This is a physics question. Would anyone like to tackle it?

Quote:

a train wheel has a diameter of 30 inches to the rim, which rests on the track. the flange,which keeps the wheel from slipping off the track, projects 1 inch beyond the rim. when the train is traveling 60 mph, what is the linear velocity in mph of a point on the outer edge of the flange?

fun question

Ok there are basically two parts to solving this.

First, given that the train moves forward at 60 mph we have to figure out how fast the wheel is spinning. This is called it's angular frequency and is measured in revolutions/second.

Then given it's angular frequency and the distance to a point on the flange d the linear velocity is

point on flange linear speed = 2*pi*d*(angular frequency)

We have to assume there is no slippage between the wheel and the track.

The wheel makes one circumference for every complete revolution.

In this case this is 2*pi*30 inches = 60 pi inches/revolution

60 mph = 60*5280*12/3600 inches/second = 1056 inches/second

So the wheel has angular velocity of

1056 inches/second / (60 pi inches/revolution) = 88/(5 pi) revolutions/second

Now the flange is 31 inches from the center. So it will have linear velocity of

88/(5 pi) (revolution/second) * 2*pi*31 inches/revolution = 88*2*31/5 = 5456/5

and to put this back into mph we have

5456/5 inches/second * 3600/(12*5280) sec/hr /(inches/mile) = 5456*3600/(5*12*5280) = 62 mph.

I should note that there is a simpler way to solve this.

Since the point on the flange is spinning at the same angular rate as the wheel itself you can just do this

flange linear speed = (distance to flange)/(distance to wheel) * (train speed) = 31/30 * 60 mph = 62 mph