Explanation Given information: The mass of the panel is 48 kg , the length of the panel is 2.4 m , the width of the panel 1.6 m , the panel complete full revolution in 6 s and the frame complete full revolution in 12 s . Write the expression of angular velocity in x direction perpendicular to panel. ω x = ω 2 sin θ ) Here, the angular velocity of the frame is ω 2 and the angle is θ . Write the expression of angular velocity in y direction perpendicular to panel. ω y = ω 2 cos θ Write the expression of angular velocity in z direction perpendicular to panel. ω z = ω 1 Here, the angular velocity of the panel is ω 1 . Write the expression of angular momentum about point G . H G = I x ω x i ′ + I y ω y j ′ + I z ω z k ′ ....... (I) Here, the moment of inertia about x axis is I x , the angular velocity in x direction perpendicular to panel is ω x , the unit vector in x direction is i ′ , the moment of inertia about y axis is I y , the angular velocity in y direction perpendicular to panel is ω y , the unit vector in y direction is j ′ , the moment of inertia about z axis is I z , the angular velocity in z direction perpendicular to panel is ω z , and the unit vector in z direction is k ′ . Write the expression of moment of inertia about x axis. I x = 1 3 m ( a 2 + b 2 ) Here, the mass of the panel is m , the half length of the panel is a and the half width of the panel is b . Write the expression of moment of inertia about y axis. I y = 1 3 m a 2 Write the expression of moment of inertia about z axis. I z = 1 3 m b 2 Substitute 1 3 m ( a 2 + b 2 ) for I x , 1 3 m a 2 for I y , 1 3 m b 2 for I z , ω 2 sin θ for ω x , ω 2 cos θ for ω y and ω 2 for ω z in Equation (I). H G = 1 3 m ( a 2 + b 2 ) ω 2 sin θ i ′ + 1 3 m a 2 ω 2 cos θ j ′ + 1 3 m b 2 ω 1 k ′ ...... (II) Write the expression of unit vector in x direction to revert the frame. i ′ = cos θ i + sin θ j Here, the unit vector in x direction to revert the frame is i and the unit vector in y direction to revert the frame is j . Write the expression of unit vector in y direction to revert the frame. j ′ = − sin θ i + cos θ j Write the expression of unit vector in z direction to revert the frame. k ′ = k Substitute cos θ i + sin θ j for i ′ , − sin θ i + cos θ j for j ′ and k for k ′ in Equation (II). H G = [ 1 3 m ( a 2 + b 2 ) ω 2 sin θ ( cos θ i + sin θ j ) + 1 3 m a 2 ω 2 cos θ ( − sin θ i + cos θ j ) + 1 3 m b 2 ω 1 k ] = [ ( 1 3 m a 2 + 1 3 m b 2 ) ( ω 2 sin θ cos θ i + ω 2 sin 2 θ j ) − 1 3 m a 2 ω 2 cos θ sin θ i + 1 3 m a 2 ω 2 cos 2 θ j + 1 3 m b 2 ω 1 k ] = [ 1 3 m a 2 ω 2 sin θ cos θ i + 1 3 m a 2 ω 2 sin 2 θ j + 1 3 m b 2 ω 2 sin θ cos θ i + 1 3 m b 2 ω 2 sin 2 θ j − 1 3 m a 2 ω 2 cos θ sin θ i + 1 3 m a 2 ω 2 cos 2 θ j + 1 3 m b 2 ω 1 k ] = 1 3 m [ a 2 ω 2 sin θ cos θ i + a 2 ω 2 sin 2 θ j + b 2 ω 2 sin θ cos θ i + b 2 ω 2 sin 2 θ j − a 2 ω 2 cos θ sin θ i + a 2 ω 2 cos 2 θ j + b 2 ω 1 k ] H G = 1 3 m [ a 2 ω 2 j + b 2 ω 1 k + b 2 ω 2 sin θ cos θ i + b 2 ω 2 sin 2 θ j ] = 1 3 m [ b 2 ω 1 k + b 2 ω 2 sin θ cos θ i + ( a 2 + b 2 sin 2 θ ) ω 2 j ] ...... (III) Differentiate Equation (III) with respect to z by assuming frame is G x y z . d d z ( H G ) = d d z [ 1 3 m { b 2 ω 1 k + b 2 ω 2 sin θ cos θ i + ( a 2 + b 2 sin 2 θ ) ω 2 j } ] d H G d z = 1 3 m [ d d z ( b 2 ω 1 k ) + d d z ( b 2 ω 2 sin θ cos θ i ) + d d z ( a 2 + b 2 sin 2 θ ) ω 2 j ] d H G d z = 1 3 m [ 0 + b 2 ω 2 { ( sin θ × ( − sin θ ) ) d θ d z + ( cos θ × cos θ ) d θ d z } i + { 0 + b 2 2 sin θ cos θ d θ d z } ω 2 j ] d H G d z = 1 3 m ω 2 b 2 d θ d z [ ( cos 2 θ − sin 2 θ ) i + 2 sin θ cos θ j ] ..... (IV) Write the expression of the angular velocity of the panel. ω 1 = d θ d z Substitute ω 1 for d θ d z and ( H G ) G x y z for d H G d z in Equation (IV). ( H G ) G x y z = 1 3 m ω 2 b 2 ω 1 [ ( cos 2 θ − sin 2 θ ) i + 2 sin θ cos θ j ] ...... (V) Write the expression of rate of change of angular momentum about point G . H ˙ G = ( H G ) G x y z + ω 2 H G j ...... (VI) Substitute 1 3 m ω 2 b 2 ω 1 [ ( cos 2 θ − sin 2 θ ) i + 2 sin θ cos θ j ] for ( H G ) G x y z and 1 3 m [ b 2 ω 1 k + b 2 ω 2 sin θ cos θ i + ( a 2 + b 2 sin 2 θ ) ω 2 j ] for H G in Equation (VI)

12th Edition

ISBN: 9781259977237

Author: BEER

Publisher: MCG

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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 18, Problem 18.154RP

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The dynamic reaction exerted on column CD by its support at D.

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Chapter 18, Problem 18.153RP

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The dynamic reaction exerted on column C D by its support at D .

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The dynamic reaction exerted on column CD by its support at D.

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Chapter 18 Solutions