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Unformatted text preview: DATE: October 1, 2011 DUE: October 7, 2011 ENG102, Solution to homework set #2. 1. After Sample Problem 2/8 in Text: a. Do the problem, answering questions ad, but start by converting all units into sensible units, e.g., mi → km, ft → m, etc. This should be done while maintaining the physics of the problem intact. Using the conversions, 1 statute mile = 1.609 km and 1 ft = 30.48 cm, we have the given information horizontal acceleration = 6.096 m/s 2 g = 9.144 m/s 2 speed along trajectory = 19,308 km/h = 10,919 m/s. The resulting answers are thereby a n = 7 . 25 m/s 2 , a t = 8 . 26 m/s 2 ; (a) ρ = 3 , 965 km; (b) ˙ v = 8 . 26 m/ 2 ; (c) ˙ β = 1 . 354 × 10 − 3 s − 1 ; (d) ¯ a = 7 . 25 m/s 2 ¯ e n +8 . 26 m/s 2 ¯ e t . Suppose the rocket is experiencing an airdrag that contributes to the total accelera tion with α ¯ v , where α is a nonnegative constant. Let us denote the angle of the rocket relative to horizontal with θ (that angle is zero as the problem is stated in the book – a positive angle θ will mean that the nose of the rocket will point up relative to its center of mass; this is not necessarily the trajectory of the rocket). b. For α = 0 calculate, as a function of θ ∈ [0 , π 2 ] the total acceleration ¯ a on the rocket. Display the result in the two coordinate systems ¯ e n , ¯ e t and ¯ e x , ¯ e y . We denote the acceleration of the thrust with a th . The acceleration is then ¯ a = a th cos θ ¯ e x + ( a th sin θ g ) ¯ e y (1) = [( g a th sin θ ) cos 15 ◦ a th cos θ sin 15 ◦ )] ¯ e n (2) + [( g a th sin θ ) sin 15 ◦ + a th cos θ cos 15 ◦ )] ¯ e t (3) = [ g cos 15 ◦ a th sin( θ + 15 ◦ )] ¯ e n + [ g sin 15 ◦ + a th cos( θ + 15 ◦ )] ¯ e t (4) c. For α = 0 . 04 s −...
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This note was uploaded on 01/19/2012 for the course ENG 102 taught by Professor Eke during the Fall '08 term at UC Davis.
 Fall '08
 Eke
 Dynamics

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