I think after watching some gymnast do twists in air and landing, I've figured something out.
It definitely has to do with hips and arms. He can stop his twist near the end before landing, because he has turned most of his angular momentum around the transverse axis, where somersaulting takes...
Ok, so let's say my calculation came to:
120 degrees=+2.02 rad/s
150 degrees=-0.60 rad/s
Does this mean that in between this point, I stopped my movement in the positive and now am traveling in the opposing direction? So angular acceleartion doesn't really tell me if it has changed...
Ok using excel, and using time values not through the quadratic method(hope that still worked) I got the folliwn values
angle radian Torque ang acc. d time ang vel(w1) angular(w2)
0.00 0.00 506.98 4.77 0.00 0.00 2.22
30.00 0.52 435.04 4.09 0.10...
Wait, couldn't I just use time as 0.1s for the first sec. then 0.2s for the second, then 0.3 s for the third. It'd be so much more simpler and less calcuations...
Also, one critical question. It says when it stops or changes direction, stop calculations...Doesn't that mean when the torque is...
Ok,
hmm. I guess I could have used a(ang acc)=w(final)- w(initial)/time.
Ok, just let me see if I can the second one then:
So with the second torque, the ang accel is 4.07rad/s^2 (which decreased, makes sense starting to slow down)
For @(ang displacement) I'm now using 60degrees...
Thanks for the response: I tried it immediately:
So, to your question about angular acceleration. At that position I'd have a positive ang acceleration or one that can be calculated as:
a(angular acc)=T/I, I=mk^2 (k=radius of gyration)
So, 506.9Nm/(68x1.25^2)=4.77rad/s^2.
Now to find...
Ok I might have figured something out,
I tried to rework it...I still get the same values for the torques. Now using cons. of mechanical energy path to find ang velocity:
angular velocity for the first at 180 degrees is 0.
But for the the one at 60 degrees or 30 degrees depending on...