Explanation Given information: The system consists of disk A which rotates about the vertical spindle. Disk D connected to disk A when spring is released. Disk D rolls without slipping. Refer figure. Mass of disk A, m A = 30 k g Mass of disk D, m D = 5 k g Initial angular velocity of disk D, ω D 1 = 440 r p m Final angular velocity of disk D, ω D 2 = 176 r p m Concept used: Impulse momentum principle. Calculation: Moment of inertia of disk A, I A = m A r A 2 2 = 1.35 k g . m 2 Moment of inertia of disk D, I D = m D r D 2 2 = 0.033 k g . m 2 Angular velocity of disk D, ω D 1 = 440 r p m = 14.667 π r a d / sec Angular velocity of disk D, ω D 2 = 176 r p m = 5.86 π r a d / sec Free body diagram of the system. P is the contact point on disk D, v p = r D ω D − v D Q is the contact point on disk A, v Q = r A ω A For rolling of the disk without slipping, r D ω D − v D = r A ω A 0.115 × 5.86 π = 0.3 ω A + v D 2.12 − 0.3 ω A = v D Applying the impulse-momentum principle,

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ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

Publisher: Mcgraw-hill Education,

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Chapter 17.2, Problem 17.87P

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The angular velocities of bar BC and disk A.

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Chapter 17.2, Problem 17.86P

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The angular velocities of bar BC and disk A.

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