106_Dynamics 11ed Manual

106_Dynamics 11ed - Engineering Mechanics Dynamics Chapter 12 Given r = 60 m v = 20 v = 3 m s m 2 s Solution 2 ar = v r ar = 6.667 m 2 s a = v a =

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Engineering Mechanics - Dynamics Chapter 12 Given: r 60 m = v 20 m s = v' 3 m s 2 = Solution: a r v 2 r = a r 6.667 m s 2 = a θ v' = a 3 m s 2 = Problem 12-146 A particle is moving along a circular path having radius r such that its position as a function of time is given by = c sin bt . Determine the acceleration of the particle at = 1 . The particle starts from rest at = 0 °. Given: r 6in = c 1 rad = b 3s 1 = 1 30 deg = Solution: t 1 b asin 1 c = t 0.184 s = c sin bt () = 'c b cos = '' cb 2 sin = ar ' 2 2 r '' 2 + = a 48.329 in
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This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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