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 . The acceleration due to gravity ( g ) is 9 .81 m/s 2 . Calculation: Show the free body diagram equation for length OG as in Figure (1). 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 ¨ + <

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Chapter 19.2, Problem 19.52P

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Show that the period of oscillation of a compound pendulum is equal to the period of simple pendulum of length OA, where the distance from A to the mass center G is GA is equal to k¯2r¯.

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Chapter 19.2, Problem 19.51P

Chapter 19.2, Problem 19.53P

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Show that the period of oscillation of a compound pendulum is equal to the period of simple pendulum of length OA , where the distance from A to the mass center G is GA is equal to k ¯ 2 r ¯ .

Solution Summary: The author shows the free body diagram equation for length OG as in Figure (1).

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Show that the period of oscillation of a compound pendulum is equal to the period of simple pendulum of length OA, where the distance from A to the mass center G is GA is equal to k¯2r¯.

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Chapter 19 Solutions