Sears Tower Penny ```Name: Gail S. Status: educator Age: 50s Location: N/A Country: N/A Date: 2001-2002 ``` Question: My 7th grade classes are currently studying Newton's Laws of Motion and we got into a discussion about falling objects. How do we go about calculating the force that a penny would have when it hit the pavement if dropped from the top of the Sears Tower? The tower is about 443 meters high and the penny has a mass of about 3 grams. We figured that by the time it hits the ground (pretending no air resistance) it is falling at a nearly 88 m/sec, but from there we are not sure how to calculate (what formula to use) for the actual force that it will have when it hits. Help. Thank you so much. Gail S. Replies: If you know the velocity of the penny and its mass, calculating it momentum is easy; p = mv, where p is momentum, m is mass, and v is velocity. Force is the change in momentum with time. So, to be able to determine the force of impact, you need to know how long the impact lasts. This depends on a number of factors, including the orientation of the penny and the hardness of the surface it hits. More relevant quantities to calculate would be the penny's momentum and its kinetic energy. Momentum you already know how to calculate; kinetic energy can be calculated in two ways: from the velocity, by E = mv^2/2, or from the height fallen, by E = mgh, where g is the acceleration due to gravity ( = 9.8 m / s^2) and h is the height. This last formula can be used because the kinetic energy of the penny at the bottom of its fall is equal to the potential energy at the beginning. In fact, that's the method I would use to calculate the velocity: first find the kinetic energy, and then calculate the velocity from that. Or, more easily, you could factor out the redundant mass terms ahead of time: ``` E = mv^2 / 2 E = mgh mgh = mv^2 / 2 gh = v^2 / 2 v^2 = 2gh v = sqrt(2gh) ``` Incidentally, when I use this formula, I calculate that the terminal velocity is 93 m/s. Richard E. Barrans Jr., Ph.D. Assistant Director PG Research Foundation, Darien, Illinois Gail, You need to know the acceleration that the penny will experience. The exact value for this depends on what the penny hits. You may get a good estimate by deciding how much time is required to stop the penny. A piece of concrete will stop the penny in much less time than will dirt. In fact, this is part of why landing on concrete hurts much more than landing in dirt. The average acceleration is change of velocity divided by time elapsed to stop the penny. I expect it will be less than a second for each. This is a good opportunity to learn how to estimate. Dr. Ken Mellendorf Illinois Central College Using energy conservation, I got the velocity to be 66 m/s, though I could be wrong. If you assume that the penny hits a hard surface and deforms by 1mm, the force would be 13,000 N. If you assume that the penny hits on its edge, and that the average contact area with the ground is 10mm square, then the "Pressure*Travel" will be 1,300,000 N/m. If the ground has a yield strength of, say, 20 KPSI (my estimate for a human head), then the impact would leave a dent 6cm deep. -Wil Lam Hi, Gail !! Let us start calculating the velocity of the penny. The potential energy at the top of the Tower is : Ep = m.g.h = (0,003 kg)(10 m/s2)(443 m) = 13,29 Joules The kinetic energy at the bottom is : Ec = (1/2) mv^2 = (1/2)(0,003).v^2 Assuming that Ep = Ec then 13,29 joules = (1/2)(0,003).v^2 and finally : v = 94 m/s. (how did you find 88 m/s ??) When the body hits the soil some of the energy will be used to change the physical structure of both the body and the surface of the ground. Heat will be developed. But, let us assume that nothing like mentioned happens. Let us assume that a hole with a depth of 1 cm is formed !! If it is so, than how big was the force to stop the penny ?? We know that : Vf^2 = Vi^2 - 2.a.X 0 = 94^2 - 2.a.(0,01) .: a = 441.800 m/s^2 or F = m.a = (0,003)( 441.800 ) = 1325 N. Well, the correct answer depends on so many factors, that only a more detailed study can solve this question. I really do not know any formula that could help us. regards Alcir Grohmann Click here to return to the Physics Archives

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