The \"Texas\" locomotives of the old T&P railway had a mass of 200,000kg, of whi

Question

The "Texas" locomotives of the old T&P railway had a mass of 200,000kg, of which 136,000 kg rested on the driving wheels. What Maximum acceleration could sucha locomotive attain(withoutslipping) when pulling a train of 100 boxcars of mass 18,000kg each on a level track? Assume that the coefficient of static friction between the driving wheels and the track is 0.25 ( for a rolling wheel, the point where the rotating wheel touches the ground is instantaneousl at rest, thus static friction is the appropriate one)?

Explanation / Answer

the linear acceleration of the wheels must be equal to r x alpha where r is the radius of the wheel (don't worry that you don't have a value for r) and alpha is the angular acceleration

if there were slipping, then the linear accel would exceed r alpha, so that a = r alpha is a condition of rolling without slipping

In order for the wheels to turn, the frictional force acting along the track must generate a torque on the wheels such that

torque to due friction = I alpha where I is the moment of inertia of the wheels and alpha is the angular acceleration, which we know must equal a/r

since the frictional force is always tangential to the wheel, the torque due to friction is f r where f is the frictional force, and we have

f r = I alpha = I (a/r) or f = I (a/r^2)

now we need an expression for I, the moment of inertia of the wheels, and here I think there is a little ambiguity...if the wheels are to be considered solid disks, the moment of inertia is 1/2 M r^2, if they are more like hoops, then I = M r^2...let't choose the latter, and we get

f = M r^2(a/r^2) = Ma (see, no need to worry about r)

now, we also know that the frictional force = u N = u M g, so we have

u M g = M a or a = u g

or the maximum a = 0.25*9.8m/s/s = 2.45m/s/s

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