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A 2.00-kg rock has a horizontal velocity of magnitude 12.0

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

Solution for problem 37E 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 37E

A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in ?Fig. E10.35?. (a) At this instant, what are the magnitude and direction of its angular momentum relative to point ?O?? (b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant? Fig. E10.35

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Solution 37E The numerical expression for angular momentum is given by, L = mvrsin …..(1) Here, m is the mass of the object, v is its velocity and rsin is the perpendicular distance of the axis of rotation to the object. The rate of change of angular momentum is torque. Given in the question, m = 2.00 kg v = 12.0 m/s r = 8.00 m 0 Therefore, the perpendicular distance is = rsin = 8.00 m × sin 36.9 = 4.80 m (a) Substituting these values in equation (1), Angular momentum L = 2.00 kg × 12.0 m/s × 4.80 m × 4.80 m 2 L = 115 kg.m /s This is the required angular momentum and will be directed into the page. (b) Let us have a look at the following figure. Therefore the weight of the stone makes an angle of 53.1 with the line passing through the point O. As already mentioned above that the rate of change of angular momentum is torque. Therefore, torque due to the weight of the stone = mgrsin 53.1 0 W 2 0 2 2 W = 2.00 kg × 9.8 m/s × 8.00 m × sin 53.1 = 125 kg.m /s The torque due to the weight of the stone is 125 kg.m /s2 2 and will be directed out of the page.

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Chapter 10, Problem 37E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The answer to “A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in ?Fig. E10.35?. (a) At this instant, what are the magnitude and direction of its angular momentum relative to point ?O?? (b) If the only force acting on the rock is its weight, what is the rate of change (magnitude and direction) of its angular momentum at this instant? Fig. E10.35” is broken down into a number of easy to follow steps, and 68 words. The full step-by-step solution to problem: 37E from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 37E from 10 chapter was answered, more than 4998 students have viewed the full step-by-step answer. This full solution covers the following key subjects: its, magnitude, Rock, direction, angular. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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A 2.00-kg rock has a horizontal velocity of magnitude 12.0