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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, . Our other possible generalized coordi- nates r, , 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, lets 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 + rsin e 1 + r cos e 2 Then the velocity becomes v = v O + v rel + e 3 r rel = r e 1 + (- ) e 3 ( r sin e 1 + r cos e 2 ) + e 3 ( r sin e 1 + r cos e 2 ) = r e 1- r sin e 2 + r cos e 1 + r sin e 2- r cos e 1 = ( r + ( - ) r cos ) e 1 + ( - ) r sin e 2 Finally, the magnitude of the velocity becomes v 2 = v v = 2 r 2 + 2 r ( - ) r cos + ( - ) 2 r 2 cos 2 + ( - ) 2 r 2 sin 2 = 2 r 2 + 2 r ( - ) r cos + r 2 ( - ) 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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