What is the average time that a golf driver is in contact with a golf for for the average drive shot?

Question

What is the average time that a golf driver is in contact with a golf for for the average drive shot?

Answer 1

Use Cof M on clubhead and ball and assume elastic collision

>M(V-V')=mv ..(1)

>M(V^2-V'^2)=mv^2...(2)

>mv^2=mv(V+V')

>v=V+V'

insert v=V+V' in equn 1..

>v=2*MV/(M+m)..where M=clubhead mass, m= ball mass and V=swing speed.

If u r ave PGA golfer ur swing speed =100-120mph (TW has 130mph, WR=200mph)

=44m/s..m=.05kg, M=0.2Kg

>v=1.6*44=70m/s

Delta(Mom) ball=mv=aveF*t

Also WD on ball =aveF*d=mv^2/2

where d= impact def of ball (from photographs)=0.5cm approx

>t=2*d/v=0.143mS (1/7th of a millisec)

This will vary according to the mass of the clubhead, and the YM of both it and the ball

Answer 2

Here, figure it out for yourself.

Driving distance D = vx T; where vx is the X vector velocity of the golfball.  T is the total time the ball is in the air.  Although vx is not constant in reality, it is close enough to constant to discount the air drag factor.  Measure both D in meters and T in seconds to find vx in mps.

Total energy at the max height of the drive is TE = ke + pe = 1/2 mvx^2 + mgH; m is ball mass in kg, g = 9.81 m/sec^2, and H is max height of the trajectory above ground in meters. And total energy off the driver head is TE = KE = 1/2 mV^2.

From the conservation of energy, we have 1/2 mvx^2 + mgH = 1/2 mV^2; so that vx^2 + 2gH = V^2.  Solve for V = sqrt(vx^2 + 2gH).  Note H = 1/2 g(T/2)^2; so the 2gH term can be substituted with 2gH = (gT/2)^2 where both g and T are known.

Once you have V, the amount of work (WE) done by the driver head becomes the kinetic energy KE = 1/2 mV^2 of the ball with mass m.  From the conservation of energy, we can write WE = Fc = 1/2 mV^2 = KE; where F is the tangential force on the driver head as it travels c distance during the swing up to and including impact with the ball.  Solve for F.

F = (1/2)mV^2/c = (1/2)mV^2/(theta L).  c = theta L; where L is the length of the driver plus the length of the golfer's arm (in other words, this is the radius of rotation for the swing).  Theta, in radians, is the arc angle of the swing from the top down to impact with the ball...that's the arc wherein the work is done.  You need to measure theta and L for a specific golfer.

Now, to your question.  Note the change in the ball's momentum is from mV = m0 = 0 before it's hit to mV > 0 when it is struck.  So we can write the impulse mV = F dt; where dt = ? is the length of contact time you are looking for.  Thus, dt = mV/F = ?

You measured the ball's mass, you calculated V the velocity of the ball off the driver head, and you calculated F.  Now plug that in and you'll have your answer.

Answer 3

Well I dont know it is a Physics question or sports.

Depends on the player  handicap & how crucial is the hole to win.

Generally on average 1-2 mins.

Well . May I ask how you get time to think such questions ?

Answer 4

golf ball in contact with the club?

I can imagine it would be difficult to find out.  Maybe you should set up the experiment.  its not too hard.

make the ball/club combo into a switch on an electronic circuit, and have a decent timer time how long the switch is closed

if you're part of a University, they should have the stuff and just ask nicely.  a school physics department might have the stuff too

i did a similar thing with ball bearings at uni years ago.  the time was somewhere around 1/10 or 1/100 of a second

steel is much more rigid than golf balls though, so i'd expect the contact time to be longer for the golf ball

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I have to disagree with all models assuming it's an elastic collision.  since it clearly is not one.  If the working does not include the Young's modulus of the equipment you will get the same answer for a steel ball as you would a ball made of Jelly, which is obviously not correct.  

I would try looking for a better model if you need any accuracy...

like this

http://www.oxfordcroquet.com/tech/gugan/...

which talks you through what's going on.

you can see it in action here

http://www.britannica.com/EBchecked/topi...

but measurement would be the best way