Amit & Deepali Sinha Foundation

A Turn Loads the Outer Rail

Irodov 1.289DifficultyUniqueness

Problem Irodov 1.289

A locomotive is driven by a turbine whose rotor turns on an axle that is horizontal and lies across the vehicle, parallel to the wheel axles, and the rotor turns in the same sense as the wheels do. The rotor’s moment of inertia about its own axle is I=240I = 240 kg⋅\cdotm2^2 and it runs at n=1500n = 1500 revolutions per minute. The locomotive travels a horizontal circular arc of radius R=250R = 250 m at the speed v=50v = 50 km/h, and the two rails are l=1.5l = 1.5 m apart. Find how much the force pressing on the rails is changed by the gyroscopic effect of the spinning rotor, and say which rail gains and which loses.

The turbine seen from behind the locomotive, which is rounding a left-hand curve. The rotor spins at ω about its transverse axle, carried by two bearings on the locomotive frame, and the vehicle as a whole turns about the vertical at Ω.

The turbine seen from behind the locomotive, which is rounding a left-hand curve. The rotor spins at ω\vect\omega about its transverse axle, carried by two bearings on the locomotive frame, and the vehicle as a whole turns about the vertical at Ω\vect\Omega.

Given
  • I=240I = 240 kg⋅\cdotm2^2
  • n=1500n = 1500 rpm =25= 25 rev/s
  • R=250R = 250 m
  • v=50v = 50 km/h =13.9= 13.9 m/s
  • l=1.5l = 1.5 m
Find
the additional force on the rails, and which rail carries it

Solution

Step 1How fast the track turns the locomotive

Rounding the curve at a steady speed, the locomotive swings about the vertical through the curve’s center at the same rate at which its position angle advances, and that rate is the speed divided by the radius. Everything bolted to the vehicle, the turbine’s bearings included, is carried round at that angular velocity.

Ω=vR=13.9 m/s250 m=5.6×10−2 rad/s\Omega = \frac{v}{R} = \frac{13.9\ \text{m/s}}{250\ \text{m}} = 5.6\times 10^{-2}\ \text{rad/s}
Step 2The angular momentum the rotor carries

The rotor turns about its own axle at nn revolutions per second, which is 2πn2\pi n radians per second, and its angular momentum is that rate times its moment of inertia, directed along the axle in the sense the right-hand rule gives. Because the axle is transverse, that vector points horizontally across the track.

L=2πnIL = 2\pi n I
(1)
Step 3Turning that vector sideways demands a moment

The transverse axle is carried round the vertical with the vehicle, so the vector of (1) keeps its length and turns at Ω\vect\Omega. Since Ω\vect\Omega is vertical and L\vect L is horizontal, the two are perpendicular, and the moment the mounting must supply is their product in magnitude, directed along the third perpendicular, which is the direction of travel.

N=dLdt=Ω×LN=ΩL\begin{align*}\vect N = \frac{d\vect L}{dt} &= \vect\Omega\times\vect L \\ N &= \Omega L\end{align*}
N=vR⋅2πnIN = \frac{v}{R}\cdot 2\pi n I
(2)
The same view with the moment of added. The moment N the mounting must apply to the rotor points along the track, out of the page here, and is drawn as a circled dot.

The same view with the moment of (2) added. The moment N\vect N the mounting must apply to the rotor points along the track, out of the page here, and is drawn as a circled dot.

Step 4The reaction reaches the ground through the two rails

By Newton’s third law the rotor returns that moment to the locomotive, as a couple N′\vect N' about the direction of travel: the frame is being rolled sideways. The frame can only pass such a rolling couple to the ground through its two lines of wheel contact, which are ll apart, so one rail is pressed harder and the other by exactly as much less.

The two rails seen from behind on the same left-hand curve, so the outer rail is the right-hand one here. The couple N' points into the page, and the rails take it as equal opposite load changes Δ F across the gauge l.

The two rails seen from behind on the same left-hand curve, so the outer rail is the right-hand one here. The couple N′\vect N' points into the page, and the rails take it as equal opposite load changes ΔF\Delta\vect F across the gauge ll.

Writing that couple as the pair of forces a gauge apart, and putting in the moment from (2),

N=Fadd l\begin{align*}N &= F_{\text{add}}\,l\end{align*}

With the data,

Fadd=2π(25 s−1)(240 kg⋅m2)(13.9 m/s)(250 m)(1.5 m)F_{\text{add}} = \frac{2\pi(25\ \text{s}^{-1})(240\ \text{kg}\cdot\text{m}^2)(13.9\ \text{m/s})}{(250\ \text{m})(1.5\ \text{m})}
Fadd=2πnIvRl=1.4 kNF_{\text{add}} = \frac{2\pi n I v}{R l} = 1.4\ \text{kN}
(3)
The outer rail is loaded more, the inner rail less.
Check

The units collapse to a force: (s−1)(kg⋅m2)(m/s)/m2=kg⋅m/s2(\text{s}^{-1})(\text{kg}\cdot\text{m}^2)(\text{m/s})/\text{m}^2 = \text{kg}\cdot\text{m}/\text{s}^2. The approximation behind (1) is also comfortably safe here: the rotor’s own rate is 2πn=1572\pi n = 157 rad/s against the vehicle’s Ω=0.056\Omega = 0.056 rad/s, a ratio of about 3×1033\times 10^3, so the vertical angular momentum dropped there is smaller than the spin angular momentum kept by a factor of that order, some four hundredths of a percent. It costs nothing at all, in fact: through a steady turn that dropped piece is a constant vertical vector, so it contributes nothing to dL/dtd\vect L/dt and (2) is exact rather than approximate.

Remark

The result does not depend on which way the curve bends, and that is worth seeing. Reverse the turn and the couple reverses, but the outer rail changes sides at the same time, so the outer rail is the loaded one either way. What does flip the effect is the gearing: the statement fixes the rotor to spin with the wheels, and a rotor geared to spin against them would push the extra kilonewton onto the inner rail instead. Note also that the locomotive’s own mass never enters, so this is a load a light railcar feels exactly as strongly as a heavy one.