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HW22 - Pickering Tracey Homework 22 Due noon Inst Drummond...

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Pickering, Tracey – Homework 22 – Due: Mar 27 2006, noon – Inst: Drummond 1 This print-out should have 11 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 2) 10 points Two objects, of masses 24 kg and 18 kg, are connected to the ends of a rigid rod (of negli- gible mass) that is 70 cm long and has marks every 10 cm, as shown. 24 kg 18 kg A B C D E F G H J 10 20 30 40 50 60 Which of the nine lettered points represents the center of mass of the sphere-rod combina- tion? 1. D correct 2. C 3. A 4. E 5. F 6. J 7. H 8. B 9. G Explanation: Let : = 70 cm , m 1 = 24 kg , m 2 = 18 kg . For static equilibrium, τ net = 0. Denote x the distance from the left end point of the rod to the center-of-mass point. m 1 g x - m 2 g [ - x ] = τ = 0 (24 kg) g x - (18 kg) g h (70 cm) - x i = 0 (24 kg) x - (18 kg) h (70 cm) - x i = 0 (42 kg) x - (1260 kg · cm) = 0 x = 30 cm . Therefore the point should be point D . 002 (part 2 of 2) 10 points The sphere-rod combination can be pivoted about an axis that is perpendicular to the plane of the page and that passes through one of the five lettered points. Through which point should the axis pass for the moment of inertia of the sphere-rod combination about this axis to be greatest? 1. J correct 2. H 3. E 4. D 5. A 6. F 7. G 8. C 9. B Explanation: The parallel-axis theorem is I = M h 2 + I cm , where I cm is the moment of inertia about an axis through the center of mass. Since I cm and M = (24 kg) + (18 kg) = 42 kg are fixed, maximum h = | (30 cm) - (70 cm) | = 40 cm will give the maximum moment of inertia. Therefore, the axis should pass the point J .
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Pickering, Tracey – Homework 22 – Due: Mar 27 2006, noon – Inst: Drummond 2 003 (part 1 of 1) 10 points Consider three objects of equal masses but different shapes: a solid disk, a thin ring, and a thin hollow square. The sizes of the three objects are related according to the following picture disk R ring R square 2 R Note: The ring and the square are hollow and their perimeters carry all the mass, but the disk is solid and has uniform mass density over its whole area. Compare the three objects’ moments of in- ertia when rotated around their respective centers of mass. Which of the following conditions is cor- rect? 1. I cm disk > I cm ring > I cm square 2. I cm ring > I cm square > I cm disk 3. I cm square > I cm disk > I cm ring 4. I cm square > I cm ring > I cm disk correct 5. I cm ring > I cm disk > I cm square 6. I cm disk > I cm square > I cm ring Explanation: The moment of inertia I = Z r 2 dm can be though as I = M × r 2 , where M is the object’s net mass and r 2 is the average distance 2 of massive points making up the object from the rotation axis. For a hoop or a thin ring, all massive point are at the same distance R from the axis, hence r 2 = R 2 = I cm ring = M R 2 . For the solid disk, the distance ranges from zero in the center to R at the perimeter, hence the average distance is less than R , r 2 < R 2 = I cm disk < M R 2 .
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