Explanation Given information: The mass of the circle ( m ) is 100 g or 0 .1kg . The total mass of the system ( M ) is 300 kg . The spring constant ( k ) is 600 kN/m . The speed of the exciter ( N ) is 1200 rpm . The amplitude of the total fluctuating forced ( F m ) is 160 N . The acceleration due to gravity ( g ) is 9.81 m/s 2 . Calculation: Calculate the natural frequency ( ω n ) using the formula: ω n = 2 k M Here, 2 k is the spring constant of two springs. Substitute 600 kN / m for k and 300 kg for M . ω n = 2 ( 600 kN / m ) ( 300 kg ) = 2 ( 600 × 1000 ) ( 300 kg ) = 63.2456 rad / s The expression for the relation between the unbalanced force and centrifugal force due to eccentricity of the counter-rotating eccentric mass as follows: P m = 2 m r ω f 2 Here, 2 m is the mass of two rotating masses and r is the eccentricity of the counter-rotating eccentric mass. Calculate the frequency of the periodic force ω f using the relation: ω f = 2 π N 60 Substitute 1200 rpm for N . ω f = 2 π ( 1200 rpm ) 60 = 40 π rad / s The amplitude of the fluctuating force is ± 160 N . This is given as the product of spring constant and deflection. The deflection is the amplitude of vibration and the spring constant is 2 k . The expression for the fluctuating force as follows: ± 160 N = 2 k x m The expression for the amplitude of forced vibration ( x m ) as follows: x m = P m / 2 k 1 − ( ω f ω n ) 2 2 k x m = P m 1 − ( ω f ω n ) 2 P m = 2 k x m ( 1 − ( ω f ω n ) 2 ) Substitute ± 160 N for 2 k x m

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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Free Damped Vibrations

Free vibration is a type of vibration in which the external force is removed after giving the initial displacement to a body/system. In other words, in this type of vibration, force is applied at the initial displacement of the system, and after the init…

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Chapter 19.4, Problem 19.119P

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Find the radius (r) of the circle.

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Chapter 19.4, Problem 19.118P

Chapter 19.4, Problem 19.120P

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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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Find the radius ( r ) of the circle.

Solution Summary: The author calculates the radius of the circle using the formula: underset_149.3mm

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Find the radius (r) of the circle.

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Chapter 19 Solutions