Explanation Given information: The mass of the concentrator is 30 Mg , the radius of gyration of the concentrator about its axis of symmetry is 12 m , the radius of gyration of the concentrator about any transverse axis is 10 m the angular velocity ( ω 1 ) is 0.2 rad / s and the angular velocity ( ω 2 ) is 0.25 rad / s and the angle ( θ ) is 60 ° . Write the expression for the concentrator. W = m g ...... (I) Here, mass of the concentrator is m and the acceleration due to gravity is g . Write the expression for the angular velocity about x axis by using principal axes G x y z . ω x = − ω 2 cos θ Write the expression for the angular velocity about y axis by using principal axes G x y z . ω y = ω 2 sin θ Write the expression for the angular velocity about z axis by using principal axes G x y z . ω z = ω 1 Write the expression for the angular momentum about point G . H G = I x ω x i + I y ω y i + I z ω z k ...... (II) Substitute I ′ for I x ω 2 cos θ for ω x , ω 2 sin θ for ω y and ω 1 for ω z in Equation () H G = − I ′ ω 2 cos θ i + I ω 2 sin θ i + I ′ ω 1 k Write the expression for the momentum at point G attached to the steel frame work. H G = [ − I ′ ω 2 cos θ ( i sin θ − j cos θ ) + I ω 2 sin θ ( ( i sin θ − j cos θ ) ) + I ′ ω 1 k ] = [ ( I − I ′ ) ω 2 sin θ cos θ i + ( I ′ cos 2 θ + I sin 2 θ ) ω 2 j + I ′ ω 1 k ] ...... (III) Write the expression for the rate of change of angular velocity. H ˙ G = ( H ˙ G ) G x y z + Ω H G ...... (IV) ( I − I ′ ) ω 2 sin θ cos θ i + ( I ′ cos 2 θ + I sin 2 θ ) ω 2 j + I ′ ω 1 k for H G , ω 2 j for Ω in Equation (IV). H ˙ G = [ ( ( I − I ′ ) ω 2 ( cos 2 θ − sin 2 θ ) θ i + 2 ( I − I ′ ) ω 2 sin θ cos θ j ) + ( ω 2 j ) ( ( I − I ′ ) ω 2 sin θ cos θ i + ( I ′ cos 2 θ + I sin 2 θ ) ω 2 j + I ′ ω 1 k ) ] H ˙ G = [ ( I − I ′ ) ω 1 ω 2 ( cos 2 θ i + sin 2 θ ) − 1 2 ( I − I ′ ) ω 2 2 sin 2 θ k + I ′ ω 1 ω 2 i ] H ˙ G = [ [ I ′ + ( I − I ′ ) cos 2 θ ] ω 1 ω 2 i + ( I − I ′ ) ω 1 ω 2 sin 2 θ j − 1 2 ( I − I ′ ) ω 2 2 sin 2 θ k ] ...... (V) Write the expression for the moment of inertia. I = m k 2 ...... (VI) Here the mass is m and the radius of gyration about the axis is k . Write the expression for the moment of inertia about the transverse axis. I ′ = m ( k ′ ) 2 ...... (VII) Here, the radius of gyration about the transverse axis is k ′ . Write the expression of force acting at A . A = A x i + A y j ...... (VIII) Write the expression for the net force. A + B − W j = 0 B = W j − A ...... (IX) Calculation: Substitute 30 Mg for m and 9.81 m / s 2 for g in Equation (I). W = ( 30 Mg ) ( 9.81 m / s 2 ) = ( 30 Mg ) ( 1000 kg 1 Mg ) ( 9.81 m / s 2 ) = 294.3 × 10 3 kg ⋅ m / s 2 ( 1 N 1 kg ⋅ m / s 2 ) ( 1 kN 1000 N ) = 294.3 kN Substitute 30 Mg for m and 12 m for k in Equation (VI). I = ( 30 Mg ) ( 12 m ) 2 = ( 30 Mg ) ( 1000 kg 1 Mg ) ( 12 m ) 2 = 4.32 × 10 6 kg ⋅ m 2 Substitute 30 Mg for m and 10 m for k in Equation (VII). I ′ = ( 30 Mg ) ( 10 m ) 2 = ( 30 Mg ) ( 1000 kg 1 Mg ) ( 10 m ) 2 = 3 × 10 6 kg ⋅ m 2 Substitute 3 × 10 6 kg ⋅ m 2 for I ′ , 4.32 × 10 6 kg ⋅ m 2 for I 0.2 rad / s for ω 1 and 0.25 rad / s for ω 2 in Equation (V). H ˙ G = [ [ 3 × 10 6 kg ⋅ m 2 + ( 4 To determine (b) The couple.

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Author: BEER

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Chapter 18.2, Problem 18.100P

To determine

(a)

The force exerted at A.

The force exerted by B.

To determine

(b)

The couple.

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Chapter 18.2, Problem 18.99P

Chapter 18.2, Problem 18.101P

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Chapter 18 Solutions

Vector Mechanics For Engineers

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