old-midterm3 - Castaneda(ic4477 oldmidterm 03 Turner(58220...

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Castaneda (ic4477) – oldmidterm 03 – Turner – (58220) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A rotating bicycle wheel has an angular speed of 62 / s at 4 . 4 s and a constant angular acceleration of 37 / s 2 . With the center of the wheel at the origin, the valve stem is on the positive x -axis (horizontal) at t 0 = 1 . 4 s. Through what angle has the valve stem turned between 1 . 4 s and 7 . 6 s? Correct answer: 407 . 34 . Explanation: Let : ω 1 = 62 / s , t 1 = 4 . 4 s , α = 37 / s 2 , θ 0 = 0 , t 0 = 1 . 4 s , and t = 7 . 6 s . The initial angular velocity can be deter- mined from ω 1 = ω 0 + α ( t 1 - t 0 ) ω 0 = ω 1 - α ( t 1 - t 0 ) = 62 / s - (37 / s 2 ) (4 . 4 s - 1 . 4 s) = - 49 / s . This θ 2 = ω 0 ( t 2 - t 0 ) + 1 2 α ( t 2 - t 0 ) 2 = ( - 49 / s) (7 . 6 s - 1 . 4 s) + 1 2 (37 / s 2 ) (7 . 6 s - 1 . 4 s) 2 = 407 . 34 . 002 (part 1 of 3) 10.0 points A beetle takes a joy ride on a pendulum. The string supporting the mass of the pendulum is 153 cm long. If the beetle rides through a swing of 33 , how far has he traveled along the path of the pendulum? Correct answer: 88 . 1217 cm. Explanation: Let : r = 153 cm , and θ = 33 . Arc length is defined as s = r θ = (153 cm)(33 ) · π 180 = 88 . 1217 cm . 003 (part 2 of 3) 10.0 points What is the displacement experienced by the beetle while moving theough the same angle 33 ? Correct answer: 86 . 9087 cm. Explanation: Using the law of cosines, we have R = r 2 + r 2 - 2 r r cos θ = r 2 (1 - cos θ ) = (153 cm) 2(1 - cos 33 ) = 86 . 9087 cm . Alternative Solution: Divide the isosce- les triangle in half. Then R = 2 . 0 r sin θ 2 = 2 . 0 (153 cm) sin 16 . 5 = 86 . 9087 cm . 004 (part 3 of 3) 10.0 points If the pendulum at some instant is swinging at 1 . 2 rad / s, how fast is the beetle traveling? Correct answer: 183 . 6 cm / s. Explanation:
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Castaneda (ic4477) – oldmidterm 03 – Turner – (58220) 2 Let : ω = 1 . 2 rad / s . Linear and angular velocity are related by v = r ω = (153 cm)(1 . 2 rad / s) = 183 . 6 cm / s . keywords: 005 10.0 points At t = 0, a wheel rotating about a fixed axis at a constant angular deceleration of 0 . 82 rad / s 2 has an angular velocity of 2 . 8 rad / s and an angular position of 6 . 3 rad. What is the angular position of the wheel after 5 s? Correct answer: 10 . 05 rad. Explanation: Let : α = - 0 . 82 rad / s 2 , ω 0 = 2 . 8 rad / s , θ 0 = 6 . 3 rad , and t = 5 s . The angular position is θ f = θ 0 + ω 0 t + 1 2 α t 2 = 6 . 3 rad + (2 . 8 rad / s) (5 s) + 1 2 ( - 0 . 82 rad / s 2 ) (5 s) 2 = 10 . 05 rad . 006 10.0 points A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end from the mass m , the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity as the rod swings through its lowest (vertical) position? 1. 6 g 5 L 2. g 2 L 3. 3 g 2 L 4. 2 g 3 L 5. g 4 L 6. 4 g 5 L 7. 2 g L 8. 12 g 5 L correct 9. 5 g 6 L 10. g L Explanation: The mechanical energy of the system is con- served. Measuring heights from the point at the bottom of the rod when it is verti- cal, the initial potential energy of the sys- tem is U i = (2 m ) g L , the potential energy at the bottom of the swing is U f = m g L 2 and the moment of inertia of the system is I = m L
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