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OH-10-preview

# OH-10-preview - Version PREVIEW OH-10 Bradley(191 This...

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Version PREVIEW – OH-10 – Bradley – (191) 2 = (8 m) 2 + (9 m) 2 4 (18 kg) = 652 . 5 kg · m 2 . 004 (part 2 of 2) 5.0 points Find the rotational energy of the system about the z axis. Correct answer: 5220 J. Explanation: Let : ω = 4 rad / s . The rotational energy of the system is K = 1 2 I z ω 2 = 1 2 (652 . 5 kg · m 2 ) (4 rad / s) 2 = 5220 J . Swinging Rod 02 005 10.0 points A massless rod of length L has a mass 2 m fastened at its center and another mass m fastened at one end. On the opposite end from the mass m , the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the maximum angular velocity as the rod swings through its lowest (vertical) position?
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OH-10-preview - Version PREVIEW OH-10 Bradley(191 This...

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