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Unformatted text preview: AE 352: Aerospace Dynamics II, Fall 2008 Homework 6 Due Friday, October 17 Problem 1. Derive equation 688 in Greenwood, and describe each term. In other words, say as much as you can about the qualitative meaning of each term in the general equations of motion. Problem 2. Greenwood 613 Solution. First note that we have 1 DOF, φ . Our other possible generalized coordi nates r,θ , are known because of the two holonomic constraints which specify their values over time. So, as usual, first find the velocity of the mass. First, let’s define a coordinate system e 1 , e 2 , e 3 attached to the hoop such that e 2 points from O to O and e 3 points out of the page. Then the position of the mass is r = r O + rsinφ e 1 + r cos φ e 2 Then the velocity becomes v = v O + v rel + ω e 3 × r rel = ωr e 1 + ( ˙ φ ) e 3 × ( r sin φ e 1 + r cos φ e 2 ) + ω e 3 × ( r sin φ e 1 + r cos φ e 2 ) = ωr e 1 ˙ φr sin φ e 2 + ˙ φr cos φ e 1 + ωr sin φ e 2 ωr cos φ e 1 = ( ωr + ( ˙ φ ω ) r cos φ ) e 1 + ( ω ˙ φ ) r sin φ e 2 Finally, the magnitude of the velocity becomes v 2 = v · v = ω 2 r 2 + 2 ωr ( ˙ φ ω ) r cos φ + ( ˙ φ ω ) 2 r 2 cos 2 φ + ( ω ˙ φ ) 2 r 2 sin 2 φ = ω 2 r 2 + 2 ωr ( ˙ φ ω ) r cos φ + r 2 ( ˙ φ ω ) 2 Now, we can write down the kinetic energy T = 1 2 mv 2 For potential energy, let the datum be O . Then the potential energy is V = mgh where h is the height of the mass above point O , which is also given by the component...
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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.
 Spring '09
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