Explanation Given information: The radius of the disk is r . The weight of the disk is W . When β is zero the period of small oscillation is given as τ 0 . 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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Chapter 19.2, Problem 19.44P

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The angle β for which the period of small oscillation is 2τ0.

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Chapter 19.2, Problem 19.43P

Chapter 19.2, Problem 19.45P

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The angle β for which the period of small oscillation is 2 τ 0 .

Solution Summary: The author shows the free body diagram equation as in Figure (1). The angle is beta and the weight of the disk is W.

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The angle β for which the period of small oscillation is 2τ0.

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