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# HW25 - Pickering Tracey Homework 25 Due Apr 3 2006 noon...

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Pickering, Tracey – Homework 25 – Due: Apr 3 2006, noon – Inst: Drummond 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points A string is wrapped around the stem of a Yo- Yo which has a mass M and a moment of inertia I about its center axis. The radius of the stem is r 1 and the radius of the disk is r 2 . Denote the tension of the string T and the descending acceleration a . h r 2 r 1 M ω The correct torque equation is given by 1. T r 1 = a I r 1 . correct 2. T r 2 = 2 a I r 1 + r 2 . 3. T r 1 = 2 a I r 1 + r 2 . 4. T r 1 + r 2 2 = 2 a I r 1 + r 2 . 5. T r 1 + r 2 2 = 2 a I r 2 . 6. T r 2 = a I r 2 . 7. T r 2 - r 1 2 = 2 a I r 1 + r 2 . 8. T r 1 + r 2 2 = 2 a I r 2 - r 1 . 9. T r 1 + r 2 2 = 2 a I r 1 . Explanation: The torque created by the tension is τ = T r 1 = I α = I a r 1 = a I r 1 . 002 (part 2 of 3) 10 points The correct vertical force equation is given by 1. M g + T = M a . 2. T - M g = M a . 3. M g - T = M a . correct Explanation: The net force on the Yo-Yo is F net = X F : M g - T = M a . 003 (part 3 of 3) 10 points Release the Yo-Yo from rest. It reaches a center of mass speed v when it falls for a height Δ h . Using the same notation as in the previous questions, the correct work-energy equation is given by 1. 1 2 M v 2 = 1 2 Iv 2 . 2. 1 2 M v 2 = 1 2 I v r 1 2 . 3. M g Δ h = 1 2 I v r 1 2 . 4. M g Δ h = 1 2 M v 2 + 1 2 I v r 1 2 . correct 5. 1 2 M v 2 = 1 2 I v r 2 2 . 6. M g Δ h = 1 2 I v r 2 2 . 7. M g Δ h = 1 2 M v 2 . 8. M g Δ h = 1 2 M v 2 + 1 2 I v r 2 2 . Explanation: The work done by the gravity is W = M gh . The combined (translational and rotational) kinetic energy is K = 1 2 M v 2 + 1 2 I ω 2 = 1 2 M v 2 + 1 2 I v r 1 2 . Thus the work energy theorem gives M g Δ h = 1 2 M v 2 + 1 2 I v r 1 2 .

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Pickering, Tracey – Homework 25 – Due: Apr 3 2006, noon – Inst: Drummond 2 004 (part 1 of 2) 10 points A hollow cylinder has length a 2 . 4 m, mass 0 . 6 kg, and radius 0 . 1 m. The cylinder is free to rotate about a vertical axis that passes through its center and is perpendicular to the cylinder’s axis. Inside the cylinder are two masses of 0 . 2 kg each, attached to springs of spring constant k and unstretched lengths 0 . 4 m. The inside walls of the cylinder are frictionless. 2 . 4 m 0 . 8 m 0 . 2 kg 0 . 2 kg 0 . 6 kg Determine the value of the spring constant k if the masses are located 0 . 6 m from the cen- ter of the cylinder when the cylinder rotates at 20 rad / s. Correct answer: 120 N / m. Explanation: Let : m = 0 . 6 kg , Δ x = 0 . 6 m , and ω = 20 rad / s . Applying Newton’s second law, the net in- ward force acting on each of the 0 . 2 kg masses is X F radial = k Δ x = m ( x + Δ x ) ω 2 . Thus k = m ( x + Δ x ) ω 2 Δ x = (0 . 6 kg) (0 . 6 m) (20 rad / s) 2 0 . 4 m = 120 N / m . 005 (part 2 of 2) 10 points How much work was needed to bring the sys- tem from rest to 20 rad / s?
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HW25 - Pickering Tracey Homework 25 Due Apr 3 2006 noon...

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