302 MISCELLANEOUS METHODS AND APPLICATIONS. [CHAP. XII. 



In time dt a portion mxdt is set in motion with velocity x. We therefore 

 have Tdt=mxdt.x, 



or T=mx 2 . 



The equation of motion of the falling portion is therefore 



mxx = mxg - mx 2 . 

 Writing v for x, this is 



dv 



or 



c 



Integrating, and observing that v and x vanish together, we have 



giving the velocity of the falling portion when its length is x. 

 The time until the length is x is 



dx 



/Ox 



~ V g' 



We note that in this and similar problems energy is dissipated in the im- 

 pulsive action at the place where the discontinuous change of motion occurs. 



II. A chain, one end of which is held fixed, is initially held with the other 

 end close to the fixed end, and the other end is then let go. To find the motion. 



Let 21 be the length of the chain, m the mass per unit length, l+x the 

 length of the part that has come to rest at time t, T the tension 

 at its lower end. 



The free end has fallen through 2# under gravity, so that 



2. = \g&, 

 and the falling portion is free from tension. 



An element, of mass mgtdt, passes from motion with velocity 

 gt to rest in an interval dt, so that the momentum destroyed by 

 the impulse Tdt is mg 2 t 2 dt. 



ft Hence T = mg 2 t 2 . 



Fig. 81. Thus the motion and the tension at any time are determined. 



267. Constrained motion of a chain under gravity. For 



a chain moving in a tube or in a groove on a surface, the condition 

 of inextensibility has the same form as before, i.e. every particle of 

 the chain has the same velocity along the tangent to the chain at 

 the point occupied by the particle. 



Let m be the mass per unit length of the chain, T the 

 tension at any particle distant s from a chosen particle, v 



