to the horizontal. Calculate the work done by (a) gravity, (b) the normal force, and (c) the frictional force.

Short Answer

Expert verified
Based on the provided solution, calculate the work done by the gravitational force, normal force, and frictional force on a cylinder rolling without slipping down an inclined plane with the following characteristics: 1. Angle of inclination (胃) = 30 degrees 2. Mass (M) of the cylinder = 5 kg 3. Radius (R) of the cylinder = 0.5 meters 4. The distance (s) the cylinder rolls = 10 meters Work done by gravity (Wg) = M * g * sin(胃) * s = 5 * 9.81 * sin(30) * 10 = 245.25 J Work done by normal force (Wn) = 0 Work done by frictional force (Wf) = 0 The work done by the gravitational force is 245.25 J, while the work done by normal force and frictional force are both 0 J.

Step by step solution

01

Calculate the Force Components

First, let's identify the force components based on the given information: - Gravitational force, Fg = M * g - Normal force, Fn, acts perpendicular to the inclined plane and is equal in magnitude to the component of gravitational force acting perpendicular to the inclined plane. - Frictional force, Ff, opposes the motion and acts tangentially to the inclined plane.
02

Calculate the Work Done by Gravity (Wg)

To calculate the work done by gravity, we need to find the component of gravitational force acting parallel to the inclined plane and multiply it with the displacement (s). The angle between the displacement and the force component will be 0 degrees. - Gravitational force component parallel to inclined plane: Fg_parallel = Fg * sin(胃) - Work done by gravity: Wg = Fg_parallel * s * cos(0) - Wg = M * g * sin(胃) * s
03

Calculate the Work Done by Normal Force (Wn)

Since the normal force acts perpendicular to the inclined plane, the angle between the displacement vector and the normal force vector is 90 degrees. - Work done by normal force: Wn = Fn * s * cos(90) - Wn = 0 The work done by the normal force on the cylinder is 0.
04

Calculate the Work Done by Frictional Force (Wf)

The frictional force acts tangentially to the point of contact between the cylinder and the inclined plane. Since the cylinder is rolling without slipping, the frictional force will not affect the linear displacement. Consequently, the work done by frictional force is 0. Wf = 0

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 91影视!

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

Determine the moment of inertia for three children weighing \(60.0 \mathrm{lb}, 45.0 \mathrm{lb}\) and \(80.0 \mathrm{lb}\) sitting at different points on the edge of a rotating merry-go-round, which has a radius of \(12.0 \mathrm{ft}\).

In another race, a solid sphere and a thin ring roll without slipping from rest down a ramp that makes angle \(\theta\) with the horizontal. Find the ratio of their accelerations, \(a_{\text {ring }} / a_{\text {sphere }}\)

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.

A space station is to provide artificial gravity to support long-term stay of astronauts and cosmonauts. It is designed as a large wheel, with all the compartments in the rim, which is to rotate at a speed that will provide an acceleration similar to that of terrestrial gravity for the astronauts (their feet will be on the inside of the outer wall of the space station and their heads will be pointing toward the hub). After the space station is assembled in orbit, its rotation will be started by the firing of a rocket motor fixed to the outer rim, which fires tangentially to the rim. The radius of the space station is \(R=50.0 \mathrm{~m},\) and the mass is \(M=2.40 \cdot 10^{5} \mathrm{~kg} .\) If the thrust of the rocket motor is \(F=1.40 \cdot 10^{2} \mathrm{~N},\) how long should the motor fire?

A uniform solid cylinder of mass \(M=5.00 \mathrm{~kg}\) is rolling without slipping along a horizontal surface. The velocity of its center of mass is \(30.0 \mathrm{~m} / \mathrm{s}\). Calculate its energy.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Medical Physics

Read Explanation

Physics of Motion

Read Explanation

Dynamics

Read Explanation

Quantum Physics

Read Explanation

Space Physics

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.