hw02sols - AE 352 Aerospace Dynamics II Fall 2008 Homework...

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AE 352: Aerospace Dynamics II, Fall 2008 Homework 2 Due Friday, September 12 Problem 1. Problem 2.6 in Greenwood. Problem 2. Problem 2.8 in Greenwood. Solution. We use Eqn. 2-106 in Greenwood to find the absolute acceleration of point P, which is in motion relative to the moving coordinate system ( r z ). 1. This equation is: a = ¨ R + ˙ ω × ρ + ω × ( ω × ρ ) + 2 ω × ( ˙ ρ ) r + ( ¨ ρ ) r We shall explain and determine each of these terms in turn. 2. a is the absolute acceleration of point P, the tack at the highest point of the wheels rotation. This is what we want to find. 3. ¨ R is the absolute acceleration of point O’, relative to the inertial frame OXYZ. We have R = R ˆ e r = 30ˆ e r ˙ R = 30 d dt ˆ e r = 30( ˙ φ ˆ e φ ) ¨ R = 30 ¨ φ ˆ e φ + 30 ˙ φ d dt ˆ e φ = 30 ¨ φ ˆ e φ - 30 ˙ φ 2 ˆ e r But ˙ φ is the angular velocity of O’ about O, which is (for a circular motion of the bike about the origin) ˙ φ = | ω | = v R = 10 m / s 30 m = . 33 rad / s Then, since ω = ˙ φ is constant, ¨ φ = 0 , and hence ¨ R = ( - 3 . 33 m / s 2 e r We note that this term is essentially the centripetal acceleration of a point at O’, due to the velocity of the bike around the track. 4. The second term on the right hand side is zero, since | ˙ ω | = ¨ φ = 0 Page 1 of 4
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AE 352: Aerospace Dynamics II, Fall 2008 5. The third term on the right hand side is the centripetal acceleration of the tack due to the angular velocity of O’ about O. Thus, we use the angular velocity ω = ˙ φ ˆ e z = ( . 33 rad / s e z , and the position vector of the tack relative to O’, ρ = ( . 8 m )(cos 15ˆ e z - sin 15ˆ
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This note was uploaded on 09/16/2009 for the course AE ae352 taught by Professor Sri during the Spring '09 term at University of Illinois at Urbana–Champaign.

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hw02sols - AE 352 Aerospace Dynamics II Fall 2008 Homework...

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