Explanation Given information: The mass ( m A B ) of the rod AB is 800 g. The mass ( m 0 ) of the disk is 1.2 kg. The spring constant ( k ) is 12 N/m . The length ( l ) of the rod is 600 mm. The radius ( r ) of the disk is 250 mm. Assume the acceleration due to gravity ( g ) as 9.81 m/s 2 . Calculation: Show the position 1 and position 2 of the rod AB with disk as in Figure (1). For position 1: Calculate the mass moment of inertia of disk ( I G ) d i s k : ( I G ) d i s k = 1 2 m 0 r 2 Substitute 1.2 kg for m 0 and 250 mm for r . ( I G ) d i s k = 1 2 × 1.2 × ( 250 mm × 1 m 1000 mm ) 2 = 0.0375 kg ⋅ m 2 Calculate the mass moment of inertia of rod ( I G ) A B : ( I G ) A B = 1 12 m A B l 2 Substitute 800 g for m A B and 600 mm for l . ( I G ) A B = 1 12 × ( 800 g × 1 kg 1000 g ) × ( 600 mm × 1 m 1000 mm ) 2 = 0.024 kg ⋅ m 2 Calculate maximum velocity ( v m ) A B of the rod AB : ( v m ) A B = ( l 2 − r ) θ m Substitute 600 mm for l and 250 mm for r . ( v m ) A B = ( ( 600 mm × 1 m 1000 mm ) 2 − ( 250 mm × 1 m 1000 mm ) ) θ m = 0.05 θ m Calculate maximum velocity ( v m ) 0 of the disk: ( v m ) 0 = r θ m Substitute 250 mm for r . ( v m ) 0 = ( 250 mm × 1 m 1000 mm ) θ m = 0.25 θ m Express the term ( θ ˙ m ) : θ ˙ m = ω n θ m Write the expression for the kinetic energy ( T 1 ) : T 1 = 1 2 ( I G ) A B θ ˙ m 2 + 1 2 m A B ( v m ) A B 2 + 1 2 ( I G ) disk θ ˙ m 2 + 1 2 m 0 ( v m ) 0 2 Substitute 800 g for m A B , 1.2 kg for m 0 , 0.0375 kg ⋅ m 2 for ( I G ) d i s k , 0.024 kg ⋅ m 2 for ( I G ) A B , 0.05 θ m for ( v m ) A B , and 0.25 θ m for ( v m ) 0 . T 1 = { 1 2 × 0.0375 × θ ˙ m 2 + 1 2 × ( 800 g × 1 kg 1000 g ) × ( 0.05 θ ˙ m ) 2 + 1 2 ( 0.0375 ) θ ˙ m 2 + 1 2 × 1.2 × ( 0.25 θ m ) 2 } = 1 2 [ 0.024 θ ˙ m 2 + 0

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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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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The period (τn) of small oscillation of the system.

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Chapter 19.3, Problem 19.82P

Chapter 19.3, Problem 19.84P

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Chapter 19 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.2P Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - Prob. 19.6P Ch. 19.1 - Prob. 19.7P Ch. 19.1 - Prob. 19.8P Ch. 19.1 - Prob. 19.9P Ch. 19.1 - Prob. 19.10P

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The period ( τ n ) of small oscillation of the system.

Solution Summary: The author shows the position 1 and position 2 of the rod AB with disk as in Figure (1).

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The period (τn) of small oscillation of the system.

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Chapter 19 Solutions