From: "Rob McDonald" Subject: Re: Terminal Velocity of a Penny Date: Thu, 16 Sep 1999 01:55:48 -0400 Newsgroups: sci.math You are correct that an object in a steady fall will reach a terminal velocity. Aerodynamic drag is proportional to velocity squared. Terminal velocity is the velocity where the aerodynamic drag force exactly cancels out the gravitational force acting on an object. The tough part of this problem comes in when you consider that a penny will tumble as it falls. This tumbling will result in the profile presented to the relative wind changing with time. This unsteady problem would be quite difficult to model. This material would be covered in any fluids/aerodynamics text. An online aerodynamics text, including an online standard atmosphere calculator, can be found at: http://www.desktopaero.com/appliedaero/appliedaero.html Consider a penny falling normal to the flow field. (While this is obviously the most unstable orientation for our penny, it also presents the most conservative steady state calculation we can perform.) The airflow will reliably separate around the circumference of the disk, and as a consequence, the drag coefficient will be nearly invariant with respect to Reynolds number. (This is a very handy simplifying condition for this problem.) The drag coefficient (C_d) for a disk oriented normal to the flow is approximately 1.18. Drag (D) is equal to one half the product of the drag coefficient, cross sectional area (S), local density (rho), and velocity squared (V^2). The density of air at sea level on a standard day is 2.37e-3 slug/ft^3 The area of a penny is 3.068e-3ft^2 (http://www.usmint.gov/circulating/specifications.cfm) The force due to gravity on the penny is equal to its mass (m) times the acceleration due to gravity (g) F=m*g The mass of a penny is 1.713044e-4 slug Acceleration due to gravity is 32.2 ft/s^2 Equating the drag force and the weight of the penny, we can solve for its terminal velocity. A simple matlab code follows: C_d=1.18; % Drag coefficient rho=0.00237; % Sea level density (slug/ft^3) r=0.5*0.750/12; % Radius of a penny (ft) m=2.5*6.58217e-5; % Mass of a penny (slug) g=32.2; % Accel due to gravity (ft/s^2) S=pi*r^2; % Area of the penny W=m*g; % Weight of the penny V=(W/(0.5*rho*S*C_d))^0.5 Our penny would fall at 35 ft/s. An expression for the acceleration of our penny is: a=g-0.5*(rho/m)*S*C_D*V^2 Numerically integrating this expression for 100 seconds in Matlab, [t, x]=ode45('xprm',[0, 100],[0, 0]); where the file 'xprm.m' exists and contains: ********************************** function xprime=xprm(t,x) C_D=1.18; % Drag coefficient rho=0.00237; % Sea level density (slug/ft^3) r=0.5*0.750/12; % Radius of a penny (ft) m=2.5*6.58217e-5; % Mass of a penny (slug) g=32.2; % Accel due to gravity (ft/s^2) S=pi*r^2; % Area of the penny V=x(2); a=g-0.5*(rho/m)*S*C_D*V^2; xprime=[V; a]; % x=[x xdot] % xprime=[xdot xdbldot] ********************* EOF***** You will see the penny reaches its terminal velocity after about 3.5 seconds, and about 100 ft of freefall. Dropping a penny from any higher would not increase its velocity. Repeating this exercise for a penny falling edge on would bound the problem in question. Drag coefficient for this orientation is a bit more tricky to determine, but a conservative estimate follows. The new drag coefficient: C_D=2.0 The new planform area: S=3.1783e-4 ft^2 All other aspects of the problem remain the same. The edge on terminal velocity would be about 84 ft/s (57mph), requiring about eight seconds of acceleration in about 500 ft. The momentum of these two pennies can be calculated as p=m*V 0.0057 slug*ft/s (face on) 0.0138 slug*ft/s (edge on) Assuming an impulsive collision with a pedestrian, with the time of interaction taken as 0.015 seconds (approximately the time a baseball is in contact with the bat) The impulsive force is p=F*dt 0.3855 lbf (face on) 0.9200 lbf (edge on) Spread over an area as pressure P=F/S 18096 psi (face on) 416870 psi (edge on) Looking up the ultimate compressive strength of some common materials, (ksi=1000psi) Oak 7ksi Granite 15-26 ksi Steel 30-60 ksi So unless I have slipped up somewhere along the way, I sure wouldn't want to be under that penny..... Sorry for the jumbled, rambling responce. Rob > Discussing projectile motion in a calculus class, I was asked if it were true > that > a penny, if dropped from a significantly high location, would achieve a > velocity > such that it could do "serious damage" to someone's skull. > > I explained how, in reality, air resistance would reduce the acceleration to > the point where the velocity would remain relatively constant (terminal > velocity). > > Does anyone have an idea as to the terminal velocity of a penny? That is, > is it dangerous (or just plain stupid) to throw coins off of the Empire State > Bldg? ============================================================================== From: "Rob McDonald" Subject: Re: Terminal Velocity of a Penny Date: Thu, 16 Sep 1999 02:06:05 -0400 Newsgroups: sci.math Oops, I spread the force of the edge on penny over its projected area. It would actually impart its force over the VVsmall(ideally infnitesimal) area of the curved edge of the penny. This would drive the pressure WAY up. > Spread over an area as pressure > > P=F/S > > 18096 psi (face on) > 416870 psi (edge on) >