Energy conservation

rcgldr

The only forces involved when the club is not in contact with the ball is aerodynamic drag and gravity. After contact is completed, the ball is decelerating (both backwards and downwards, one it reaches peak height, it starts accelerating downwards).

Look at that video again, the ball is long gone before the club gets about 1/3rd over the tee. The club isn't going to exert any aerodynamic push on the ball.

Perhaps what you're referring to is the visual illusion that makes a golf ball seem to accelerate well after contact. This is due to a tracking problem by a person. The person overshoots the visual tracking then slows it down, and the ball appears to accelerate as it returns to the center of vision.

newTonn

Then tell me when does the ball accelerated,due to the force exerted by club?

rcgldr

Yes it's accelerated by the force from the club. As posted before this force only last for about 1/200th of a second, but it's enough to accelerate the golf ball from 0 to 170mph. In case you're wondering, that's an average acceleration of 1550 g's, it's no wonder the ball deforms so much from the rate of acceleration.

Doc Al

Where in the world did you get the idea that the ball attains its maximum velocity (from being hit with the club) without moving or in zero time? Of course there is a displacement of the ball as it's being smashed by the club.

HallsofIvy

The face of the golf club deforms a very slight amount, the golf ball deforms much more and the shaft of the golf club bends. All of that takes place in a very short time (according to Jeff Reid, above, about 1/200 sec.) and it is during that time that the acceleration takes place- while the face of the golf club is in contact with the ball.

newTonn

Sorry if anybody got frustrated on my arguments.But please go through some points from one of your previous post to find the answers for your question.(text was made bold-by me)

the club was in contact with the ball ,when ball was at rest.ball moved a millionth of an inch means no contact and hence force on it is zero.
consider the ball tooks a millionth of second to move millionth of an inch,(acceleration from zero to final velocity).what is the force on body in between this interval? to be specific,the interval between billionth of a second after club left its contact and billionth of second before ball reaches its final velocity-For acceleration,of course ball has to change its velocity from zero to final velocity-it has to be displaced-may be fractions of our standard units,but those fractions are again divisible.

Doc Al

I'm not sure what you are saying here. When the club first makes contact with the ball, of course the ball is at rest. During the interaction, both club and ball move a bit. Once the ball shoots off the face of the club, there is no longer any force exerted on the ball by the club. What is so hard to understand about this?


Again you repeat the same misconception. For some reason you think the ball reaches maximum speed some time after it breaks contact with the club. No! As soon as the club loses contact with the ball, the only forces on the ball are gravity and air drag: these forces act to slow down the ball.

Ariste

I don't understand why you keep saying this. In those couple of hundredths (not billionths) of a second that the ball is in contact with the club, the ball is accelerated from rest to around 170 mph. It attains ALL of this velocity in the time that it is in contact with the club. After the ball leaves the club, it is no longer accelerating forward. At the instant that the ball leaves the club, it is traveling at 170mph. There is no additional acceleration afterwards, except in the backwards direction due to air resistance.

newTonn

I am saying again that ,as i understand,velocity is the rate of displacement.
So while the club was in contact with ball,since ball didn't displaced.how somebody can say there is a velocity for ball ,without being displaced?
If you corrected your statement as ,while the ball is in contact with the club,it is attaining a potential(kinetic energy),to move at 170mph,i will agree.yes. throughout the path of the ball this potential is reduced until it becomes zero,due to the interaction of other forces.
Any how if you can represent this as a vector.you can clearly draw a freebody diagram of the ball,at any point in the path of ball,justifying the movement of ball (yes i understand ,it is accelerating ,negatively because the net force(resultant is in forward direction,but getting reduced)
My point is that if you consider this potential/kinetic energy as force,you have to consider same for a body moving with uniform velocity also.

Ariste

The ball is displaced while it is in contact with the club. It's only displaced by a small amount before it leaves the club, but it is definitely displaced.

Here, I'll give you some mathematical figures. Assume that the ball weighs .05kg and that it reaches a final velocity of 80 m/s. Thus the club gives the ball .5(.05)(80)^2 = 160J of kinetic energy. This means that the club did 160J of work on the ball. Let's assume that the club exerts an average force of 3200N on the ball. Using the formula W=Fd we conclude that the ball was in contact with the club for 160/3200 = .05m.

