Explanation Given information: The radius of the half section cylinder is r . The mass of the cylinder is m . The radius of the cylinder A and B is r 4 . The mass of the cylinder is m 4 . Calculation: Show the position 1 and position 2 of the system with velocity as in Figure (1). For position 1: Write the expression for the displacement ( h ): h = 4 r 3 π ( 1 − cos θ m ) = 4 r 3 π ( 1 − cos θ m ) ( ∵ ( 1 − cos θ m = 2 sin 2 θ m 2 ) ) = 4 r 3 π 2 sin 2 θ m 2 For small oscillation 2 sin 2 θ m 2 ≈ θ m 2 2 . h = 4 r 3 π θ m 2 2 = 2 r θ m 2 3 π Write the expression for the potential energy ( V 1 ) : V 1 = m g h Substitute 2 r θ m 2 3 π for h . V 1 = 2 m g r θ m 2 3 π Write the expression for the kinetic energy ( T 1 ) : T 1 = 0 For position 2: Calculate the mass moment of inertia of the cylinder A ( I A ) and B ( I B ) using the relation: I A = I B = 1 2 ( m 8 ) ( r 4 ) 2 = m r 2 256 Calculate the mass moment of inertia of the half section cylinder ( I h ) using the relation: I h = 1 2 m r 2 The natural frequency of the cylinder A ( ω A ) and B ( ω B ) : ω A = ω B = 4 ω Write the expression for the kinetic energy ( T 2 ) : T 2 = 1 2 I A ω A 2 + 1 2 I B ω B 2 + 1 2 I h Substitute m r 2 256 for I A , m r 2 256 for I B , 4 ω for ω A , 4 ω for ω B , and 1 2 m r 2 for I h

12th Edition

ISBN: 9781264095032

Author: BEER

Publisher: MCGRAW-HILL HIGHER 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 19.3, Problem 19.92P

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The frequency (fn) of small oscillations.

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Chapter 19.3, Problem 19.91P

Chapter 19.3, Problem 19.93P

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The frequency ( f n ) of small oscillations.

Solution Summary: The author shows the position 1 and position 2 of the system with velocity as in Figure (1).

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The frequency (fn) of small oscillations.

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Chapter 19 Solutions