Explanation Given information: Mass of rod 2 k g . Length of rod L = 250 m m = 0.25 m θ = 0 Concept used Kinetics principle Conservation of energy Calculation: Moment of inertia, I = 1 12 m L 2 = 1 12 × 2 × 0.25 2 = 0.0104 k g . m 2 AD = Radius = 0.125 m AB = Length = 0.25 m sin β = R L = 0.5 β = 30 0 B G = A G = L 2 = 0.125 m Hence, v G = L 2 ω = 0.125 ω V A = ( L cos 30 ) ω = 0.1265 ω V B = ( L sin 30 ) ω = 0.125 ω For position 1: θ = 0 T 1 = 0 v 1 = − m g L 2 = − 2.4525 J o u l e . For position 2: θ = 90 0 T 2 = 1 2 I G ω 2 + 1 2 m v G 2 = 5.2 × 10 − 3 ω 2 + 0.0156 ω 2 = 0.0208 ω 2 v 2 = − m g ( R + L 2 cos β ) v 2 = − 2.830 J o u l e . By conservation of energy principle, T 1 + V 1 = T 2 + V 2 0 − 2.4525 = 0.0208 ω 2 + ( − 2.830 ) ω = 4.26 r a d / sec . V A = 0.2165 ω = 0.9223 m / s V G = 0.125 ω = 0.5325 m / s We know that, point A moves in a curved slot. a A = ( a c ) x i + V A 2 R j = ( a c ) x i + 6.805 j For rod, α = α k a n d V B = V B j r A B = − L sin 30 0 i + L cos 30 j = − 0.125 i + 0.2165 j r G B = r A B 2 = − 0.0625 i + 0.1082 j a A = a B + α × r A B − ω 2 r A B = a B j + α k ( − 0.125 i + 0.2165 j ) − ( 4

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ISBN: 9781259977237

Author: BEER

Publisher: MCG

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Chapter 17, Problem 17.140RP

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The reactions at A and B when θ=900 as shown in figure.

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Chapter 17, Problem 17.139RP

Chapter 17, Problem 17.141RP

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The reactions at A and B when θ = 90 0 as shown in figure.

Solution Summary: The author shows the reactions at A and B when theta =900.

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The reactions at A and B when θ=900 as shown in figure.

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Chapter 17 Solutions