hw7 - r = 3 = constant =-3 ² 50 3 ³ 2 ˆ e r 3(16 1 ˆ e...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 7 (1.114, 1.119) 1.114 We are given ˙ θ = kt . So θ = kt 2 2 + ± ±² 0 C 1 , ( C 1 = 0, since @ t = 0 , θ = 0) At release, the speed of the rock is 50 f t / s . This corresponds to a tangential velocity of ˙ θ = 50 3 f t / s , as the length of the sling is 3 f t . So, 50 3 = kt r (where t r represents the time at release) Also, at release the angular distance travelled = 360 + 135 = 495 = 8 . 64 rad . So, 8 . 64 = kt 2 r 2 Solving, k = 16 . 1 s - 2 and t r = 1 . 04 s a = ± ³ ³ ´ 0 ¨ r - r ˙ θ 2 ! ˆ e r + ± r ¨ θ + 2 ³ ³ ´ 0 ˙ r ˙ θ ! ˆ e θ since, while on circle,
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Unformatted text preview: r = 3 = constant =-3 ² 50 3 ³ 2 ˆ e r + 3(16 . 1) ˆ e θ =-833ˆ e r + 48 . 1ˆ e θ f t / s 2 1.119 We are given r OA = 2cos t 50 ˆ i + 2sin t 50 ˆ j + t 50 ˆ k m ∴ v A =-2 50 sin t 50 ˆ i + 2 50 cos t 50 ˆ j + 1 50 ˆ k m / s So, r OA | t = 30 s = 1 . 65 ˆ i + 1 . 13 ˆ j + . 60 ˆ k m v A | t = 30 s =-. 0226 ˆ i + . 0330 ˆ j + . 0200 ˆ k m / s 1...
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This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

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