View Full Version : How many "G's" produced here?
Galveston1
Nov 10, 2008, 04:21 PM
This may be the wrong thread, BUT:
How much "G" force is generated by a spoke 4,000 miles long, the outer end moving at 1,000 miles per hour? I'm not a good enough mathematician to figure this.
My curiosity was piquéd yesterday by something the Pastor said.
Scleros
Nov 10, 2008, 06:49 PM
See if this is correct:
The centripetal force, F, required to keep a mass, M, at end of spoke of radius, R, moving at a velocity, V, in a circular path = (MV^2)/R. But Newton's second law states F = (Mass)(Acceleration), so F= MA = (MV^2)/R. Solving for A = V^2/R = (1000 miles/hour)^2 / (4000 miles) = 250 miles/sec^2 = [250 miles * (5280 feet/1 mile)]/sec^2 = 1,320,000 feet/sec^2.
If 1 G = 32.174 feet/sec^2, then (1,320,000 feet/sec^2)/(32.174 feet/sec^2) = 41,027 Gs
My gut reaction to the value is that it seems high, although I have no collection of working knowledge for a reference. (*shrug*)
ebaines
Nov 11, 2008, 07:29 AM
You were fine right up to A = V^2/R. That is the correct formula. But it seems you messed up on converting units - the units your formula yields is mile/hr^2, not mile/s^2. Multiply your 250mi/hr^2 by (1 hr/3600sec)^2 *5280 ft/mile, and you get an acceleration of a= 0.102 ft/s^2, which is about 0.003g's. Virtually nothing. You can see this seems right if you consider that your spoke is the distance from the surface to the center of the earth. Moving at 1000 MPH is not quite twice the velocity of a typical jet airplane. So your question is basically: how many g's does a airline passenger feel in cruising flight? Obviously the answer should be close to nothing. Hope this helps.
Capuchin
Nov 11, 2008, 07:50 AM
Yes - thinking about it quickly and simply - the entire circle swept out by the arm is 25000 miles in circumference - after an hour you've only moved around 1/25th of the circle - i.e. you're moving so slowly compared to the size of the circle that it's comparable to moving in a straight line.
Capuchin
Nov 11, 2008, 07:53 AM
On second thought, seeing that 1/25th, these figures are comparable to the radius of the Earth and the speed at which someone on the Equator moves with respect to the center of the Earth. What's the real question galve? :)
(moving to physics.. )
Scleros
Nov 12, 2008, 09:13 PM
But it seems you messed up on converting units
LOL, some things never change. Same problem I had taking tests in school - always had to go through them twice.
harum
Nov 13, 2008, 10:51 PM
a=v^2/r; a= (1000 mi/hr * (1/3600) hr/sec * 1609 m/mi )^2 / (4000 mi * 1609 m/mi) = (1.609/3.6)^2 / 6.4 = 0.2/6.4 = 0.03 m/(sec^2); g = 9.81 m/(sec^2) Answer: a/g = 0.03/9.81 = 0.003. What does your Pastor say about evolution?