Explanation Given information: Period τ = 6 s Time Period of sphere τ s = 3.8 s Diameter of sphere d = 1.25 i n . Weight of steel γ = 490 l b / f t 3 Weight of rotor W r = 4 − o z Concept used: First we take the moment of the sphere; ∑ M = ( ∑ M ) e f f − K θ = I θ ¨ I θ ¨ + K θ = 0 θ ¨ + K I θ = 0 Compare the above equation with un-damped equation of vibration; M θ ¨ + ω n 2 θ = 0 Natural frequency: ω n 2 = K I ω = K I Now, time period is given by the equation: τ = ω n 2 π τ = 1 2 π K I _ _ _ _ _ _ _ _ ( 1 ) Now, from the above equation radius of gyration is calculated as: K = 4 π 2 I τ 2 _ _ _ _ _ _ _ _ ( 2 ) Also, the equation for radius of gyration is written as: I = K τ 2 4 π 2 _ _ _ _ _ _ _ _ ( 3 ) Calculation: Firstly we calculate the moment of inertia of the sphere attached with the rotor: The radius of sphere r = d 2 r = 1.25 2 r = 0.625 i n . o r , r = 0.625 12 r = 0.0521 f t Now, volume of sphere V = 4 3 π r 3 V = 4 3 × 3.14 ( 0.0521 ) 3 V = 0.0005921 f t 3 Now, weight of sphere W = γ V W = 490 × 0

12th Edition

ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

Publisher: Mcgraw-hill Education,

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Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Gen…

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Chapter 19.2, Problem 19.67P

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The radius of gyration of the rotor.

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Chapter 19.2, Problem 19.66P

Chapter 19.2, Problem 19.68P

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The radius of gyration of the rotor.

Solution Summary: The author compares the above equation with the un-damped equation of vibration.

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The radius of gyration of the rotor.

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