Force or Kinetic Energy?

Factual Questions
1

I was once taught that I should remember the following formula while golfing: 1/2M * V^2. The idea was to remember that faster club head speed, which was squared, would bring longer drives. However, I read a question somewhere on this page about heavy versus light baseball bats, and the answere included this formula: mass * acceleration.

What is the difference between the two formulas? Obviously they both indicate that bat or club head speed is more important than the mass of the bat / club. But what does each formula measure? Which one really matters to me when I’m golfing?

2

Man - you must be a better golfer than me! All I think about is, “don’t slice, don’t slice, don’t slice.” :smiley:

Seriously though, neither matters much. In golf, the key is to get the hell out of the way and let the club and ball do their thing. I don’t really think that most of us are capable of producing enough clubhead speed to really make a difference. Even the guys that hit it forever in long drive contests swing at 85% on the course.

3

Bro, that you?

4

To answer the physics part of the OP: F = ma is the more “fundamental” concept. Any classical physical problem can, in principle, be solved using that law alone. However, in many situations, the math would be hellish. Hence, physicists have come up with a number of other concepts, that are sometimes useful and often make the math a lot simpler. One of these concepts is energy, and the formula for one type of energy (i.e., kinetic) is K = 1/2 mv[sup]2[/sup] . My golfing experience being strictly limited to put-put, I can’t say which would be more useful for golf. Probably, though, knowledge of physics won’t help you nearly so much as knowledge of the rules of thumb used by golfers, such as sdimbert’s “don’t slice”.

5

I hope I get this right.

The kinetic energy formula is the time differential of momentum (which is mass times velocity). So it’s all about how fast you’re transferring momentum from the club to the ball. That means time is a factor here as well, even though it doesn’t show in your two equations. The time that matters is the time of contact between the club and the ball. The longer that is, the more of the club’s kinetic energy gets turned into the ball’s momentum. Lot’s of momentum gets you that longer drive.

That contact time is, I suspect, purely determined by form. The accuracy of your form is more important than the speed of the club, because you need to get the energy into the ball, and a poor swing will bounce off the ball without allowing as much momentum transfer.

I’m not sure if that really added anything to the topic, but I enjoyed going through it.

Have you looked up Cecil’s profile here on the SDMB? He lists his interests as “The universe, except golf.” Maybe topics like this will bring him around to being well-rounded.

6

. I think that’s reversed, no?

7

KarlGuass, you’re right. Force, not Kinetic Energy, is the time derivative of momentum:

F = p’ = (mv)’

If m is constant with respect to time (and why wouldn’t it be?) then this becomes:

F = m(v’) = ma

which is broofrhythmdvl’s second equation. Kinetic Energy is defined in terms of Work, the Work needed to get an object going a certain speed. (Kinetic Energy, unlike Force, is a scalar quantity, so speed would be the correct term here, rather than velocity.) The fundamental formula for it is:

KE = W = integral of (F·dx)

This takes a bit more work to simplify, but it eventually boils down to the OP’s original formula:

KE = ½mv²

If you want a better graphical representation, or more in-depth explanation of these formulae, you can look here:

http://www.treasure-troves.com/physics/Force.html
http://www.treasure-troves.com/physics/KineticEnergy.html

Anyway, these are two of the most common formulae in Classical Physics, so I don’t think the OP was having a problem with them. The question seemed to be asking, Does the distance a golf ball travel depend on the golf club’s speed, or on the square of its speed? On this point, Saltire is entirely correct, in that it becomes all complicated and ugly because you have to take impulse and collisional things into account. However, if I make the assumption that the speed of the golf club V is directly proportional to the speed of the golf ball v, then I can answer the question. I didn’t remember the formula, so I did a two-minute calculation and found out that the distance D that a projectile will travel with an initial velocity of v is:

D = v²×sin(2T)/g

where g is the acceleration due to gravity and T is the angle with the horizontal. The important thing here is that D is proportional to v² rather than just v.

Now, as to how important the mass of the golf club is… That’s where it gets complicated. I’m guessing here, but I’d think that in general, the momentum that the golf ball will get will be related to the momentum the golf club has, or:

mv = kMV

where k is some constant. (Somebody please correct me if I’m off in this assumption.) Now then, these formulae together give:

D = (kMV/m)²sin(2T)/g

or D is proportional to M²V². Thus, the club speed and the club mass are equally important. If you have a club that weighs twice as much, but you can only swing it half as fast, then you’ll get the same distance on your drives as if you’d stuck with your original club. However, if you double the mass of your club but keep the swing speed the same, it’ll quadruple the distance of the drive. A far-fetched example, admittedly, but nice to know. As to which formula you remember, Chronos is right. F = ma, as long as you remember your Vector Calculus, and feel like deriving Physics equations on the golf course, will let you figure out anything you want to know.