Explanation Given information: The radius ( r ) of the disk is 100 mm. The weight ( W ) of the disk is 1.5 kg. The weight ( w ) attached at A and B is 40 g. The angle ( β ) is 60 ° . The acceleration due to gravity ( g ) is 9.81 m/s 2 . Calculation: Show the free body diagram equation as in Figure (1). The angle α is θ ¨ , Write the expression for tangential acceleration ( a t ) : a t = r α Substitute θ ¨ is α . a t = r θ ¨ Write the expression for moment of inertia ( I ¯ D ) using the relation: I ¯ D = 1 2 W g r 2 Here, g is the acceleration due to gravity. The external forces in the system are tension in the belt and force due to the mass of the disk. The effective couple in the system is I ¯ θ ¨ . Take moment about C for external force, ∑ M C = w r sin ( β − θ ) − w r sin ( β + θ ) Take moment about C for effective force, ∑ ( M C ) e f f = 2 w g r a t + I ¯ α Equate the moment about C in the system for external and effective forces. ∑ M C = ∑ ( M C ) e f f w r sin ( β − θ ) − w r sin ( β + θ ) = 2 w g r a t + I ¯ α w r [ sin ( β − θ ) − sin ( β + θ ) ] = 2 w g r a t + I ¯ α ( ∵ cos A sin B = 1 2 sin ( A + B ) + sin ( A − B ) ) − ( 2 w r sin θ cos β ) = 2 w g r a t + I ¯ α ( ∵ sin θ ≈ 0 ) − ( 2 w r cos β ) θ = 2 w g r a t + I ¯ α Substitute θ ¨ for α , 1 2 W g r 2 for I ¯ D and r θ ¨ for a t

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Author: BEER

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 19.2, Problem 19.45P

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The frequency (fn) of small oscillation when β is 60°.

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Chapter 19.2, Problem 19.44P

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The frequency ( f n ) of small oscillation when β is 60 ° .

Solution Summary: The author shows the free body diagram equation as in Figure (1). The radius of the disk is 100 mm, the weight (w) attached at A and B is 40 g.

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The frequency (fn) of small oscillation when β is 60°.

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