00:02
So let's first write down the masses of the two collars, a and b, the mass of a and the mass of b are identical.
00:12
So m -a is equal to m -b, and this is their weight 2 .6 pounds over g, which is 32 .2 feet per square second.
00:23
And so their masses work out to 0 .08 -075.
00:31
So we'll keep as many decimal points as possible and that's pound second squared per foot now the first thing we'll do is apply the conservation of angular momentum for collar a and this means that the angular momentum in state two when the chord breaks should be equal to the angular momentum of the system in state one which is the initial state of the system so h02 must be equal to h o one so so you write this in equation form.
01:26
This is an equation form, this is m .a.
01:38
R1 v theta 1 is equal to m .a .r2 and the transverse component of the velocity v theta in state 2.
01:53
And so we can rearrange this and we can find v theta in state 2.
02:00
So v theta and state 2 is equal to r1.
02:05
V theta 1 over r2 and this is equal to r1 squared we'll replace v theta by r theta dot so that's r1 squared theta dot 1 over r2 and so this is we can substitute our values in here 0 .6 squared multiplied by the radial speed 12 radians per second over r2 1 .2 and hence we get that the radial component of velocity is 3 .6 feet per second.
02:59
So that's the final radio component of the velocity.
03:07
Now from here since we have v theta we can calculate theta dot.
03:17
So theta dot 2 is equal to v3 theta in the second state over ra and this is equal to 3 .6 which you just calculated over 1 .2 so that's 3 radiance per second and we'll use that later on so now for part b let y be the position coordinate of b and we'll take it as positive upward with a region at o a constraint of the rod is that r minus y is equal to a constant and this implies that the second derivative of y with respect to time is simply equal to the second derivative of r with respect to time so using some kinematics we can see that the y component of the acceleration of b is obviously y double dot and this is r double dot we also know that the r component or the radial component of the acceleration of a is equal to r double dot minus r theta dot squared.
05:02
So now that we have the accelerations, let's apply them to each color.
05:09
Let's apply newton's second law.
05:11
So for color b, we must have that all the forces in the y direction, when summed, must give the mass of b times its final acceleration a b.
05:27
And so this is the tension in the chord t minus the weight of b must equal to the mass of b times y double dot, which we said is the mass of b times r double dot.
05:48
And we'll call this equation 1...