Explanation Given information: The radius of the brake drum is r = 10 in . . The total mass moment of inertia of the flywheel and the drum is I ¯ = 16 lb ⋅ ft ⋅ s 2 . The coefficient of kinetic friction between the drum and the brake shoe is μ k = 0.40 . The initial angular velocity is ω 1 = 240 rpm ( anti-clockwise ) . The number of revolutions required to stop the hydraulic cylinder is θ = 75 rev . Calculation: Show the free-body diagram of the brake drum as in Figure 1. Find the moment at point D by taking moment about point D . ∑ M D = 0 M D = F × 10 in . × 1 ft 12 in . = 5 F 6 Find the total kinetic energy ( T 1 ) using the relation. T 1 = 1 2 I ¯ ω 1 2 Substitute 16 lb ⋅ ft ⋅ s 2 for I ¯ and 240 rpm for ω 1 . T 1 = 1 2 × 16 × ( 240 rpm × 2 π rad 1 rev × 1 min 60 s ) 2 = 5053.237 ft ⋅ lb Find the final total kinetic energy ( T 2 ) using the relation. T 2 = 1 2 I ¯ ω 2 2 Substitute 16 lb ⋅ ft ⋅ s 2 for I ¯ and 0 for ω 2 . T 2 = 1 2 × 16 × 0 2 = 0 Find the work done ( U 1 → 2 ) for the rigid body using the relation; U 1 → 2 = − M D θ Substitute 5 F 6 for M D and 75 rev for θ

12th Edition

ISBN: 9781260265453

Author: BEER

Publisher: MCG

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Chapter 17.1, Problem 17.10P

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Find the force exerted by the cylinder if the system is to stop in 75 revolutions.

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The 10-in.-radius brake drum is attached to a larger flywheel which is not shown. The total mass moment of inertia of the flywheel and drum is 22 lb ⋅ ft ⋅ s 2 and the coefficient of kinetic friction between the drum and the brake shoe is 0.41. Knowing that the initial angular velocity is 255 rpm clockwise, determine the force which must be exerted by the hydraulic cylinder at point B if the system is to stop in 85 revolutions.

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Find the force exerted by the cylinder if the system is to stop in 75 revolutions.

Solution Summary: The author calculates the force exerted by the hydraulic cylinder if the system is to stop in 75 revolutions.

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