Explanation Given information: The weight ( W ) of the gyroscope rotor is 4 oz. The period ( τ 1 ) is 6 sec when the gyroscope rotor is 4 oz. The period ( τ 2 ) is 3.80 sec when the diameter ( d ) of the steel sphere is 1.25 in . . Assume the acceleration due to gravity ( g ) is 32.2 ft/s 2 . The specific gravity ( γ ) of the steel is 490 lb/ft 3 . Calculation: Show the free body diagram as in Figure (1). Refer Figure (1), Take moment about center of the system for external forces. ( ∑ M ) e x t e r n a l = − K θ Take moment about center of the system for effective forces. ( ∑ M ) e f f e c t i v e = I ¯ θ ¨ Equate the moment about O in the system for external and effective forces. ( ∑ M ) e x t e r n a l = ( ∑ M ) e f f e c t i v e − K θ = I ¯ θ ¨ − I ¯ θ ¨ − K θ = 0 I ¯ ( θ ¨ + K θ I ¯ ) = 0 θ ¨ + K θ I ¯ = 0 (1) Compare the differential Equation (1) with the general differential equation of motion ( x ¨ + ω n 2 x = 0 ) and express the natural circular frequency of vibration: ω n 2 = K I ¯ = K I ¯ Write the expression for period ( τ ) of oscillation: τ n = 2 π ω n Substitute K I ¯ for ω n . τ = 2 π K I ¯ = 2 π I ¯ K (2) Rewrite the above expression for the moment of inertia ( I ¯ ) . I ¯ = K τ 2 4 π 2 (3) Rewrite the above expression for the torsional spring constant ( K ) . K = 4 π 2 I ¯ τ 2 (4) Calculate the radius ( r ) using the relation: r = d 2 Substitute 1.25 in . for d . r = 1.25 2 = 0.625 in . × 1 ft 12 in . = 52.0833 × 10 − 3 ft Calculate the volume ( V s ) of the sphere: V s = 4 3 π r 3 Substitute 52.0833 × 10 − 3 ft for r . V s = 4 3 π ( 52.0833 × 10 − 3 ft ) 3 = 591.81 × 10 − 6 ft 3 Calculate the weight ( W s ) of the sphere using the relation: W s = γ V s Substitute 490 lb/ft 3 for γ and 591.81 × 10 − 6 ft 3 for V s

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 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 centroidal radius of gyration (k¯) of the rotor.

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Chapter 19.2, Problem 19.66P

Chapter 19.2, Problem 19.68P

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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 centroidal radius of gyration ( k ¯ ) of the rotor.

Solution Summary: The author explains the centroidal radius of gyration of the rotor, the acceleration due to gravity, and the specific gravity.

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The centroidal radius of gyration (k¯) of the rotor.

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