Just one thing, I've read this book and the whole of chapter 2, which surprisingly also includes '2-A resilience' so much I can just about quote it all by memory. When someone says 'it may help to read 2-A' it really bugs the hell out of me...
I actually mean't 2-C-1#3.... it was accidental... I did not mean the lob shot pictures of 2-C-3....
I know the line of flight and the line of compression are seperate. You must take me to be a real idiot or something... That was not what I asked.....
My question is why is the clubhead force going down the angle of approach like in picture 2-C-1#3 instead of off a tangent of the circular clubhead orbit.
Matthew I definately do not think you are an idiot.I was merely responding to your post,which was, I thought, most interesting as this aspect of the book is seldom discussed.I am sorry you took offence as none was meant.I am not trying to make an excuse but originally my post was much longer,but when Itried to preview it ,a note came up saying it was an "invalid thread "or something like that (maybe admin can look into that as it has happened before)I was a little pushed for time so I posted the short thread which does sound a bit condescending in hindsight .I assure you I was only trying to help.
Matthew I definately do not think you are an idiot.I was merely responding to your post,which was, I thought, most interesting as this aspect of the book is seldom discussed.I am sorry you took offence as none was meant.I am not trying to make an excuse but originally my post was much longer,but when Itried to preview it ,a note came up saying it was an "invalid thread "or something like that (maybe admin can look into that as it has happened before)I was a little pushed for time so I posted the short thread which does sound a bit condescending in hindsight .I assure you I was only trying to help.
Its ok, its probably not just you, I had just woken up also so I was probably 'crabby' mood anyways.
Anyways looks like my sleep after the nightshift isn't happening so I might ask some questions just now....
I didn't do physics at school past standard grade and just passed that and no more just using common sence and pretty much no study other than knowing speed=distance/time...lol. I was good with maths though which helped some....but im a REAL newbie at this...
Quote:
To maintain a certain velocity of anything, it is always accelerating ?
No...a constant velocity means NO acceleration. Acceleration causes a change in velocity.
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 ?
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.
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.
Hybrid Mode
