Homework 08 - Bautista Aldo Homework 8 Due 4:00 am Inst...

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Bautista, Aldo – Homework 8 – Due: Oct 25 2005, 4:00 am – Inst: Maxim Tsoi 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points A large wheel is coupled to a wheel with halF the diameter as shown. r 2 How does the rotational speed oF the smaller wheel compare with that oF the larger wheel? How do the tangential speeds at the rims compare (assuming the belt doesn’t slip)? 1. The smaller wheel has twice the rotational speed and the same tangential speed as the larger wheel. correct 2. The smaller wheel has halF the rotational speed and halF the tangential speed as the larger wheel. 3. The smaller wheel has twice the rotational speed and twice the tangential speed as the larger wheel. 4. The smaller wheel has Four times the ro- tational speed and the same tangential speed as the larger wheel. Explanation: v = r ω The tangential speeds are equal, since the rims are in contact with the belt and have the same linear speed as the belt. The smaller wheel (with halF the radius) rotates twice as Fast: µ 1 2 r (2 ω ) = r ω = v 002 (part 1 oF 2) 10 points A grinding wheel, initially at rest, is ro- tated with constant angular acceleration oF 7 . 71 rad / s 2 For 8 . 56 s. The wheel is then brought to rest, with uniForm deceleration, in 5 . 94 rev. ±ind the magnitude oF the angular deceler- ation required to bring the wheel to rest. Correct answer: 58 . 3525 rad / s 2 . Explanation: Let : α 1 = 7 . 71 rad / s 2 , t = 8 . 56 s , and θ 1 = 37 . 3221 rad . Basic Concepts ω = ω 0 + α t ω 2 = ω 2 0 + 2 α θ θ = ¯ ω t . Solution: ±irst fnd the speed attained beFore the wheel begins to slow down. With ω 0 = 0 rad / s , we have ω 1 = α 1 t 1 = (7 . 71 rad / s 2 ) (8 . 56 s) = 65 . 9976 rad / s . Next, consider the wheel as it comes to rest. Now ω 1 = 65 . 9976 rad / s, ω 2 = 0 rad / s , so ω 2 2 = 0 = ω 2 1 + 2 α 2 θ 1 α 2 = - ω 2 1 2 θ 1 (1) = - (65 . 9976 rad / s) 2 2 (37 . 3221 rad) = - 58 . 3525 rad / s 2 k 2 k = 58 . 3525 rad / s 2 . 003 (part 2 oF 2) 10 points Determine the time needed to bring the wheel to rest. Correct answer: 1 . 13102 s.
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Bautista, Aldo – Homework 8 – Due: Oct 25 2005, 4:00 am – Inst: Maxim Tsoi 2 Explanation: Since ω 1 = 65 . 9976 rad / s , ω 2 = 0 rad / s , and α 2 from Eq. 1, we have ω 2 = ω 1 + α 2 t t = - ω 1 α 2 = 2 θ 1 ω 1 = 2 (37 . 3221 rad) (65 . 9976 rad / s) = 1 . 13102 s . 004 (part 1 of 3) 10 points The turntable of a record player rotates ini- tially at the rate 33 rev / min and takes 29 . 2 s to come to rest. What is the angular acceleration of the turntable, assuming the acceleration is uni- form? Correct answer:
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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Homework 08 - Bautista Aldo Homework 8 Due 4:00 am Inst...

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