A Problem in Dynamics


 * AN inextensible heavy chain
 * Lies on a smooth horizontal plane,
 * An impulsive force is applied at A,
 * Required the initial motion of K.




 * Let ds be the infinitesimal link,
 * Of which for the present we've only to think;
 * Let T be the tension and T + dT
 * The same for the end that is nearest to B.
 * Let a be put, by a common convention
 * For the angle at M 'twixt OX and the tension;
 * Let V$t$ and V$n$ be ds's velocities,
 * Of which V$t$ along and V$n$ across it is;
 * Then $$\frac{V_t}{V_n}$$ the tangent will equal,
 * Of the angle of starting worked out in the sequel.


 * In working the problem the first thing of course is
 * To equate the impressed and effectual forces.
 * K is tugged by two tensions, whose difference dT
 * [1] Must equal the element's mass into V$t$.
 * V$n$ must be due to the force perpendicular
 * To ds's direction, which shows the particular
 * Advantage of using da to serve at your
 * Pleasure to estimate ds's curvature
 * For V$n$ into mass of a unit of chain
 * [2] Must equal the curvature into the strain.


 * Thus managing cause and effect to discriminate,
 * The student must fruitlessly try to eliminate,
 * And painfully learn, that in order to do it, he
 * Must find the Equation of Continuity.
 * The reason is this, that the tough little element,
 * Which the force of impulsion to beat to a jelly meant,
 * Was endowed with a property incomprehensible,
 * And was "given", the the language of Shop, "inextensible."
 * It therefore with such pertinacity odd defied
 * The force which the length of the chain would have modified,
 * That its stubborn example may possibly yet recall
 * These overgrown rhymes to their prosody metrical.
 * The condition is got by resolving again,
 * According to axes assumed in the plane.
 * If then you reduce to the tangent and normal,
 * [3] You will find the equation more neat tho' less formal.
 * [4] The condition thus found after these preparations,
 * When duly combined with the former equations,
 * Will give you another, in which differential
 * [5] (When the chain forms a circle), become in essentials
 * No harder than those that we easily solve
 * [6] In the time a T totum would take to revolve.


 * Now joyfully leaving ds to itself, a—
 * Ttend to the values of T and of a.
 * The chain undergoes a distorting convulsion,
 * Produced first at A by the force of impulsion.
 * In magnitude R, in direction tangential,
 * [7] Equating this R to the form exponential,
 * Obtained for the tension when a is zero,
 * It will measure the tug, such a tug as the "hero
 * Plume-waving" experienced, tied to the chariot.
 * But when dragged by the heels his grim head could not carry aught,
 * [8] So give a its due at the end of the chain,
 * And the tension ought there to be zero again.
 * From these two conditions we get three equations,
 * Which serve to determine the proper relations
 * Between the impulse and each coefficient
 * In the form for the tension, and this is sufficent
 * To work out the problem, and then, if you choose,
 * You may turn it and twist it the Dons to amuse.


 * $$\frac{V_t}{V_n} = \tan \beta = - \frac {e^{(a_1 - a)} - e^{-(a_1 - a)}} {e^{(a_1 - a)} + e^{-(a_1 - a)}}$$


 * 1: $$dT = m V_t ds$$


 * 2: $$m V_n = T \frac{da}{ds}$$


 * 3: $$\frac{dV_t}{ds} = V_n \frac{da}{ds}$$


 * 4: $$\frac{d^2 T}{ds^2} = V_n \frac{da}{ds} = 0$$


 * 5: $$\frac{d^2 T}{da^2} = 0$$


 * 6: $$T = C_1 e^a + C_2 e^{-a}$$


 * 7: $$R = C_1 + C_2$$


 * 8: $$0 = C_1 e^{a_1} + C_2 e^{-a_1}$$