×
Log in to StudySoup
Get Full Access to Physics - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Physics - Textbook Survival Guide

A thin, light string is wrapped around the outer rim of a

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

Solution for problem 27E Chapter 10

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 0 303 Reviews
12
5
Problem 27E

A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.75 kg having inner and outer radii as shown in ?Fig. E10.25?. The cylinder is then released from rest. (a) How far must the cylinder fall before its center is moving at 6.66 m/s? (b) If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the distance in part (a)? (c) Why do you get two different answers when the cylinder falls the same distance in both cases?

Step-by-Step Solution:

Solution 27E Step 1: Initially, the kinetic energy of the system was zero since, the cylinder was at rest. Then, the energy will be completely potential and it will be, PE initialmgH The kinetic energy of the system can be calculated when the speed is 6.66 m/s. The kinetic energy at a speed v can be given by, KE = KE translational Rotational 2 KE translational½ mv Where, m is th mass of the system. 2 KE rotational½ I Where, I - moment of inertia of the system and - Angular velocity Step 2: In order :to calculate the moment of inertia of a hollow cylinder having inner radius r 1 and outer radius r , we2an use the equation I = ½ m (r + r ) 2 cylinder 1 2 Provided, mass of the cylinder, m = 4.75 kg r = 20 cm = 0.20 m and r = 35 cm = 0.35 m 1 2 2 2 2 2 Icylinder ½ × 4.75 kg (0.20 m + 0.35 m ) = v/r Where, v - linear velocity of the cylinder, r - outer radius of the cylinder Then, = 6.66 m/s / 0.35 m 2 2 2 KE translational= ½ × 4.75 kg × 6.66 m /s KE = ½ × ½ × 4.75 kg (0.20 m + 0.35 m ) × (6.66 m/s / 0.35) m 2 2 2 rotational KE = ½ {(4.75 kg × 6.66 m /s ) + ½ × 4.75 kg ([0.20 m / 0.35 m ]+ [0.35 m / 0.35 2 2 2 2 2 2 2 m ]) × 6.66 m 2 2 KE = ½ {(4.75 kg × 6.66 m /s ) + ½ × 4.75 kg ([0.20 m / 0.35 m ]+ 1 ) × 6.66 m 2 2 2 2 2 2 2 2 2 2 2 2 Or, KE = ½ × 4.75 kg × 6.66 m { 1 + ½ × ([0.20 m / 0.35 m ]+ 1) } Step 3: a) Suppose, it reaches a height h after it reaches a speed 6.66 m/s. Then, the change in PE = Change in KE of the system mgH - mgh = ½ × 4.75 kg × 6.66 m { 1 + ½ × ([0.20 m / 0.35 m ]+ 1) } - 0 2 2 2 2 2 2 2 2 2 2 mg {H-h} = ½ × 4.75 kg × 6.66 m { 1 + ½ × ([0.20 m / 0.35 m ]+ 1) } 2 2 2 2 2 2 Rearranging, H - h = [½ × 4.75 kg × 6.66 m { 1 + ½ × ([0.20 m / 0.35 m ]+ 1) } ] / mg = [4.75 kg × 6.66 m { 1 + ½ × ([0.20 m / 0.35 m ]+ 1) } ] / (2 × 4.75 kg × 9.8 m/s ) 2 Cancel out the term 4.75 kg in numerator and denominator 2 2 2 2 2 2 2 H - h = [ 6.66 m { 1 + ½ × ([0.20 m / 0.35 m ]+ 1) } ] / (2 × 9.8 m/s ) H - h = [ 44.36 m { 1 + ½ × (0.326+ 1) } ] / 19.6 m/s 2 H - h = 2.263 { 1 + ½ × (0.326+ 1) } H - h = 2.263 { 1 + 0.663 } H - h = 3.76 m The cylinder should move distance of 3.76 m before it reaches a speed of 6.66 m/s.

Step 4 of 5

Chapter 10, Problem 27E is Solved
Step 5 of 5

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

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

A thin, light string is wrapped around the outer rim of a

×
Log in to StudySoup
Get Full Access to Physics - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Physics - Textbook Survival Guide
×
Reset your password