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Unformatted text preview: AE 352: Aerospace Dynamics II, Fall 2008 Homework 6 Due Friday, October 17 Problem 1. Derive equation 6-88 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 6-13 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