COAM and the role of hands

yeah, I see how it is kind of like a slap shot. now if you are talking about gregs RH drill, ask him on that. I believe that the right hand must cup first then cock. I dont know if that is what is described in gregs drill, but he can explain that. I feel that the right wrist cup is taken care of in a good waggle.

It's not really my speciality either but I guess the magnitude of the angular momentum in this case is L = mvr, where L is the angular momentum, m is the mass of the small object, v is the magnitude of its velocity, and r is the separation between the objects. The formula can be rearranged to give v = L/(mr) and L is a constant for an isolated system, the velocity v and the separation r are inversely correlated. Thus, conservation of angular momentum demands that a decrease in the separation r be accompanied by an increase in the velocity v, and vice versa. In this case the ice skater rotates much faster with their arms held in than pushed out, but I guess they don't think this much about why.

Anyway, where were we

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Best Regards
Brian

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Funny o'l game!

OK thanks, I have asked him now

The "thrusting" applies another force to the system (ice skater).

If you thrust the arms straight out from the origin of rotation, then I would add nothing and would only get the hands out quicker.

If you thrust the hands in the same direction as you are spinning then you will actually slow the rotation down (arms move faster at the expense of the rest of the body slowing down because there is no friction for the arm action to act against).

I will grant you that if you thrust your arms and there is friction that you would be able to increase the speed of the hands, but remember in a golf swing this would be at the cost of lead upper arm separation from the chest (if you think that is important) because you are using your shoulders to provide the additional power.

Ice skaters always think about the why. They know that if the go into a spin or a spinning jump that in order to 2, 3 or 4 rotations they need to be spinning as fast and as wide as they can before, because this will maximize number of rotations they can achieve.

I have no idea where we are , but you brought us here so you get us out.

I think there is a misunderstanding of the principle of COAM as it pertains to the skater and the velocity of their hands. Once the rotation is started, the hands will be moving at the same velocity, whether they are extended or whether they are against the body. What changes is the speed of the rotation. Say for instance that, when extended, the skaters hands are moving at 6 meters per second and say the distance of one complete circular path of the hands is 6 meters (the circumference of the circle traced by the hands, when extended). This means the hands (and the skater) will make one revolution every second. Now if the skater pulls the hands in to a smaller circular path, say 3 meters in total length ((the circumference of the circle traced by the hands, when close to the body) and the hands are still moving at 6 meters per second, the hands (and the skater) will make two revolutions in one second. Revolving twice as fast, but the same hand speed.

Momentum is defined as (mass x velocity). This is what is being conserved. IOW, the mass of the hands is constant, therefore the velocity must remain constant, in order to "conserve" momentum.

May be one of the experts at GTO can clear it up, but I think rather than focusing on the cupping of the right wrist, the focus should be on lag. If there is lag, then the right wrist will automatically be in the cupped position at impact. Right?

Not the way I see it. Read my post #9 in this thread please.

other way around, if the right wrist is cupped there will automatically be lag

Yes but more like a karate chop than a slap and it allows you to hit as hard as you want with your right hand without loosing control and alignment

Yes and no. I didn't want to have to do this, but...

Angular momentum: L = rXp (r=radius, X=cross product and p=momentum)
Linear momentum: p = m*v (m=mass and v=linear velocity)
Angular velocity: w = v/r

If you have two particles orbiting the same point and they both have identical mass and angular momentum, but one is twice as far from the point as the other, then:

L1 = L2

r1*m*v1 = r2*m*v2

v1 = r2/r1*v2

assuming: r1 = 2*r2

v1 = v2/2

So the particle that is further out will have half the linear velocity as the particle that is closer.

Looking at angular velocity:

w2 = v2/r2
w1 = v1/r1 = v2/(2*r1) = v2/(4*r2) = w2/4

So the particle in close orbits 4 times for every one orbit of the other particle (or RPM is 4 times greater for the in close particle).

Note that linear velocity is not conserved between the two orbiting particles.

The area of the segment is conserved (assume one revolution of the far away particle).

Particle 2 Area: A2 = pi*r2*r2
Particle 1 area: A1 = pi*r1*r1 = 4*pi*r2*r2 = 4*A2

So the circle area of one revolution of the far away particle is equal to 4 circle areas of the particle that is in close.

Hope this clears it up.



I probably should have had a disclaimer at the begining.

Thats what I said

could you put that in simple terms
take gravity for example, gravity may have a lot of numbers/equations, but I dont know those. what I do know is if I drop something it falls

This was stupid, I should know better.....of course r1xv1 equals r2xv2 so
when r2 equals 1/2 of r1, v2 should equal 2xr1. Apologies all around.

If you are spinning, when you pull your hands in you will rotate faster AND your hands will be moving faster.

Isn't the "puck" method exactly what golf "experts" tell you to avoid or have I been listening to the wrong experts?