The Physics of the Shotput
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Nick Trotter Mr. Croom Academic Physics 30 May 2011 Hot Shots, a Study of Physics in Terms of the Shotput The ancient Greek Olympics used to use a weighted stone toss as a measure of strength, the stronger man could obviously throw the stone farther than a weaker man. Today, the stone is replaced with a metal sphere, and you don't necessary have to be the strongest man to throw the farthest. All it takes today is a deep understanding of the physics involved, and then enough strength to use those principles. The most important principles to understand are: gravity, velocity, and momentum With those three basic principles understood, mastery of the shotput is within reach. In order to really understand the sport however, a certain amount of background knowledge in needed. The shotput originated in the ancient Greek Olympics as a contest of strength. Contestants threw heavy stones while standing on a rectangular wooden board. If a thrower was to step off of the board than his throw would not count. Today's shotput events are largely the same, albeit with a metal sphere instead of a stone. Today's shotputs are weighted at 6, 8, 12, and 16 pounds depending on gender and competition level. A professional and collegiate male will use a 16 pound shotput, a woman at the same level would use an 8 pound put. Instead of a wooden rectangle, competitors today throw from a 7-foot circle with a board at the front end called a toe-board. The thrower must not step over or on the toe -board while throwing, and must leave the circle under his or her own power after a throw has been completed. Then after the shot is thrown it must land within the vector lines, two white lines that start at the toe board and extend outward, the distance between them expanding as they get farther from the circle. There are other, smaller, rules that dictate a throw, but they are small details. With this basic understanding, it comes time for the physics of the shotput. The major deciding factor of the shotput is gravity, and momentarily defying gravity is the key to success in the shotput. Remember that gravity pulls objects to the earth at a rate of 9.8 m/s2. So in order to find the weight of a shotput, you must take the mass and multiply it by 9.8. So a 5.44 kg shot is a 12 lb shot, which is the weight used by male high-school participants. Or you could look at the little numbers etched into the surface, they're usually right Due to the shot's density and mass it is virtually unaffected by air-resistance and drag (Hokin) . With that understood, it is important to note how that affects a shot. When a shotput is thrown it becomes a case of projectile motion, that is to say that the sphere's motion is a combination of both horizontal motion and downward vertical acceleration (Hokin). So in any given throw, a shotput will have a constant change in horizontal motion every .01 seconds, while being pulled to the earth at the constant 9.8 m/s2 from the earth's gravitational field. That leads to a need to compete with gravity by choosing correct angles. Many professional shot putters and sports scientists agree that the best throws come from angles ranging from 40-45 degrees. That is if everybody had the same strength and height to use behind the throws. Instead given the varying heights of athletes and their strength levels, many throw between 35-40 degrees, angles that their bodies are comfortable with, as to put the most power behind the throw (Physics of Shotput). The angle of the throw is important because the right angle can give a throw more distance. Beyond the angle however, to determine the distance of a shot the velocity is needed. Velocity is distance traveled divided by time, measured in meters per second. It is commonly confused with speed, but velocity also includes direction, making it a vector quantity. In terms of a shotput, velocity is created from the sudden push of force that comes from the thrower's body. So, a thrower will use the muscles in his or her body to do work, which in turn creates a force inside the body of the thrower. That force is finally used to push the shotput up and away from the thrower's body. That push creates the velocity of the shot. The measure of the velocity when the shot is first thrown is the initial velocity. To show a mathematical way of equating distance, imagine a thrower who releases the shot at a 40 degree angle, with an initial velocity of 15 m/s. The distance can be determined with this formula: Distance = 1/2 * (acceleration) * (time squared) + (Velocity) * (Initial time) + (Initial Distance) The angle of release is not a variable in this problem; the velocity of the ball as it is released is also not specifically in this problem. To find the correct velocity to insert to the equation, one must use the Pythagorean theorem to find the horizontal velocity. Hypotenuse (the measured velocity) / Adjacent (horizontal velocity) = cos (angle of release 40 degrees). So the algebra ends up being: hyp * cos(40) or: 15 m/s * cos(40). The answer is 11.5 m/s for horizontal velocity. This determines, in combination with angle and time, how far the shot will go. From here, the vertical velocity is simple. Use algebra for the Pythagorean theorem. The answer should be 9.6 m/s. To find the time, we must use the distance formula and plug in all values. The distance is zero because we are measuring the distance relative to the ground. 0 = 1/2 * -9.8 * (t^2) + (9.6) * t + 2. The two represents the initial height. Now use the quadratic equation to solve for time.(-9.6 - Square Rt. (9.6^2 - 4(4.9)(2))) / -9.8 The answer comes out to be 2.15 seconds. This number multiplied by the horizontal velocity turns out to be the distance. 2.15 * 11.5 = 24.7 meters (Physics of Shotput). The throw would be around 81 feet then. Sadly nobody has ever thrown that far, the best throw in the world being 23.12 meter, or 75' 10”. Even that throw was helped by a transfer of momentum, the third area of the physics of shotput. Many throwers, especially at high-school competitions will use a technique called the glide. This technique requires a thrower to leap, or “glide”, across the circle to create a larger amount of force to put behind the throw. Given the law of conservation of momentum, the momentum of the moving thrower is transferred to the shot as it is released. If the glide is executed properly then a serious amount of force can be transferred. Of course, all humans make mistakes, be it releasing to early or late, throwing too low, etc. As much as the physics is a vital part of understanding the sport, it is rarely what goes through a thrower's head before a throw. Sure a thrower might think about releasing at a good height and at the right time, but not because they are calculating the angles interference with their throw. Sure gravity, velocity, and momentum are vital to success in the circle, few throwers fret about calculating them. Physics is everywhere, including the world of shotput. Understanding the rules of physics may be a huge advantage for an athlete, so it's a shame they're so confusing.
Works Cited Hokin, Sam. “The Physics of Everyday Stuff.” Bsharp.org. IMS, 2011. Web. 30 May 2011. "The Physics of Shotput." N.p., 2006. Web. 30 May 2011