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# test 3 - Overton Mays Midterm 3 Due Apr 6 2005 10:00 pm...

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Overton, Mays – Midterm 3 – Due: Apr 6 2005, 10:00 pm – Inst: Turner 1 This print-out should have 20 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 1) 10 points Two objects, of masses 25 kg and 10 kg, are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown. 25 kg 10 kg A B C D E F G 10 20 30 40 50 60 If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord? 1. F 2. D 3. E 4. A correct 5. B 6. C Explanation: Let : = 70 cm , m 1 = 25 kg , m 2 = 10 kg . For static equilibrium, τ net = 0. Denote x the distance from the left end point of the stick to the point where the cord is attached. m 1 g x - m 2 g [ - x ] = τ = 0 ( m 1 - m 2 ) x = m 2 x = m 2 m 1 - m 2 = (10 kg) (70 cm) 25 kg - 10 kg = 20 cm . Therefore the point should be point A . 002 (part 1 of 1) 10 points A 3 kg bicycle wheel rotating at a 2368 rev / min angular velocity has its shaft supported on one side, as shown in the figure. When viewing from the left (from the positive x -axes), one sees that the wheel is rotating in a clockwise manner. The distance from the center of the wheel to the pivot point is 0 . 6 m. The wheel is a hoop of radius 0 . 5 m, and its shaft is horizontal. Assume all of the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9 . 8 m / s 2 . y z x 0 . 6 m m g ω 2368 rev / min 0 . 5 m radius Find the change in the precession angle after a 1 . 6 s time interval. Correct answer: 8 . 695 . Explanation: Let : m = 3 kg , ω = 2368 rev / min = 2 π (2368 rev / min) (60 s / min) = 247 . 976 rad / s , b = 0 . 6 m , and R = 0 . 5 m .

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Overton, Mays – Midterm 3 – Due: Apr 6 2005, 10:00 pm – Inst: Turner 2 Basic Concepts: ~ τ = d ~ L dt Solution: The magnitude of the angular momentum I of the wheel is L = I ω = m R 2 ω , and is along the negative x -axis. From the figure in Part 2, we get Δ φ = Δ L L . Using the relation, Δ L = τ Δ t , where τ is the torque, ~ τ = ~ b × m~g . The precession angle Δ φ is Δ φ = Δ L L = τ Δ t L = m g b Δ t m R 2 ω = g b Δ t R 2 ω = (9 . 8 m / s 2 ) (0 . 6 m) (1 . 6 s) (0 . 5 m) 2 (247 . 976 rad / s) = 0 . 151756 rad = 8 . 695 . ~ τ = ~ b × m~g , where ~ b is along the positive y -axis and ~g is into the page, in the figure below. Therefore the direction of the torque ~ τ is along the positive y -axis. + x - x L L Δ L Δ φ τ wheel Viewed from Above + y L We can see the direction of precession is clockwise, due to ~ τ = -→ Δ L Δ t , that is Δ L is in the direction of the torque τ and is producing clockwise motion ( ~ L × ~ τ produces clockwise motion). 003 (part 1 of 2) 10 points A small puck of mass 27 g and radius 14 cm slides along an air table with a speed of 1 . 7 m / s. It makes a glazing collision with a larger puck of radius 34 cm and mass 66 g (ini- tially at rest) such that their rims just touch.
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