00:01
All right, so first for figure 285, the mass is on the inner radius r, so no slip.
00:09
So x is equal to r theta, which means theta is x over r.
00:16
And so theta dot would be x dot over r.
00:20
The spring is on the outer radius for r, so the stretch is delta equals 4r times theta, which would be 4r times x over r, which would be 4x.
00:35
Now the energies.
00:36
The kinetic energy, t is one -half -m -x -dott -squared plus one -half j -not -theta -dot squared, which is one -half -m -dot, not m -dot, but m -x dot squared, plus one -half j -k -not over x -over r squared.
00:59
The potential energy from the spring would be v equals one -half -k, delta squared, which is one half k times 4x squared, which is one half k times 16x squared, which is 8kx squared.
01:18
And then we'll use ddt, which is times dt dx dot plus dvdx equals zero.
01:34
And we'll compute dt dx dot would be mx dot plus j.
01:40
R squared times x dot which would be m plus j not over r times x dot so d d t that would be m plus j not over r squared times x double dot and d v d x is d d d x times 8 kx squared which would be 16 kx so m plus j knot over r squared times x double dot plus 16 k x equals 0 and now for figure 286 when the moving pulley goes down by x the two rope segments around at each change by x giving 2x total change also the left spring extension e1 takes up rope in the opposite sense and the right spring extension e5 uses rope length so the small change constrained is negative e1 plus 2x minus e5 equals 0.
02:55
So e5 equals 2x minus e1.
03:00
The same rope tension t everywhere.
03:03
It's a massless rope and frictionless pulle...