# Golf Questions Answered by a Physicist, from Charles Pennington

April 21, 2010 |

I was recently asked by a golfer:

Here's a basic physics question for you. One golfer's clubhead speed is 100mph at impact with the ball and 105mph immediately after. He has an accelerating swing. Another golfer has 100mph clubheadspeed at impact but only 95mph immediately after. He has a decelerating swing. Assuming everything else is constant (type of ball, wind speed, etc), does one drive go farther than the other?

The question is a surprisingly hard one, and I've thought about this kind of question in tennis as well. Here are two extreme scenarios that are easy to understand:

1) If the club (racquet) is much, much, much heavier than the ball (hitting a golf ball with a sledgehammer), then the speed of the club is all that matters.

2) On the other hand, if the club (racquet) is much, much lighter than the ball (hitting a bowling ball with a badminton racquet ), then the speed prior to collision hardly matters, and all that matters is how much you shove through during the actual collision.

In any sports/ bat/ ball/racquet situation, you're always somewhere in between the two extreme scenarios described above because that's the best way to design the bat/racquet/club. (It is analogous to having your car in the right gear for the speed that you're driving.) So there's not any simple answer except that it's somewhere "in between". It's neither 1st gear nor overdrive, but something in between. You need high speed, but also a somewhat "firm foundation". (A quibble — it will be absolutely impossible for the swing to be moving faster immediately after compared to immediately before. [I'm modeling the club-ball collision as essentially instantaneous.] Would require you to apply an infinite torque or force or whatever applied by the golfer to the club.)

This seems an obvious application of the basic principle taught to someone learning how to hit a baseball: follow-through on your swing for a better hit. It must be the case that an accelerating swing in baseball, golf, tennis or anything else will impart more force (for a longer movement of the ball), because the moment of contact is longer than instantaneous.

## Sushil Kedia writes:

You have to make an assumption that the movement of the club is following a harmonic motion as in a pendulum. Highest Kinetic Energy at the point of equilibrium and zero at the extremes. Acceleration will have to be negative from equilibrium onwards.

Case 1:

Velocity of the club after impact > velocity of the club before impact: This will be possible only when the point of impact is reached before the point of equilibrium.

Assume mass of the ball is B and the acceleration it achieves on impact is A1. Then the amount of force transferred in this case is B*(A1)^2. Since there is an equal and opposite force it exerts on the club. The net acceleration of the club at the point of impact will be calculated by adjusting the existing harmonic form of the kinetic energy equation MINUS B*(A1)^2.

If this value is 105 mph it only tells us that without knowing the length of the club AND NOT just the effective weight (Center of gravity adjusted leveraged weight for the length of the club) it would be impossible to know at which angle the impact happened.

So either it is not possible for this to happen of there is inadequate data for comparison since you mention everything else is constant.

Case 2:

Velocity of the club after impact < velocity of the club before impact: Without knowing the acceleration of the ball at the point of impact (which can be estimated by estimating the values of friction, time before which the ball stops and the distance traveled) it is not possible to calculate the net force transferred by the club into the ball at the point of impact. EVEN if one assumed that the impact happened after the equilibrium point was reached on the trajectory of the club swing. It is worth mentioning here that the swing of the club would HAVE TO HAVE A NEGATIVE acceleration beyond the point of equilibrium else it is not the point of equilibrium.

So, in both cases 1 and 2 the assumption of everything else is constant is a situation of inadequate data. If however the question is implying that the club hit both the balls at the same angle, then the question is a trick and does not have a solution.

## Stefan Jovanovich writes:

Here is a link to a fascinating site called batspeed.com — the most comprehensive study of the baseball swing on the web.

There's no way to solve this problem theoretically–only experimentally–but there is a good theoretical framework to think about it. You model it as a totally elastic head-on collision between a mass M moving with speed V and a mass m (the golf ball) moving with initial speed 0.

In that case, the final speed of the ball is:

v = V*(2M)/(M+m)

and the final speed Vf of the "club" (*) is

Vf = V*(M-m)/(M+m)

(Note that this is always less than V, and could even be negative, e.g. if a golf club hits a bowling ball.)

The problem is, what do we use for M? Do we use the mass of the club? The mass of the golfer who's connected to the club? The mass of the earth, which provides some friction so that the golfer doesn't slide backwards?

There are some simplifying cases:

If M>>m (sledgehammer hits golf ball), you'll see that the equation reduces to v=2V, i.e. the ball will fly off with twice the initial speed of the club.

If M<<m (feather hits golf ball), then the equation reduces to v~(2M/m)V.

The only way to find out the value of M is to do an experiment: get a real golfer, measure his (pre-collision) swing speed and the speed of the ball, and then use these equations to find the golfer's "effective M". A golfer who braces himself more and has a solid stance might have a bigger effective M than a golfer who uses a more floppy, wristy swing–but he also might have a lower V, so his shot might end up with less speed.

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