When a rod handle is mounted to an angle somewhere before RSP with 10g line and then "spring" is bent and line is released producing 10m/s line speed it is 0.5J
Then when you cast the rod "the lever" accelerates line to 20m/s which produces 2.0J to line and rod reaches same bend. Then when rod straightens lets assume it still accelerates line 10m/s making line launch speed 30m/s it is 4,5J
and
Of course in real life it does not happen like that because when rod bends there is line energy loss and because of various other reasons but doesn't it approaches it?
Esa,
You are correct in thinking that the SHO is a very simplified model of what is going on in casting, but your idea that the maximum PE in the rod is only about 20% of the total KE in the line is certainly in the right ballpark. I think you have the right idea about what is going on, but I might differ with you about how much energy is applied to the line with the bending and unbending of the rod. In casting the caster continues to rotate the rod forward during the deceleration phase of the cast, so you will find the tip travels a bigger distance as the rod unloads.
I like to think of the basic physics of casting in terms of work energy. To get a better understanding of the work energy principle and how it impacts the change in the KE of an object you might want to read (or re-read) this
Wikipedia article.
Back so soon? Now you should know that the work applied to an object is just equal to the integral of the force (dot) distanace curve where dot indicates a dot product. Thus the direction the force is applied relative to the path the object is moving is important in a three dimensional world.
The SHO model only has one dimension (either x or phi) so that makes it easier to compute and maybe understand. The basic thing to understand is that force applied to the line is related to the deflection in the rod. In the basic form of the model it is assumed that force is equal to:
f_spring=-k*x. The non-linear characteristics of the rod can be somewhat accommodated by adding in a cubic term to that Hook's law equation
In the real world some 30% of the deflection in the rod is just due to inertial bending. Consequently it gets complicated when you measure the deflection and then compute the actual force being applied to the line. We try to adjust for that by adding in an effective mass value for the rod to the mass of the line in the model, but that is only a first order correction.
The distance that force is applied is determined by how far the caster moves the tip of the rod from RSP0 to RSP1.
In you scenario let's assume in the clamped bow and arrow cast the rod was deflected 1 m before it was let go. Thus as the rod straightened it will apply a varying force to the line over that 1 meter distance. With no losses you would find the work energy from the force over distance integral would equal 1/2*(mo+m)*v.^2. In the loss-less case that energy would also equal the 1/2*k*x.^2 the initial PE in the rod. Thus some of that available energy went into increasing the KE of the line, and some went into increasing the KE associated with the moving mass in the rod.
Now lets assume that from RSP0 to MRF the caster moves the rod tip 3 meters and gets the same maximum deflection of 1m. The shape of the deflection vs distance curve over that path will not be an exact stretched, mirror-image replica of the force vs distance curve for the bow and arrow cast, put for our ball park estimate lets assume that was the case. That would mean the work energy done on the line would be 3 times larger than the value for the bow and arrow cast for that portion of the cast. As the rod unloads from MRF to RSP1 lets assume the caster rotates the rod such that the tip 2 meters (1 meter from rotation + 1 m for the deflection) . That will not have the exact shape of the f vs d curve for the bow and arrow either, but assuming it is close would mean the work energy applied to the rod as the it unloads in the cast would be 2 times the bow and arrow cast value.
That means the work energy applied to the line for that cast would be roughly 5 times the value for the bow and arrow cast for the same maximum deflection in rod.
The force for a given rod deflection is important to the force over distance work energy calculation, but you will notice the PE in the rod never enters directly into computing how much KE you could expect to be induced in the line.
Gordy
