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A machine part consists of a thin, uniform 4.00-kg bar

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

Solution for problem 55E Chapter 8

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 55E

A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (?Fig. E8.55?). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90° to make the entire part horizontal?

Step-by-Step Solution:

Solution 55E Introduction We have to initial position of center of mass and final position of center of mass. Using this values we can calculate the shift of center of mass in horizontal and vertical direction. Step 1 Let us consider the origin of the coordinate system is at the the hinge. Now the position of the center of mass of the individual bar will be at the center of the bar and the position of the center of mass of the ball will be at the center of the ball. Hence the horizontal and vertical position of the center of mass of each individual objects and the mass of the object are given below. For horizontal bar we have , and For vertical bar we have , and And for the ball we have , and Hence the initial horizontal position of the center of mass is And the initial vertical position of the center of mass will be Step 2 Now if we make the vertical bar horizontal then the position of the center of mass of each individual object will change, hence the new position will be For the horizontal bar, that is the 1.50 m bar, the positions are , and For the vertical bar, that is the 1.80 m bar, the position of the center of mass are , and And for the ball , and So the new horizontal position of the center of mass is And the vertical position of the new center of mass si

Step 3 of 3

Chapter 8, Problem 55E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 55E from chapter: 8 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. This full solution covers the following key subjects: bar, part, vertical, mass, counterclockwise. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 55E from 8 chapter was answered, more than 347 students have viewed the full step-by-step answer. The answer to “A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (?Fig. E8.55?). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90° to make the entire part horizontal?” is broken down into a number of easy to follow steps, and 75 words. University Physics was written by and is associated to the ISBN: 9780321675460.

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