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Unformatted text preview: Overton, Mays – Midterm 3 – Due: Apr 6 2005, 10:00 pm – Inst: Turner 1 This printout should have 20 questions. Multiplechoice 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 xaxes), 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 . 6 m mg ω 2 3 6 8 r e v / m i n ω . 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 . 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 = mR 2 ω , and is along the negative xaxis. 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 = mg b Δ t mR 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 yaxis and ~g is into the page, in the figure below. Therefore the direction of the torque ~ τ is along the positive yaxis. + x x L + Δ L Δ L Δ φ Δ L 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)....
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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