00:01
Let x -b downward displacement of the system cg at point o, and theta be a small counterclockwise rotation about o.
00:09
Let the rod length be l, let the distances from the cg to mass 1 and mass 2 be a, the distance to m1, and b the distance to m2, and a plus b is equal to l.
00:33
The cg condition is that m1 times a is equal to m2 times b.
00:40
And we solve a equals m2 over m1 plus m2 times l, and b equals m1 over m1 plus m2 times l.
00:52
Then kinematics relates x1 and x2 to x and theta.
00:58
For a small theta about the cg, the left mass goes down by a times theta, and the right mass goes up by b times theta.
01:05
So x1 equals x plus a theta and x2 equals x minus b times theta.
01:14
Each mass has two springs, okay? and so that's equivalent vertical stiffness to 2k per mass.
01:24
So the potential energy v is 1 half times 2k times x1 squared plus 1 half times 2k times x2 squared, which is k times x2 squared, which is k times x1 squared, which is k times x1 squared plus x2.
01:36
Squared.
01:39
Kinetic energy t is one -half times m1 times x.
01:42
Dot 1 squared plus one -half and two times x2...