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Name: Stephanie
Status: other
Grade: other
Location: OH
Country: USA
Date: Fall 2011

If two identical golf balls are dropped from the same height what difference (if any) would there be if one of the golf balls was spinning? (The reason why I ask this is because my daughter is a cheerleader who is tossed in the air and caught by a group of girls. They are working on a new skill where they drop her from approximately two people high. Right now, they just drop her and she seems to come down with a greater force, than when she spins as she comes down.)

I do not think the drag of a person's body will change their falling speed much over only 10 feet or so, spinning or not. It takes some hundreds of feet free-falling before a human body starts developing enough drag to counteract a significant portion of gravity's downward acceleration.

If there is a perception of greater force, start thinking about secondary reasons, such as: - the hands that catch her must stop the downwards momentum and the spinning momentum simultaneously. Bigger jerk on at least some of the arms that catch her. - it might be harder to track a spinning body to move the hands to do the good moving catch that normally stretches the deceleration time longer for skilled athletes. Poorer catch, sharper yank...

Yes, there might be a difference between a golf-ball spinning and a golf-ball not spinning. To tell you the truth I do not have any clear idea which would have more drag than the other. But I bet they would be a little different one way or the other.

You should consider applying scaling concepts, like Reynold's number, or ballistic coefficient, to the comparison of golf-ball and girl.

The prone human body averages something like 6" thick and has the density of water. So it's ballistic coefficient (think of it as frontal density that helps the object punch through air) is 15cm * 1gm/cm3 = 15 gm/cm2.

A golf-ball's density is maybe 1.5 times that of water, and its size is maybe 3cm and it is a sphere. So its ballistic coefficient is density * volume / area = (1.5gm/cm3) * ((3cm)^3*pi/6) / ((3cm)^2*pi/4)) about 3 grams/cm2.

So the golf ball has about five times less frontal density than the human body. So the body would have to fall five times farther than the golf-ball for both to experience a similar drag-to-weight ratio. How much drag would you feel that a golf-ball experiences falling only 2-3 feet? Not much, is my feeling. For that reason I think air drag slows down your cheerleader very little in only 10-12 feet. and a modest modulation of such a small portion of momentum is probably not noticeable.

Jim Swenson

Stephanie, A golf ball that is just dropped will make contact in only one spot. All of the stopping impulse is applied in one place. A spinning golf ball will roll a little during the landing. The impulse is distributed over a greater area. Also, the spinning golf ball will maintain its original orientation. The ball that is just dropped will have a tendency to turn a little in a random direction during the fall.

Dr. Ken Mellendorf Physics Instructor Illinois Central College

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