Log out Go to App
Go to App

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 (θ) = 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(θ) * 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(θ) - Work done by gravity: Wg = Fg_parallel * s * cos(0) - Wg = M * g * sin(θ) * 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.

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 round body of mass \(M\), radius \(R,\) and moment of inertia \(I\) about its center of mass is struck a sharp horizontal blow along a line at height \(h\) above its center (with \(0 \leq h \leq R,\) of course). The body rolls away without slipping immediately after being struck. Calculate the ratio \(I /\left(M R^{2}\right)\) for this body.

A wagon wheel is made entirely of wood. Its components consist of a rim, 12 spokes, and a hub. The rim has mass \(5.2 \mathrm{~kg}\), outer radius \(0.90 \mathrm{~m}\), and inner radius \(0.86 \mathrm{~m}\). The hub is a solid cylinder with mass \(3.4 \mathrm{~kg}\) and radius \(0.12 \mathrm{~m} .\) The spokes are thin rods of mass \(1.1 \mathrm{~kg}\) that extend from the hub to the inner side of the rim. Determine the constant \(c=I / M R^{2}\) for this wagon wheel.

A uniform rod of mass \(M=250.0 \mathrm{~g}\) and length \(L=50.0 \mathrm{~cm}\) stands vertically on a horizontal table. It is released from rest to fall. a) What forces are acting on the rod? b) Calculate the angular speed of the rod, the vertical acceleration of the moving end of the rod, and the normal force exerted by the table on the rod as it makes an angle \(\theta=45.0^{\circ}\) with respect to the vertical. c) If the rod falls onto the table without slipping, find the linear acceleration of the end point of the rod when it hits the table and compare it with \(g\).

A disk of clay is rotating with angular velocity \(\omega .\) A blob of clay is stuck to the outer rim of the disk, and it has a mass \(\frac{1}{10}\) of that of the disk. If the blob detaches and flies off tangent to the outer rim of the disk, what is the angular velocity of the disk after the blob separates? a) \(\frac{5}{6} \omega\) b) \(\frac{10}{11} \omega\) c) \(\omega\) d) \(\frac{11}{10} \omega\) e) \(\frac{6}{5} \omega\)

It is harder to move a door if you lean against it (along the plane of the door) toward the hinge than if you lean against the door perpendicular to its plane. Why is this so?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Physics of Motion

Read Explanation

Electromagnetism

Read Explanation

Linear Momentum

Read Explanation

Energy Physics

Read Explanation

Measurements

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