Spring 2018, Homework 03 Posted by D. Dane Quinn on Fri 02 of Feb, 2018 12:14 EST posts: 475 ☆ ☆ ☆ ☆ ☆ It’s time... for Homework 03 Lyric of the dayI’m just the same as I was Now don’t you understand That I’m never changing who I am.

Spring 2018, Homework 03 Posted by D. Dane Quinn on Fri 02 of Feb, 2018 12:14 EST posts: 475 ☆ ☆ ☆ ☆ ☆ It’s time... for Homework 03 Lyric of the dayI’m just the same as I was Now don’t you understand That I’m never changing who I am.

Posted by Garrett Brisbin on Sat 03 of Feb, 2018 16:53 EST posts: 5 ☆ ☆ What is meant by “Rewrite the (second-order) equation of motion as a pair of first-order differential equations.”? I’m not sure what is being asked.

Posted by System Administrator on Sat 03 of Feb, 2018 19:15 EST posts: 9 ☆ ☆ ☆ If you identify the velocity as v = dx/dt then the acceleration can be written as a = dv/dt Therefore the equation of motion can be written as an equation for dv/dt, together with the definition of the velocity above.

Posted by Sean Sheridan on Mon 05 of Feb, 2018 17:43 EST posts: 3 ☆ For question one part two regarding the stability of the equation of motion, just to be clear, you are looking for a range or set of values correct?

Posted by Nathan Adkins on Tue 06 of Feb, 2018 08:00 EST posts: 2 ☆ I’m having a hard time figuring out what the moment of inertia is for objects. Specifically for problem 1. All I can find for bars is ml^2/12.

Posted by Garrett Brisbin on Tue 06 of Feb, 2018 18:29 EST posts: 5 ☆ ☆ I use the equation I_o = I_g + md2 This I believe the the parallel axis theorem. So (ml2)/12 is the I_g. d the distance your are moving the axis of rotation.

Posted by Scott Toth on Thu 08 of Feb, 2018 17:05 EST posts: 1 ☆ In general, if the inertia of an object about its mass center is known (IG), the inertia about any other point may be calculated with the parallel-axis theorem as IP = IG +md*d (d squared. The carrot up symbol does something to the post), where d is the distance between G and P. IG=(mL*L)/12 (Inertia about the mass center) If the bar is fixed on the end, you are looking for the inertia from the end of the bar (IA) IA=(mL*L)/3