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Unformatted text preview: AE 352: Aerospace Dynamics II, Fall 2008 Homework 5 Due Friday, October 10 Problem 1. Using x 1 and x 2 as coordinates, obtain the kinetic and potential energy of the system of sliding blocks shown in Figure 1 (assume no friction). Figure 1: A system of sliding blocks. Solution. Begin by defining a coordinate system e 1 , e 2 that is attached to the tip of m 1 , with e 1 pointing in the direction of m 2 . Also, let’s use the standard i , j , with i pointing to the right, and j pointing up. Then r 1 = x 1 i and r 2 = x 1 i + x 2 e 1 , where r 1 and r 2 are the position vectors of the masses in the fixed frame. Now, v 1 = ˙ x 1 i , and v 2 = ˙ x 1 i + ˙ x 2 e 1 . But note that e 1 = cos θ i sin θ j , where θ = 45 ◦ . Thus we have: v 1 = ˙ x 1 v 2 = ( ˙ x 1 ˙ x 2 cos θ ) i ˙ x 2 sin θ j And taking dot products of these vectors with themselves to get magnitudes, we have: v 2 1 = ˙ x 2 1 v 2 2 = ˙ x 2 1 + ˙ x 2 2 2 ˙ x 1 ˙ x 2 cos θ = ˙ x 2 1 + ˙ x 2 2 2 ˙ x 1 ˙ x 2 √ 2 2 Finally, the kinetic energy is the sum of the kinetic energy of each block, is: T = 1 2 m 1 ˙ x 2 1 + 1 2 m 2 ( ˙ x 2 1 + ˙ x 2 2 √ 2 ˙ x 1 ˙ x 2 ) The potential energy is solely due to gravity, and is the sum of the potential energy...
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 Spring '09
 SRI
 Energy, Kinetic Energy, Mass, Potential Energy, Cos, Aerospace Dynamics II

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