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 Author: acroduster1 Time: 04/26/04 06:06 PST This is a reply to message no. 13707 by rubinho Reply | Original Message | New Topic | List Topics | List Messages on This Topic
 Current Topic:Calculating trajectory of ball rolling down slope and off launch ramp Bravo, rubinho - you did it right!  Neglecting friction, the energy equation for this type of problem will have the mass term cancel out.  I knew this from the start and was hoping you'd get it.  Here's the calc with no numbers for a solid sphere (it would work with the mixed sphere as well): Basic energy equation:    PE     =        KEt        +          KEr m*g*h = 1/2*m*v^2 + 1/2*I*omega^2 For a sphere rolling without slipping, v = omega*r omega = v/r And, I = 2/5*m*r^2 Plug those in the basic equation: m*g*h = 1/2*m*v^2 + 1/2*2/5*m*r^2*(v/r)^2 Notice at this point the m's can be completely cancelled out.  Do the math and you end up with: v = SQRT(10*g*h/7) This is your velocity no matter what the mass of the ball is.  If you are in a vacuum, and you launch at the same speed, you will land in the same place - sorry rorschach -  Newton's Laws work horizontally as well as vertically.  As for using magnets to deflect uranium atoms in a cyclotron - that's another story. As stated before, this calc does not consider friction.  If you were separating ping pong balls and ball bearings of the same size, things would be different.  I really don't believe you'd get any appreciable difference in rolling friction if using a hard surface.  Let's look at air friction: Drag equation: F = 1/2*Cd*A*rho*v^2 Where Cd is the drag coefficient (about 0.3 for a smooth sphere), A is the cross sectional area, rho is the air density (1.19kg/m^3 at 25 deg C) and v is the velocity. Let's find our velocity first.  Using the reduced equation from above, with a 2.5 meter drop in our ramp from start to launch, v = SQRT(10*9.81*2.5/7) = 5.92m/s Plug this into the drag equation for a 20mm dia. ball: F = 1/2*0.3*(.01)^2*pi*1.19*5.92^2 = 0.002N Applying this to a steel ball weighing 0.033kg or a nylon ball weighing 0.004kg will make some difference. If the force is applied constantly (we know it's not - it decreases as the ball slows), the speeds after two seconds of travel would be as follows: F = m*a, subscript s for steel, subscript n for nylon, as = .002/.029 = 0.07m/s^2 an = .002/.004 = 0.5m/s^2 vfs = vis - as*t = 5.92 - .07*2 =  5.78m/s vfn = vin - an*t = 5.92 - .5*2  =  4.92m/s This would imply that the steel ball would travel 17% farther than the nylon, but we know it's over estimated.  Let's say that taking into account for the force lessening as the ball travels, and the fact that the retarding force isn't always fully opposing horizontal travel that the difference in speeds is half as much as above, say 8%.  This would mean that, by the rough calcs above, there is a pretty small window between your two extreme cases (steel vs nylon), not to mention the steel/nylon hybrid that would fall between them.  I would concur with your conclusion that this is a poor separation technique in your situation.  You might do this write-up in your report to show the work to disprove your original idea - professors love that... As your final design, I'd go with the tipping trough used in coin-ops as previously suggested.