As you all can see and most definitely feel, it is getting awfully cold outside. While for most that mean bundling up in Canada Goose jackets and strapping on the Bean Boots, for me it is a time of sorrow. The golfing season is coming to a close. So, to pay homage to the departure of this fantastic season, I thought it would be a lovely idea to take a look at the physics of a golf swing! YEAH!
Analyzing the golf swing is best done by looking into the energy of the system, defined KE=-PE+Wnc. Since I will be analyzing the downswing of a golf club (m=.33kg, L=1m), the height of the swing is from 1m. This is further defined as ½(1/12ml2)w2=rFdeltatheta+mgh. The force is applied to the grip on the club about .012 m from the point of rotation and causes the club to swing backward 90o (if it were a chip shot). If the golfer applies a force of 50 N on the club, the angular velocity will be equal to ½(1/12(.33kg)(12)w2=(.012m)(50N)(1.5rads)+(.33kg)(9.8)(1m), so w=17 rads/s. Once club head velocity is found, I can use conservation of inertia (assuming this is an inelastic collision), to calculate the initial velocity of the ball (.045g) which is defined as M1V1+M2V2=(M1+M2)Vf. So [(.33kg)((17m/s)/(1m))+(.045kg)(0)]/(.33kg+.045)=14 m/s. The distance the ball travels varies depending on the angle of the club face used. If the club is a 52-degree wedge, then the ball will hit the ground 9.5m from the impact (calculated by kinetics). When using the same club, the higher the clubhead speed, the farther the golf shot will be.
Now there are many more factors that play into the quality of a golf shot, swing path, quality of impact, spin of the ball, and other less friendly factors like wind and weather. All of these alter where the ball lands on the golf course. So when you see Jordan Spieth drop a 200-yard iron shot with a 20-knot crosswind on a rainy day within 10 feet of the pin, give the man some credit because it isn’t easy.