/*! 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 119 The solid ball of radius \(r\) a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 119

The solid ball of radius \(r\) and mass \(m\) rolls without slipping down the \(60^{\circ}\) trough. Determine its angular acceleration.

Short Answer

Expert verified
The angular acceleration \(\alpha\) of a solid sphere rolling without slipping down a 60-degree incline is given by \(\frac{g \times \sin(60)}{r}\).

Step by step solution

01

Formulate the known quantities

First, identify the known quantities from the problem: The radius \(r\) of the solid sphere, the angle of the incline \(\theta = 60^{\circ}\), the gravitational acceleration \(g\), and the mass \(m\) of the solid sphere.
02

Formulate the equations of motion

Next, two key equations of motion are used: \n\n1. The tangential acceleration \(a_t\) is \(r \times \alpha\), where \(\alpha\) is the angular acceleration. \n\n2. The equation for linear motion down the incline due to gravity is \(a = g \times \sin(\theta)\). \n\nThese two accelerations are equal, so you get \(r \times \alpha = g \times \sin(\theta)\).
03

Solve for angular acceleration

Rearrange the equation from Step 2 to solve for the angular acceleration, \(\alpha\). This gives: \(\alpha = \frac{g \times \sin(\theta)}{r}\).

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

The \(150-\mathrm{kg}\) uniform crate rests on the \(10-\mathrm{kg}\) cart. Determine the maximum force \(P\) that can be applied to the handle without causing the crate to slip or tip on the cart. The coefficient of static friction between the crate and cart is \(\mu_{s}=0.2\).

If a horizontal forcee of \(P=100 \mathrm{~N}\) is applied to the \(300-\mathrm{kg}\) reel of cable, determine its initial angular acceleration. The reel rests on rollers at \(A\) and \(B\) and has a radius of gyration of \(k_{O}=0.6 \mathrm{~m}\).

The uniform girder \(A B\) has a mass of \(8 \mathrm{Mg}\). Determine the internal axial, shear, and bending-moment loadings at the center of the girder if a crane gives it an upward acceleration of \(3 \mathrm{~m} / \mathrm{s}^{2}\)

The \(20-\mathrm{kg}\) roll of paper has a radius of gyration \(k_{A}=120 \mathrm{~mm}\) about an axis passing through point \(A .\) It is pin supported at both ends by two brackets \(A B .\) The roll rests on the floor, for which the coefficient of kinetic friction is \(\mu_{k}=0.2\). If a horizontal force \(F=60 \mathrm{~N}\) is applied to the end of the paper, determine the initial angular acceleration of the roll as the paper unrolls.

The spool has a mass of \(100 \mathrm{~kg}\) and a radius of gyration \(k_{G}=0.3 \mathrm{~m}\). If the coefficients of static and kinetic friction at \(A\) are \(\mu_{s}=0.2\) and \(\mu_{k}=0.15\), respectively, determine the angular acceleration of the spool if \(P=600 \mathrm{~N}\)
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Modern Physics

Read Explanation

Kinematics Physics

Read Explanation

Electricity

Read Explanation

Nuclear Physics

Read Explanation

Magnetism and Electromagnetic Induction

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.