hw06sols - AE 352: Aerospace Dynamics II, Fall 2008...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 4

hw06sols - AE 352: Aerospace Dynamics II, Fall 2008...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online