Locomotive pulls a boxcar along circular track of radius R = 400 ft?

Question

Track gauge is D = 4 ft 8 1/2 inch.
http://en.wikipedia.org/wiki/Track_gauge
Radius of boxcar wheels is r = 1ft.

How many more revolutions did left wheels
of the boxcar, than its right wheels, after
locomotive copletes one loop.

Answer 1

The wheels can move slightly left or right, as the axle length allows this. Also, the contact surface of the wheel is not cylindrical, but slightly conical with the wheel diameter being slightly greater on the inner side. This means that as the boxcar moves outward from centrifugal force when the train track is curved, as circular, the outer wheel has a slightly greater diameter than the inner wheel at their respective contact points.
Since the wheels are fixed to the axle, they both turn at the same rate, quite contrary to your assumption. But given the taper of the outer ring of the wheels, there is much less sliding than your question implies. This problem was addressed quite some time ago.

Answer 2

None. What kind of a stupid question is this?

Did the left wheels take off and ditch the right wheels?

Left wheels to right wheels:
"Screw you guys, we're sick of going slower on the inner rail and all the increased wear and tear of being the inside wheel on a turn. We're out of here. We're gonna travel the same speed as you and leave you behind on the turns."

Edit--oh, you mean revolutions of the wheel, not of the track. In that case, give the outside wheels an extra 2 pi * guage of distance covered. So supposing they were free to rotate separately (which they probably are not if they are on the same axle)...

Extra revs = 2 pi * gauge / 2 pi wheelradius
= gauge / wheel radius

Answer 3

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Answer 4

So in other words, the outside wheels slid along the track a little bit since they're on the same axle.

Or the inside wheels spun out, slightly.