×
Log in to StudySoup
Get Full Access to University Physics - 13 Edition - Chapter 9 - Problem 78p
Join StudySoup for FREE
Get Full Access to University Physics - 13 Edition - Chapter 9 - Problem 78p

Already have an account? Login here
×
Reset your password

A uniform, solid disk with mass m and radius R is pivoted

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 78P Chapter 9

University Physics | 13th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

4 5 1 312 Reviews
20
1
Problem 78P

A uniform, solid disk with mass ?m? and radius ?R? is pivoted about a horizontal axis through its center. A small object of the same mass ?m? is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

Step-by-Step Solution:
Step 1 of 3

Solution 78P This question can be solved by using the principle of conservation of energy. Total initial mechanical energy of the system is equal to the total final mechanical energy. The disk is released from rest, hence initial kinetic energy is zero. Initial potential energy = mgR, where R is given to be the radius of the disk. Initial total mechanical energy = mgR…..(1) Let the angular speed be . 2 So, kinetic energy = I2 , I is the moment of inertia of the whole system. Now, I = moment of inertia of the disk + moment of inertia of the small object 1 2 2 I = 2R + mR I = mR 2 2 1 3 2 2 3 2 2 So, kinetic energy = 2 × (2mR ) = mR 4 When the small object is directly below the axis, its potential energy will be zero. 3 2 2 Therefore, total final mechanical energy = mR4 …..(2) Equating equations (1) and (2), we get 3 2 2 mgR = mR4 2 4g = 3R 4g = 3R This is the required angular speed.

Step 2 of 3

Chapter 9, Problem 78P is Solved
Step 3 of 3

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

A uniform, solid disk with mass m and radius R is pivoted