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As shown in Fig. E10.20, a string is wrapped several times

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

Solution for problem 78P Chapter 10

University Physics | 13th Edition

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

University Physics | 13th Edition

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Problem 78P

As shown in Fig. E10.20, a string is wrapped several times around the rim of a small hoop with radius 0.0800 m and mass 0.180 kg. The free end of the string is pulled upward in just the right way so that the hoop does not move vertically as the string unwinds. (a) Find the tension in the string as the string unwinds. (b) Find the angular acceleration of the hoop as the string unwinds. (c) Find the upward acceleration of the hand that pulls on the free end of the string. (d) How would your answers be different if the hoop were replaced by a solid disk of the same mass and radius?

Step-by-Step Solution:

Solution 78P Problem (a) Step 1: Radius of the hoop R = 0.08 m Mass of the hoop M = 0.18 Kg To find the tension in the string Step 2: Tension T = Mg -----(1) 2 g = 9.8 m/s (acceleration due to gravity) T = 0.18 x 9.8 T = 1.76 N Tension in the string is 1.76 N Problem (b) Step 1: To find the angular acceleration of the hoop Total Torque produced = I-----(2) Total torque = force x perpendicular distance, here the force is nothing but the tension produced in the string. And perpendicular distance is radius of the hoop R. = TR -----(3) Moment of inertia of the hoop I = MR ----(4) Step 2: From equations (1) to (4), we can find angular acceleration value = TR I = Mg.R MR 2 g = R -----(5) = 0.08 = 122.5 rad/m 2 The angular acceleration of the hoop as the string unwinds is 122.5 rad/m 2 Problem (c) Step 1: To find the upward acceleration Upward acceleration is tangential acceleration (a). a = R----(6) From equation (5) a = R. g R a = g 2 a = 9.8 m/s 2 Upward acceleration is 9.8 m/s Problem (d) Step 1: The change in T, and a if the hoop is replaced by a solid disk of same mass and radius. 1 2 Moment of inertia of the disk I = MR 2 Tension T won’t be changed because T is independent of moment of inertia. Step 2: Change in Angular acceleration ( ) = TR I = T.R MR2 2 = 2 TR MR 2 = 2 The value of angular acceleration is doubled

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Chapter 10, Problem 78P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 78P from 10 chapter was answered, more than 338 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “As shown in Fig. E10.20, a string is wrapped several times around the rim of a small hoop with radius 0.0800 m and mass 0.180 kg. The free end of the string is pulled upward in just the right way so that the hoop does not move vertically as the string unwinds. (a) Find the tension in the string as the string unwinds. (b) Find the angular acceleration of the hoop as the string unwinds. (c) Find the upward acceleration of the hand that pulls on the free end of the string. (d) How would your answers be different if the hoop were replaced by a solid disk of the same mass and radius?” is broken down into a number of easy to follow steps, and 114 words. This full solution covers the following key subjects: string, hoop, Find, unwinds, end. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 78P from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM.

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