Explanation Given information: Mass of three identical bars (each) are 3.6 kg. Bar BC is given a small displacement and released. Calculations: Displacing bar BC by a small angle θ with vertical will result into rotational motion of bars AB and CD about point A and D and the translational motion of slender bar BC. Gravitational potential energies of the bars - V A B = − m g l 2 cos θ V B C = − m g l cos θ V C D = − m g l 2 cos θ Kinetic energic of bars AB and CD due to rotation: T A B = 1 2 I A w 2 = 1 2 ( m L 2 3 ) θ ˙ 2 o r T A B = m L 2 6 θ ˙ 2 T C D = m L 2 6 θ ˙ 2 Kinetic energic of bar BC due to translation: T B C = 1 2 m v 2 = 1 2 m ( x ˙ ) 2 ∴ Total energy of the system: E = V B C + V A B + V C D + T B C + T A B + T C D E = − m g l cos θ − m g l cos θ 2 − m g l 2 cos θ + 1 2 m ( x ˙ ) 2 + 2 ( m L 2 2 θ ˙ 2 ) E = − 2 m g l cos θ + m L 2 3 θ ˙ 2 + 1 2 m x ˙ 2 Also for small rotation of bar AB ;�

12th Edition

ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Forced undamped vibrations

In mechanical engineering, vibration is expressed as the term that occurs due to an object's backward and forward movement about a point of equilibrium. Generally, there are three types of vibrations which are,

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Chapter 19.3, Problem 19.84P

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Period of vibration of the system.

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Chapter 19.3, Problem 19.83P

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Period of vibration of the system.

Solution Summary: The author explains how bar BC is given a small displacement and released. Gravitational potential energies of the bars are calculated.

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Period of vibration of the system.

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