Line of compression and tangential force

[EdZ]

Be careful with the words you use. Mass is not a force, never will be.

Interesting choice of words given their 'relative' gravity.

[ram418]

Velocity is a vector.

You can think of Velocity as a Speed with a Direction.


A rock whirling on a string is always Accelerating because it's constantly changing its Direction.

[Toolish]

I am very confused with the two. Velocity to me has always been pretty much completely interchangable with speed - is that correct ?

I have always seen it as the faster the speed and the more heavy the weight of an object hitting another object, the more it the second object moves. But with F=ma newtons second law, it confuses me....because it is not the velocity but the acceleration and like you said 'a constant velocity (speed I think...) means no acceleration which results in no force via the equation. If two rocks in space(just to get rid of other forces at the moment) are traveling at a speed or velocity but not accelerating, it appears to me that their should be force applied on to the other. Kinda like if you hit a cue ball in pool and its slowing down to a crawl where it seems to be deaccelerating, it will still give force to the object ball. Im just very confused and I know Im not right here but this is the way im thinking....where am I going wrong ?

As others have said, velocity is speed with direction. So an object moving in a circle at constant speed will have a constantly changing velocity as the direction is changing. Think of a car driving a straight bit of road. If it drive 10mph along the road, then does a U turn and drives 10mph back down the road the speed has been the same both ways, but the velocities are the exact opposite of each other.

Use the two balls floating in space, towards each other. While they are floating (at a constant velocity) there is no force on either, hence no acceleration. Now, once these two balls collide with each other, as soon as they touch, they apply a force to each other, and therefore accelerate each other. This is where the F=ma comes into it. Throughout all collision there is energy conservation. That is the energy in the initial system will carry through to the final system. The initial system will have the kinetic energy (1/2 * mass * velocity^2) of both balls, during the collision some energy is given out as sound, the final system will then have the remaining energy still as kinetic energy of the 2 balls.

As another way to think of it. A ball sitting still in space weighing 100grams, and a block weighing 1kg approaches at 10m/s. Assume a perfect collision (no energy as noise, heat etc). Initial kinetic energy in the system is all in the block (.5 * 1 * 10^2 = 50 Joules). Assume during impact the collision causes the block speed to halve to 5m/s it now has energy of (.5 * 1 * 5^2 = 12.5 Joules), this leaves 50 - 12.5 = 37.5 joules of energy for the ball.

Using the KE=.5 * m * v^2,

37.5 = .5 *.1 * v^2
v = 27.4 m/s

Not sure how this helps a golf swing, but hope it helps with the physics you are trying to understand.

[neil]

I've been really trying to get into chapter 2 with in respects to physics to really get into Homer Kelley's mind further beyond than perhaps the text of the golfing machine itself and fully appreciate it.

It is my understanding (and ive just started studying this stuff so its very probable I've got the wrong end of the stick) that if you whirl a heavy object in a consistant orbit that the velocity and its mass (force) of the object being whirled around is always tangential to the point that the object is on the circle. Like if the string was cut - the ball would fly off at 90 degrees to the line of the centrifugal pull out.....

What is puzzling me is at impact the line of compression looks more down the angle of approach in picture 2-C-3 instead of down the tangent of the orbiting sweetspot. I would of thought the line of compression would be tangential to the orbit at the the point that the sweetspot travels through the ball.

Why also is in the picture 2-C-1, the line of compression pointing downwards at low point in comparison to the picture in 2-A?

Any ideas

I am trying to understand what it is we are talking about here Matthew.Is it the clubhead or the ball?

[Mathew]

As others have said, velocity is speed with direction. So an object moving in a circle at constant speed will have a constantly changing velocity as the direction is changing. Think of a car driving a straight bit of road. If it drive 10mph along the road, then does a U turn and drives 10mph back down the road the speed has been the same both ways, but the velocities are the exact opposite of each other.

Use the two balls floating in space, towards each other. While they are floating (at a constant velocity) there is no force on either, hence no acceleration. Now, once these two balls collide with each other, as soon as they touch, they apply a force to each other, and therefore accelerate each other. This is where the F=ma comes into it. Throughout all collision there is energy conservation. That is the energy in the initial system will carry through to the final system. The initial system will have the kinetic energy (1/2 * mass * velocity^2) of both balls, during the collision some energy is given out as sound, the final system will then have the remaining energy still as kinetic energy of the 2 balls.

As another way to think of it. A ball sitting still in space weighing 100grams, and a block weighing 1kg approaches at 10m/s. Assume a perfect collision (no energy as noise, heat etc). Initial kinetic energy in the system is all in the block (.5 * 1 * 10^2 = 50 Joules). Assume during impact the collision causes the block speed to halve to 5m/s it now has energy of (.5 * 1 * 5^2 = 12.5 Joules), this leaves 50 - 12.5 = 37.5 joules of energy for the ball.

Using the KE=.5 * m * v^2,

37.5 = .5 *.1 * v^2
v = 27.4 m/s

Not sure how this helps a golf swing, but hope it helps with the physics you are trying to understand.

It will help my understanding of the physics behind the golfing machine, once I have a better understanding of physics as a whole....to understand the forces at work, I will appreciate certain perspectives that Homer talks about.

I kinda understand your answer, but I think it mostly shows I have no real knowledge and I have to spend a few days studying it - since my last post I bought 'physics for dummies'..lol for a start and will read it pretty soon.

Thank you for your answers they have been appreciated.

[neil]

I've been really trying to get into chapter 2 with in respects to physics to really get into Homer Kelley's mind further beyond than perhaps the text of the golfing machine itself and fully appreciate it.

It is my understanding (and ive just started studying this stuff so its very probable I've got the wrong end of the stick) that if you whirl a heavy object in a consistant orbit that the velocity and its mass (force) of the object being whirled around is always tangential to the point that the object is on the circle. Like if the string was cut - the ball would fly off at 90 degrees to the line of the centrifugal pull out.....

What is puzzling me is at impact the line of compression looks more down the angle of approach in picture 2-C-3 instead of down the tangent of the orbiting sweetspot. I would of thought the line of compression would be tangential to the orbit at the the point that the sweetspot travels through the ball.

Why also is in the picture 2-C-1, the line of compression pointing downwards at low point in comparison to the picture in 2-A?

Any ideas

Matthew ,Hope you've slept well! I think (without being presumtuous)that you are looking at the angle of approach without taking into account the closing of the clubface.If you look at 2-C-1#3 both the angle of approach and the arc of approach give a staightaway line of flight (tangential to the clubhead arc of approach at separation )but this is using Dual Horizontal Hinge Action ,so may not be for hitters.Hope this helps