00:02
So for this problem, i'm going to go out of order.
00:07
So i'm not going to go in a, b, c, d order.
00:09
I'm going to go in a different order.
00:11
So be sure and pay attention to that as i go through this problem.
00:15
But we're going to start off with part b.
00:18
And that is the undamped natural time period of the rotor.
00:23
So for one cycle, that's two seconds.
00:28
And so we have 80 over 5 is equivalent.
00:34
To e to times 2 t w n and that goes to t d equaling 2 pi over the square root of w n squared minus t squared w n squared and so w n squared is equivalent to 2 pi squared over t d squared plus t squared w n squared and that's equal to 4 pi squared over 4 plus 1 .3863 squared.
01:23
And that gives you 11 .7915.
01:28
And so wn, the natural frequency is 3 .4339 radians per second.
01:38
I'm going to highlight these for you, so you know, to pay attention to those.
01:43
Next, i'm going to go to part d.
01:45
So pay attention there and so part d we need to find the torsional stiffness of the spring and so since angular displacement of rotor under applied torque is 50 degrees that is 0 .8772 -7 radians and so kt is equal to torque divided by the angular displacement and that is 2 times 10 to the negative 3rd by 0 .8727...