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Unformatted text preview: pels (kcp389) – Homework 8 – hill – (666) 1 This printout should have 38 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A grinding wheel, initially at rest, is ro tated with constant angular acceleration of 7 . 29 rad / s 2 for 4 . 66 s. The wheel is then brought to rest with uniform deceleration in 13 rev. Find the angular deceleration required to bring the wheel to rest. Correct answer: 7 . 06437 rad / s 2 . Explanation: Let : ω = 0 rad / s , α 1 = 7 . 29 rad / s 2 , t 1 = 4 . 66 s , and θ 2 = 13 rev . From kinematics ω f = ω + αt. First find the speed attained before the wheel begins to slow down. Since ω = 0 rad / s for the acceleration, the final speed ω 1 = α 1 t 1 for the acceleration is also the initial speed for the deceleration. Considering the wheel as it comes to rest, ω 2 2 = ω 1 2 + 2 α 2 θ 2 = 0 α 2 = ω 2 1 2 θ 2 = α 2 1 t 2 1 2 θ 2 = (7 . 29 rad / s 2 ) 2 (4 . 66 s) 2 2 (13 rev) · 1 rev 2 π rad = 7 . 06437 rad / s 2 . 002 (part 2 of 2) 10.0 points Determine the time needed to bring the wheel to rest. Correct answer: 4 . 80884 s. Explanation: The initial speed for the deceleration is ω 1 = α 1 t 1 , so from kinematics, ω f = ω 1 + α 2 t 2 = 0 t 2 = ω 1 α 2 = α 1 t 1 α 2 = (7 . 29 rad / s 2 ) (4 . 66 s) 7 . 06437 rad / s 2 = 4 . 80884 s . 003 10.0 points A dog on a merrygoround undergoes a 1.2 m/s 2 linear acceleration. If the merrygoround’s angular accelera tion is 1.4 rad/s 2 , how far is the dog from the axis of rotation? Correct answer: 0 . 857143 m. Explanation: Let : a t = 1 . 2 m / s 2 and α = 1 . 4 rad / s 2 . a t = r α r = a t α = 1 . 2 m / s 2 1 . 4 rad / s 2 = . 857143 m . keywords: 004 (part 1 of 4) 10.0 points A bug is on the rim of a disk of diameter 10 in . that moves from rest to an angular speed of 75 rev / min in 5 . 8 s. What is the tangential acceleration? Correct answer: 0 . 171975 m / s 2 . Explanation: Let : r = 5 in ., ω i = 0 rad / s , ω f = 75 rev / min , and t = 5 . 8 s . pels (kcp389) – Homework 8 – hill – (666) 2 The radius is r = (5 in . ) parenleftbigg 2 . 54 cm 1 in . parenrightbiggparenleftbigg 1 m 100 cm parenrightbigg = 0 . 127 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 a t = r α = r w f Δ t = (0 . 127 m) (7 . 85398 rad / s) 5 . 8 s = . 171975 m / s 2 . 005 (part 2 of 4) 10.0 points What is the tangential velocity of the bug at the end of the 5 . 8 s? Correct answer: 0 . 997456 m / s. Explanation: v = r ω = (0 . 127 m) (7 . 85398 rad / s) = . 997456 m / s ....
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This note was uploaded on 11/30/2008 for the course PHY 2048 taught by Professor Guzman during the Fall '08 term at FAU.
 Fall '08
 Guzman
 Physics, Work

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