Explanation Given information: Mass of rod AB: m A B = 800 g Mass of disk: m d i s k = 1.2 k g Spring constant: k = 12 N / m Disk rolls without sliding. Calculations: F o r P o s i t i o n 1 : K i n e t i c e n e r g y : T 1 = 1 2 ( I ¯ G ) A B θ ˙ m 2 + 1 2 m A B ( 1 2 − r ) 2 θ ˙ m + 1 2 ( I ¯ G ) d s i k θ ˙ m 2 + 1 2 m d s i k r 2 θ ˙ m 2 w h e r e , ( I ¯ G ) A B = 1 12 m l 2 = 1 12 ( 0.8 ) ( 0.6 ) 2 = 0.024 k g ⋅ m 2 m A B ( 1 2 − r ) 2 = ( 0.8 ) ( 0.3 − 0.25 ) 2 = 0.002 k g ⋅ m 2 ( I G ) d s i k = 1 2 m d s i k r 2 = 1 2 ( 1.2 ) ( 0.25 ) 2 = 0.0375 k g ⋅ m 2 m d i s k r 2 = 1.2 ( 0.25 ) 2 = 0.0750 k g ⋅ m 2 s u b s t i t u t i n g , ∴ T 1 = 1 2 [ 0.024 + 0.002 + 0.0375 + 0.0750 ] θ . m 2 T 1 = 1 2 [ 0.1385 ] θ

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ISBN: 9781259977237

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 19.83P

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Period ofsmall oscillations of the system.

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Chapter 19.3, Problem 19.82P

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Period ofsmall oscillations of the system.

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Period ofsmall oscillations of the system.

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