HW15 - schneiter(tcs639 – HW15 – Ross –(89152 1 This...

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Unformatted text preview: schneiter (tcs639) – HW15 – Ross – (89152) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A wheel rotating with a constant angular ac- celeration turns through 12 revolutions during a 5 s time interval. Its angular velocity at the end of this interval is 18 rad / s. What is the angular acceleration of the wheel? Note that the initial angular veloc- ity is not zero. Correct answer: 1 . 16814 rad / s 2 . Explanation: Let : N = 12 , t = 5 s , and ω = 18 rad / s . From kinematics α t = ω f − ω ω = ω f − α t and Δ θ = N 2 ( π ) , so Δ θ = ω t + 1 2 α t 2 2 N π = ( ω f − α t ) t + 1 2 αt 2 = ω f t − 1 2 αt 2 α = 2 ω f t − 2 N π t 2 = 2 (18 rad / s) (5 s) − 2 (12) π (5 s) 2 = 1 . 16814 rad / s 2 . keywords: 002 (part 1 of 4) 10.0 points A bug is on the rim of a disk of diameter 8 in that moves from rest to an angular speed of 75 rev / min in 2 . 6 s. What is the tangential acceleration? Correct answer: 0 . 306909 m / s 2 . Explanation: Let : r = 4 in , ω i = 0 rad / s , ω f = 75 rev / min , and t = 2 . 6 s . The radius is r = (4 in) parenleftbigg 2 . 54 cm 1 in . parenrightbiggparenleftbigg 1 m 100 cm parenrightbigg = 0 . 1016 m and the final speed is ω f = (75 rev / min) parenleftbigg 1 min 60 s parenrightbiggparenleftbigg 2 π rad 1 rev parenrightbigg = 7 . 85398 rad / s , so the tangential acceleration is...
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This note was uploaded on 04/23/2009 for the course PHYS 152 taught by Professor Button during the Spring '08 term at IUPUI.

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HW15 - schneiter(tcs639 – HW15 – Ross –(89152 1 This...

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