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A 2.20-kg hoop 1.20 m in diameter is rolling to the right

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

Solution for problem 19E 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 19E

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 rad/s. (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in Part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

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Solution 19E The moment of inertia of a hoop is I = MR , where M is the mass of the hoop and R is its radius. The top point of the hoop will have double the linear speed and the bottom point will have zero speed as it is momentarily at rest. The situation can be summarized as shown in the figure below. Given. Mass of the hoop M = 2.20 kg Diameter = 1.20 m Radius r = 1.20 /2 m = 0.60 m Angular speed = 3.00 rad/s (a) The linear speed of the center is v = r = 0.60 m × 3.00 rad/s = 1.80 m/s Therefore, the center is moving at a speed of 1.80 m/s. 2 2 2 2 (b) Moment of inertia of the hoop is I = MR = 2.20 kg × (0.60) kg.m = 0.792 kg.m Total kinetic energy of the hoop is K = Mv + I 2 1 2 2 2 K = 1× 2.20 kg × (1.80) m /s + 2 1 × 0.792 kg.m × (3.00) rad /s 2 2 2 2 K = 7.13 J (c) For a person at rest on the ground (i) he highest point on the hoop This point will...

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Chapter 10, Problem 19E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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