# Golf Physics

1. Jul 15, 2009

### cepheid

Staff Emeritus
When a golf club strikes a ball, is the ball carried along with the club face for part of the circular arc of motion (after impact)? If so, how does this occur?

For instance, I am assuming that the impact takes place at the instant that the club is exactly vertical. I am also assuming that the club face moves at a constant velocity, at least until impact. I am assuming that the force that the club face exerts on the ball is in a direction normal to the club face. This suggests that if you want to get any loft, the club face should be angled.

If the club face is angled, then a component of the normal force is vertical, and a component of it is horizontal. However, I'm still not clear on how exactly the ball remains in contact with the club face, moving in a circular arc. Its weight can be balanced by the upward component of the normal force, but what is keeping it from moving off in a straight line? Is friction involved?

Is momentum conserved in this collision? (The momentum of the club face is continuously changing anyway). Is the collision inelastic, because the two objects "stick"?

Also, what, eventually, is responsible for the ball being released and moving (presumably) tangent to the circle?

2. Jul 15, 2009

### Pengwuino

Check it out :)

Last edited by a moderator: Sep 25, 2014
3. Jul 15, 2009

### Danger

I know only enough about golf to not play it in front of witnesses.
Since a golf ball is compressible, I suspect that a fair bit of the initial impact is absorbed by deformation before the ball starts to accelerate. Also, club faces are angled specifically to alter the trajectory of the ball. A #1 driver is relatively straight, while a sand wedge is somewhere around 45° from vertical.

4. Jul 15, 2009

### Pengwuino

Yes it is angled. Infact, one way to hit a 3-iron or 2-iron if you are in a bunch of trees is to line up so that you'll hit "infront of" the ball. You line up so that the ball is behind where you normally center a ball. That allows you to hit the ball horizontally because if you're still in the downswing, there's a point that the club face is 90-degrees to the ground which results in a very low shot.

There's also the opposite shot that you use sand wedges with. You hit behind the ball so that the club hits it on the upswing and you get a huge angle on the shot.

5. Jul 15, 2009

### Danger

I see that the tuxedoed one beat me to it. Nice clip.

6. Jul 15, 2009

### cepheid

Staff Emeritus
So the answer would seem to be, "No, the ball is not carried along with the club face for any significant distance, only for a short period of time during which the club does work on the ball by simultaneously deforming it and accelerating it."

Yes? No?

By the way, I knew that there were different kinds of clubs with different lofts, I was just sort of stating the obvious for the sake of building up my premise.

So I gather that the answer to this question is:

"The collision is clearly inelastic, not because the two objects stick, but because some energy goes into deforming the ball."

Any takers?

So I gather the answer to this question is: "Newton's laws. Once the ball has finished its elastic deformation and the ball has accelerated to its final speed (which is determined by what, exactly?), the trajectories of the club face and the ball will diverge because one is moving in a circle and the other is moving in a straight line?"

In case it wasn't clear, I want people to correct my inferred answers to these questions in case my inferences are wrong.

7. Jul 15, 2009

### Danger

I'm way out of my depth here. The clip is still a bit fast for me to be sure, but it appears that the ball has rebounded to its spherical shape before parting company with the club. If so, the energy that squished it would be regained and aid in acceleration.
I don't know how much the flexibility of the club itself assists. There's a bit of a 'whiplash' effect, wherein the club shaft bends at impact and then springs forward. That, combined with the bounciness of the ball, might result in more acceleration than would be expected from two solid objects in collision at the same speed.
I'm afraid that that's all I've got, and it might not even be right.

8. Jul 15, 2009

### Staff: Mentor

And it is. A typical driver is 9.5-10.5 degrees.
The ball deforms, acting like a spring, and remaining in contact with the club face while it accelerates.
I'm not clear on what you mean here. The ball is basically reflected off the club face. If the club face is angled at 10 degrees, the ball is lofted by 20.
No, the two objects don't stick, the ball bounces off the club face in a near perfecly elastic collision (try dropping a golf ball on the ground and you'll see just how elastic it is - better than 90% I would guess). Momentum and energy are conserved.
It is a pent-up spring when fully compressed by the club.

9. Jul 15, 2009

### Staff: Mentor

For visualization, it may be easier to consider what a golf ball looks like when bouncing off the ground. The dynamics of that are exactly the same as a ball getting hit by a golf club. Since the club speed is almost constant, you can change reference frames and consider it stationary and the golf ball being bounced off it at it at 120 mph.

Just bounced a golf ball off my cement garage floor. It was less elastic than I expected: only around 80%.