Explanation Given information: Weight of rod, W = 9 l b Spring constant, k = 3 l b / i n θ = 50 0 Concept used: Energy conservation principle, Calculation: v 2 = L 2 ω 2 v B = L ω 2 x 1 = L − L cos 50 0 = 8.93 i n At position 1: v 1 = − W L 2 sin 50 0 + k × 1 2 2 = − 29.51 + 119.63 = 90.12 l b . i n v 1 = 7.509 f t

12th Edition

ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

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Chapter 17.1, Problem 17.39P

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To calculate the angular velocity of the rod and velocity at the end B when θ=0 ? Rod AB moves in a slot cut in vertical plane spring is attached at rod end A such that tension is zero. Initially θ=500.

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Chapter 17.1, Problem 17.38P

Chapter 17.1, Problem 17.40P

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To calculate the angular velocity of the rod and velocity at the end B when θ = 0 ? Rod AB moves in a slot cut in vertical plane spring is attached at rod end A such that tension is zero. Initially θ = 50 0 .

Solution Summary: The author calculates the angular velocity of the rod and velocity at the end B when theta =0.

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To calculate the angular velocity of the rod and velocity at the end B when θ=0 ? Rod AB moves in a slot cut in vertical plane spring is attached at rod end A such that tension is zero. Initially θ=500.

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Chapter 17 Solutions