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| Re: Calculation (36993) | |||
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Re: Calculation |
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Posted by Jeff H. on Sat Jan 8 01:35:34 2005, in response to Re: Calculation, posted by VictorM on Fri Jan 7 15:05:10 2005. I'm glad there are at least two other people who are interestedin this topic! I did all that in my head without the assistance of pencil and paper, so I went back to re-check. I was balancing forces in a line parallel to the axles. Now that I think about it, that is not the direction of centripetal acceleration. The train is moving in a plane which is level, i.e. normal to gravity, not a plane which is parallel to the super-elevated axle. Only a component acts parallel to the track. I should have realized that because in the limit, if the track were tilted 90 degrees, obviously none of the centripetal force would be acting parallel to the track! So, in that case, balancing along the line of the axle, centripetal "force" is cos(x)*m*g*v**2/r, the force is gravity is sin(x)*m*g, solving tan(x)=v**2/(r*g), which is what you got by solving in the other direction (normal to the track). |
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