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Unformatted text preview: kuruvila (lk5992) – HW 11 – opyrchal – (11113) 1 This printout should have 10 questions. Multiplechoice 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 . 39 m and mass 2 . 8 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 . 39 m 2 . 8 kg ω As the disc descends, calculate the tension in the string. Correct answer: 9 . 14667 N. Explanation: Let : R = 0 . 39 m , M = 2 . 8 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 . 8 kg) (9 . 8 m / s 2 ) 3 = 9 . 14667 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 . 2 m. Correct answer: 3 . 9598 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 . 2 m) 3 = 3 . 9598 m / s . 003 10.0 points Consider a wheel (solid disk) of radius kuruvila (lk5992) – HW 11 – opyrchal – (11113) 2 . 474 m, mass 5 . 2 kg and moment of iner tia 1 2 M R 2 . The wheel rolls without slipping in a straight line in an uphill direction 23 ◦ above the horizontal. The wheel starts at angular speed 43 . 4599 rad / s but the rotation slows down as the wheel rolls uphill, and even tually the wheel comes to a stop and rolls back downhill....
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This note was uploaded on 12/03/2010 for the course PHYS 111 taught by Professor Moro during the Fall '08 term at NJIT.
 Fall '08
 moro

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