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Name: Kelvin
Status: other
Grade: other
Location: HI
Country: N/A
Date: 4/1/2005


Question:
I see that many people apply the equation force equals mass times acceleration to a golf club hitting a golf ball. But isn't this related to torque since this is rotational dynamics going on here?

The follow up question to that would be if a 6'2" golfer swung at 120 mph, would his ball fly farther than a 5'6" golfer swinging at the same speed? All other things being equal, is this correct?


Replies:
No, the mass and acceleration in "mass times acceleration" is the mass and acceleration of the club. All things being equal, a 120 mph club swung by anybody is a 120 mph club to the ball regardless of the person hitting it. The velocity of the ball is related to how well the momentum of the club is transferred to the ball. However, in order to accelerate the club, a 6'2" person will use less muscle to get the club up to speed compared to the 5'6" person. That is the mechanical advantage of the moment arm that you mention. Hope this helped. Thanks for using NEWTON.

Chris Murphy, P.E.
Air Force Research Lab


Kelvin,

The situation actually involves both rotational and translational dynamics. Most of the rotation is in the golfer bringing the club up to speed. The "120mph" is the translational speed of the golf club's head, the end of the club.

The club is in contact with the ball for only a short time. Torque from the golfer's hands on the club, during the contact time, has essentially no effect on the ball. The club head collides with the ball. Energy is transferred from the club head to the ball. If most of the club mass were near the handle, you would have to use torque. Since almost all the mass of a driver is in the club's head, using force works just as well. If both golfers' clubs have the same head mass, then they will both send the ball to about the same distance.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College


Yes, torque is involved. However, the mass of a golf club is concentrated in its head, so the approximation that linear momentum of the head governs the collision is a reasonable place to start. (If *all* the mass of the club were in the head, you could use either linear or angular momentum and get the same answer. The approximation just ignores the angular momentum and kinetic energy of the shaft.)

The 6'2" and 5'6" golfers would hit the ball the same distance if they both made the club head move at 120 mph, either in the approximation that linear momentum of the head is all you have to consider, or with all other things being equal.

However, all other things are not likely to be equal. The 5'6" golfer's shaft will likely be rotating slightly faster, and will thus have slightly more angular momentum (it depends on how much of the club speed comes from arm motion, and how much from wrist snap). The shaft with the greater angular momentum will transfer more energy and momentum to the ball, assuming the shaft is rigid and all remaining things are equal.

However, all remaining things are not equal, because the 6'2" golfer must use a longer club (or be a gorilla), which will have a greater moment of inertia. At the moment of impact, the ball will try to make the golf club rotate "backwards" about its center of mass, and the club with the larger moment of inertia will rotate less -- all other things being equal.

Tim Mooney



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