Explanation Given information: The distance from A to the mass center G is GA is equal to k ¯ 2 r ¯ . The length of OG is r ¯ . The length of OA is l . Calculation: Take moment about O in the system for external forces. ( ∑ M O ) e x t e r n a l = W r ¯ sin θ Substitute − m g for W . ( ∑ M O ) e x t e r n a l = − m g r ¯ sin θ Take moment about O in the system for effective forces. ( ∑ M O ) e f f e c t i v e = I α + m a ¯ t r ¯ The effective restoring couple I ¯ α is equal to the product of the centroidal radius of gyration at which the mass of the pendulum and angular acceleration ( m k ¯ 2 θ ¨ ) . The tangential component of acceleration a ¯ t is equal to the product of radius and angular acceleration ( m r ¯ 2 θ ¨ ) . Thus, express the moment about O due to the effective forces: ( ∑ M O ) e f f e c t i v e = m k ¯ 2 θ ¨ + m r ¯ 2 θ ¨ Equate the moment about O i n the system for external and effective forces. ( ∑ M A ) e x t e r n a l = ( ∑ M A ) e f f e c t i v e − m g r ¯ sin θ = m k ¯ 2 θ ¨ + m r ¯ 2 θ ¨ − m g r ¯ sin θ = m ( k ¯ 2 θ ¨ + r ¯ 2 θ ¨ ) − k ¯ 2 θ ¨ + r ¯ 2 θ ¨ − g r ¯ sin θ = 0 ( k ¯ 2 + r ¯ 2 ) θ ¨ + g r ¯ sin θ = 0 θ ¨ + g r ¯ k ¯ 2 + r ¯ 2 sin θ = 0 For small oscillation sin θ ≈ θ

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Author: BEER

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Chapter 19.2, Problem 19.53P

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Show that the smallest period of oscillation occurs when the distance r¯ from point O to the mass center G is equal to k¯.

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Chapter 19.2, Problem 19.52P

Chapter 19.2, Problem 19.54P

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Show that the smallest period of oscillation occurs when the distance r ¯ from point O to the mass center G is equal to k ¯ .

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Show that the smallest period of oscillation occurs when the distance r¯ from point O to the mass center G is equal to k¯.

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