Obviously these numbers are all made up, and I think they're only valid when the ball is hit perfectly horizontally, but you can see that there most certainly is a displacement while the ball is in contact with the club, and thus a velocity.

newTonn

Really i am confused by by your last statement.Any how i think you are explaining that ball was in contact with the club until it reach 0.05m (in your example).
if you assume that the club exerts an average force of 160N(instead of 3200N) on the ball,the club will be in contact with the ball until it reach 1 m.This i feel has no meaning.

Let us look at the problem in another way.
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 10 kg m/s2.
So acceleration a =f/m = 10/.05 = 200 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;
now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 200 = 5 meter (does it make sense that the club was in contact with the ball for 5m?)
Please check .i am not sure about my calculations

Note:Logic and concept of the problem is Not mine-it belongs to Ariste.
Approach and new figures are the only contribution from my side

Cthugha

Your example is just nonsense.

Usual forces acting on golf balls range between 3000N and 15000N.
http://hypertextbook.com/facts/2001/...ccamando.shtml

Hitting a ball with just 10N will never get the ball to 200m/s...unless the force works on the ball for a long time, which means long contact.

In reality you will just have much less than 200 m/s. Using 10N, you won't even reach 15 m/s, I suppose.

Using silly numbers produces ridiculous results.

newTonn

Yes may be practically you have to use more force.because here in my calculations,i didnt consider any other forces like gravity or air resistance.(Moreover my unit is meter/second -not miles per second).But i think for force acting on the ball (due to club),i don't have to bother about other forces.

Also at the end of article,i find somebody has calculated a mere 22.5 N of force but the author ignored it in a democratic way.(here any body can get a doubt that wheather the force was calculated by scientist's or golf player's?- I am just joking)

Even Using some relevent figures ,i am getting some ridiculous results,here it is.

if you use the fundamental equation of net force ,F = ma; for a force of 3000N,the ball should accelerate by, a = F/m = 3000/.05 =60,000 m /sec2 or we can say this force can accelerate a shot put ball(15 kg) to 200 m/s2.

And to be frank,I never read(ignorance) anywhere in physics text books ,and i din't find in any equations for force about the relevence of contact period of force.
If you can help me on this ,giving me an equation for force,in which,we can use contact period as a variable,then may be i can agree with you.

Cthugha

Why not? This sounds ok to me.
An accelaration of 60 km/sec^2 means a velocity of 60 m/sec, if the contact time is 1 ms (using v=at). That sounds sensible.

An acceleration of 200 m/sec^2 concerning a 15 kg ball sounds ok as well. Using 1 ms as contact time again gives a velocity of 20 cm/s. I don't know the exact contact time in real situations, but the ms time range seems plausible.

This page provides some further reading about the physics of baseball, including citations of some reviewed publications:
http://www.kettering.edu/~drussell/bats-new/impulse.htm

K.J.Healey

You did that calculation correct, just bad numbers.
Lets change that force the club exerts to 6000N which is more reasonable.
Why? Well imagine the deformation the ball goes through. What sort of "weight" would be needed in a static world to get it to deform that much? You couldn't put 10N of force on it and expect it to deform. It takes quite a bit.

So with 6k as the new number.
"
Let the weight of ball remains same-0.05 kg.
Let the club exerts a force of 6000 kg m/s2.
So acceleration a =f/m = 6000/.05 = 120,000 m/s2.
since a = (v-0)/t ; Final velocity v= 200m/s;
now kinetic energy = 0.5 (0.05) * 200^2 =1000 kg m2/s2
Now d = W/F = 1000/ 120,000 = 8.3mm (does it make sense that the club was in contact with the ball for 8.3mm?)
"
Yes, with a realistic force the distances becmoe much more realistic for the displacement of the ball during acceleration.
The time would then be
dt = dv/a
dt = 200 / (6000/0.05) = 200/120000 = 0.00166 seconds

So it accelerates to 200m/s in 0.0016 seconds over a distance of 8.3mm

You'll see that the more force you give, for a given final velocity, the less distance the force can be applied. This makes sense because the object accelerates only when it has a force, so its very "touchy" where a huge force has to be applied for only a split second for it to accelerate to that speed.

K.J.Healey

Haha, I remember all of that. I went to KU and worked/learned with/from Dr. Russell, though I was never big into acoustics.

Ariste

No, it doesn't, but you chose terrible numbers. A club will never exert merely 10N.

Honestly, though, the numbers are irrelevant. You can see that, choosing ANY number that you like, the ball is in contact with the club for a finite displacement. I have no idea what you are trying to argue anymore.