Explanation Given: A uniform slender rod AB of length L = 600 mm is placed with its center equidistant from two supports that are located at a distance b = 100 mm from each other and end B of the rod is raised a distance h 0 = 80 mm and released; the rod then rocks on the supports as shown. Assuming that the impact at each support is perfectly elastic and that no slipping occurs between the rod and the supports. Calculation: Fig 1. We know that moment of inertia for slender rod, I = m L 2 12 At position 1, Potential Energy, V 1 = b h 0 L + b V 1 = m g b h 0 L + b Kinetic energy T 1 = 0 At position 2 Potential energy V 2 = 0 Velocity, v 2 = 1 2 b ω 2 Kinetic Energy, T 2 = 1 2 m v 2 2 + 1 2 I ω 2 2 ……………………..(1) Substituting the value of v 2 in equation (1) T 2 = 1 24 ω 2 2 ( 3 b 2 + I 2 ) By the conservation energy, T 1 + V 1 = T 2 + V 2 0 + m g b h 0 L + b = 1 24 m ω 2 2 ( 3 b 2 + L 2 ) + 0 ω 2 2 = C 1 h 0 Where C 1 = 24 g b ( L + b ) ( L 2 + 3 b 2 ) ………………….(2) Impact at D To determine (b) To calculate the height h2 reached by end B after the second impact as shown in the figure.

12th Edition

ISBN: 9781259977237

Author: BEER

Publisher: MCG

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Chapter 17.3, Problem 17.117P

To determine

(a)

The calculate the height h1 reached by end A after the first impact as shown in the figure.

To determine

(b)

To calculate the height h2 reached by end B after the second impact as shown in the figure.

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Chapter 17.3, Problem 17.116P

Chapter 17.3, Problem 17.118P

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also the reaction at B thanks! Need only a handwritten solution only (not a typed one).

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(a) The calculate the height h1 reached by end A after the first impact as shown in the figure.

Solution Summary: The author calculates the height h1 reached by end A after the first impact as shown in the figure.