Thus in the case of a railway truck travelling with velocity u the kinetic energy is 1⁄2 (M + mκ2/α2)u2, where M is the total mass, α the radius and κ the radius of gyration of each wheel, and m is the sum of the masses of the wheels; the reduced inertia is therefore M + mκ2/α2. Again, take the system composed of the flywheel, connecting rod, and piston of a steam-engine. We have here a limiting case of three-bar motion (§ 3), and the instantaneous centre J of the connecting-rod PQ will have the position shown in the figure. The velocities of P and Q will be in the ratio of JP to JQ, or OR to OQ; the velocity of the piston is therefore yθ̇, where y = OR. Hence if, for simplicity, we neglect the inertia of the connecting-rod, the kinetic energy will be 1⁄2 (I + My2)thetȧ2, where I is the moment of inertia of the flywheel, and M is the mass of the piston. The effect of the mass of the piston is therefore to increase the apparent moment of inertia of the flywheel by the variable amount My2. If, on the other hand, we take OP (= x) as our variable, the kinetic energy is 1⁄2 (M + I/y2)ẋ2. We may also say, therefore, that the effect of the flywheel is to increase the apparent mass of the piston by the amount I/y2; this becomes infinite at the “dead-points” where the crank is in line with the connecting-rod.

If the system be “conservative,” we have

1⁄2Aq2 + V = const.,

(15)

where V is the potential energy. If we differentiate this with respect to t, and divide out by q̇, we obtain

Aq̈ + 1⁄2 dAq̇2 + dV= 0
dq dq

(16)

as the equation of motion of the system with the unknown reactions (if any) eliminated. For equilibrium this must be satisfied by q̇ = O; this requires that dV/dq = 0, i.e. the potential energy must be “stationary.” To examine the effect of a small disturbance from equilibrium we put V = ƒ(q), and write q = q0 + η, where q0 is a root of ƒ′ (q0) = 0 and η is small. Neglecting terms of the second order in η we have dV/dq = ƒ′(q) = ƒ″(q0)·η, and the equation (16) reduces to

Aη̈ + ƒ″ (q0)η = 0,

(17)