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Originally Posted by DavidNJ
I actually used kilograms and coverted back...never could keep the English units straight. Matcad does the unit conversions for you though, In this case the constant, g, is the same on both sides so the ratio holds.
However, you are correct, I forgot to convert diameter to radius, so I was off by a factor of 4. This would make the energy equivalent to around 2mph. I knew the number should have been around 4-6 lb*ft²...so I must have been asleep at the wheel here. There is probably another 1-1.5 lb*ft² from the crank, rods, pistons, pins, rings, and bearings.
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David,
Yes, kilograms work, as this is the SI unit of mass. Yet, the masses of the two are an order of magnitude apart, so you would expect a significant variation in their velocities in order for them to maintain the same energy level. Yes, about 2 mph works:
Set the two equations equal to each other, and solve for the velocity of the car.
KE(flywheel) = KE (car)
457.78ft-lb = 0.5(2800/32.2)(v)²
(457.78)/[0.5(2800/32.2)] = v²
(457.78/43.478) = v²
(10.529) = v²
v = (10.529)^(1/2) = 3.24 ft/s = 2.21 mph
You lost me on the units in your reply above (lb*ft²)... if this is energy/torque, then I am thinking that there should not be an exponant, i.e. no squared for the
ft units... is that correct?
I agree that the crank contributes, though it would be a bear to do anything but aproxiamte it
! Don't forget the balancer, as these are typically about 8-inches in diameter and weigh in the 10-14 lb range, with most of the mass out at the perimeter of the disk.
By the way, how did you get the superscript on the exponant? (I cut and pasted, except for the 1/2 power (i.e. squareroot) at the end
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