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solution_pdf11 - billing(cab4763 hw 11 opyrchal(11109 This...

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billing (cab4763) – hw 11 – opyrchal – (11109) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A string is wound around a uniform disc of radius 0 . 55 m and mass 2 kg . The disc is released from rest with the string vertical and its top end tied to a fixed support. The acceleration of gravity is 9 . 8 m / s 2 . h 0 . 55 m 2 kg ω As the disc descends, calculate the tension in the string. Correct answer: 6 . 53333 N. Explanation: Let : R = 0 . 55 m , M = 2 kg , and g = 9 . 8 m / s 2 . Basic Concepts summationdisplay vector F = mvectora summationdisplay vector τ = I vectorα Δ U + Δ K rot + Δ K trans = 0 Solution summationdisplay F = T M g = M a and (1) summationdisplay τ = T R = I α = 1 2 M R 2 parenleftBig a R parenrightBig . (2) Solving for a in (2), a = 2 T M . (3) Using a from Eq. (3) and solving for T in (1), T = M ( g a ) = M parenleftbigg g 2 T M parenrightbigg = M g 2 T 3 T = M g T = M g 3 (4) = (2 kg) (9 . 8 m / s 2 ) 3 = 6 . 53333 N . 002 (part 2 of 2) 10.0 points Calculate the speed of the center of mass when, after starting from rest, the center of mass has fallen 1 . 7 m. Correct answer: 4 . 7131 m / s. Explanation: From conservation of mechanical energy we have Δ U + Δ K rot + Δ K trans = 0 1 2 parenleftbigg 1 2 M R 2 parenrightbigg ω 2 + 1 2 M v 2 = M g Δ h . When there is no slipping v = R ω , so 1 4 R 2 ω 2 + 1 2 v 2 = g Δ h 1 4 v 2 + 1 2 v 2 = g Δ h 3 4 v 2 = g Δ h v = radicalbigg 4 g Δ h 3 = radicalbigg 4 (9 . 8 m / s 2 ) (1 . 7 m) 3 = 4 . 7131 m / s . 003 10.0 points Consider a wheel (solid disk) of radius 1 . 23 m,

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billing (cab4763) – hw 11 – opyrchal – (11109) 2 mass 13 kg and moment of inertia 1 2 M R 2 . The wheel rolls without slipping in a straight line in an uphill direction 33 above the hor- izontal. The wheel starts at angular speed 21 . 2195 rad / s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops? The acceleration of gravity is 9 . 8 m / s 2 .
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solution_pdf11 - billing(cab4763 hw 11 opyrchal(11109 This...

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