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solutions 14

# solutions 14 - Patel Kinal Homework 14 Due 7:00 pm Inst D...

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Patel, Kinal – Homework 14 – Due: Oct 26 2007, 7:00 pm – Inst: D Weathers 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 1) 10 points A wheel starts from rest at t = 0 and rotates with a constant angular acceleration about a fixed axis. It completes the first revolution in 6 . 2 s. How long after t = 0 will the wheel com- plete the second revolution? Correct answer: 8 . 76812 s. Explanation: In the conventional notation θ = s r ω = d θ dt = v r α = d ω dt = a t r , we use “ θ - θ 0 = ω 0 t + 1 2 α t 2 ” at time t 1 to find the acceleration. The wheel starts from rest ( ω 0 = 0) θ 1 = 0 + 1 2 α t 2 1 α = 2 θ 1 t 2 1 = 0 . 326909 rad / s 2 . If we use the same equation again at time t 2 θ 2 = 0 + 1 2 α t 2 2 t 2 = r 2 θ 2 α = s 2 θ 2 t 2 1 2 θ 1 = r θ 2 θ 1 t 1 = 2 t 1 , because we know θ 2 = 2 θ 1 (two revolutions). Notice that the result does not depend on how many revolutions the wheel went through, only the relative number of revolutions θ 2 θ 1 . keywords: 002 (part 1 of 1) 10 points A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel of radius r = 0 . 4 m and observes that drops of water fly off tangentially. She measures the height reached by drops moving vertically. A drop that breaks loose from the tire on one turn rises h 1 = 59 . 2 cm above the tangent point. A drop that breaks loose on the next turn rises h 2 = 54 . 3 cm above the tangent point. The height to which the drops rise decreases because the angular speed of the wheel decreases. The acceleration of gravity is 9 . 8 m / s 2 . Assume: The angular deceleration is con- stant. h 0 . 4 m radius ω Find the magnitude of the angular deceler- ation of the wheel. Correct answer: 0 . 477664 rad / s 2 . Explanation: The first drop has a velocity at the point of leaving the wheel given by 1 2 m v 2 1 = m g h 1 or v 1 = p 2 g h 1 = 3 . 40635 m / s . In the same way the second drop has velocity v 2 = p 2 g h 2 = 3 . 26233 m / s .

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Patel, Kinal – Homework 14 – Due: Oct 26 2007, 7:00 pm – Inst: D Weathers 2 From ω = v r we obtain ω 1 = v 1 r = 8 . 51587 rad / s ω 2 = v 2 r = 8 . 15583 rad / s . Using the kinematic equation 2 α ( θ f - θ i ) = ω 2 f - ω 2 i the angular acceleration is α = ω 2 2 - ω 2 1 2 (2 π ) = g [ h 2 - h 1 ] 2 π r 2 = (9 . 8 m / s 2 ) [(54 . 3 cm) - (59 . 2 cm)] 2 π (0 . 4 m) 2 = - 0 . 477664 rad / s 2 , and | α | = 0 . 477664 rad / s 2 . keywords: 003 (part 1 of 1) 10 points The speed of a moving bullet can be deter- mined by allowing the bullet to pass through two rotating paper disks mounted a distance 85 cm apart on the same axle. From the angular displacement 42 . 7 of the two bul- let holes in the disks and the rotational speed 729 rev / min of the disks, we can determine the speed of the bullet. 42 . 7 v 729 rev / min 85 cm What is the speed of the bullet?
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