Explanation Given information: The period ( τ ) of its small oscillation is observed to be 1.03 sec. The distance ( r a ) is 6 in. The mass of the rod is m . The acceleration due to gravity is g . Calculation: Convert the unit of the distance r a from inches to feet. r a = 6 in . = 6 in . × 1 ft 12 in . = 0.5 ft Calculate the time period ( τ ) of oscillation: τ = 2 π ω n Here, ω n is the natural circular frequency. Substitute 1.03 s for τ . 1.03 s = 2 π ω n ω n = 2 π 1.03 ω n = 6.1002 rad / s At maximum displacement position: The kinetic energy corresponding to the maximum displacement position is zero. In the connecting rod, the displacement of the center of gravity when it is displaced about an angle θ m is r a ( 1 − cos θ m ) . Here, the distance at which the center of gravity is located is r a . Express the maximum displacement x m as: x m = r a ( 1 − cos θ m ) As the angular displacement of the rod ( θ m ) is very small, take the value of ( 1 − cos θ m ) as θ m 2 / 2 . Thus, the maximum displacement of the rod is 1 / 2 r a θ m 2 . Write the expression potential energy ( V 1 ) corresponding to the maximum displacement position: V 1 = m g x m Substitute 1 / 2 r a θ m 2 for x m . V 1 = m g x m = m g ( 1 / 2 r a θ m 2 ) = 1 2 m g r a θ m 2 (1) At maximum velocity position: The potential energy corresponding to the maximum velocity position is zero. Write the expression for the kinetic energy ( T 2 ) corresponding to the maximum displacement position: T 2 = 1 2 m v G 2 + 1 2 I ¯ θ ˙ m 2 (2) Here, v G is velocity of the center of gravity, m is the mass of the rod , I ¯ is the moment of inertia of the rod, and θ ˙ m the derivative of the angular displacement. Write the expression for the velocity ( v G ) of the center of gravity: v G = r a θ ˙ m Write the expression for the moment of inertia of the rod: I ¯ = m k ¯ 2 Here, k ¯ 2 is the radius of gyration of the connecting rod. Substitute r a θ ˙ m for v G and m k ¯ 2 for I ¯ in Equation (2). T 2 = 1 2 m v G 2 + 1 2 I ¯ θ ˙ m 2 = 1 2 m ( r a θ ˙ m ) 2 + 1 2 ( m k ¯ 2 ) θ ˙ m 2 = 1 2 m ( r a 2 + k ¯ 2 ) θ ˙ m 2 (3) Assume the resulting motion of the rod as simple harmonic motion

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

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 19.3, Problem 19.74P

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The centroidal radius of gyration (k¯) of the connecting rod.

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Chapter 19.3, Problem 19.73P

Chapter 19.3, Problem 19.75P

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Chapter 19 Solutions

Connect 1 Semester Access Card for 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 connecting rod.

Solution Summary: The author explains the centroidal radius of gyration of the connecting rod. The period of its small oscillation is observed to be 1.03 sec.

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The centroidal radius of gyration (k¯) of the connecting rod.

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Chapter 19 Solutions