77 



Now v^^ 



f * df 1 dr ^ df 1 du 1 ' , f clf^ dr df ^ dn 

 du dt J "^ I dr dt "^ "du dt 



I dr dt 

 in wiiicli t represents the time and v tlie velocity 



fdfgdr df^ dui 

 ■^ I dr dt ^ du dt J 



(6) 



dr 



dt" 



du ) 

 dt i 



dr 



Therefore, 

 dui 2 



+ lsinu-+reosu^^ 



+ 



b du I 



2^ dtj 



This is the differential equation of the motion. 

 Its integral furnishes solutions of the following: 



1. Wliat is the time of descent? 



2. What is the equation of the curve of quickest descent? 



3. What is the space passed over in a given time? 



4. Wliat is tlie velocity at any instant? 



5. What is the normal pressure on the surface? 



Problem: A wheelman rides down a lielix surface along the line of 

 pitch 30°, keeping liis wheel at a constant radial distance of 30 feet. Find 

 tlie time of descent and his velocity upon reaching the ground; the helix 

 making one complete turn. 



Since r is constant and e(iual to r,j, we have: 

 («) r = ro = 30. 



b = 2 TT r tan 300 = 3. 1416 X 60 + i v/"3 = 108.824. 

 g = 32. 

 Equation (7) now becomes, 



I b2 1 rdui = 



^"+4^2] latl 



Substituting from (8) 

 du_ f 32 X 10 8-82 

 dt 



13.1416X1199-982 

 2 /— l2T _200 

 ~ 96 

 From equation (4). 



•••^=.i6^-I 



j -l/u = .96l/u. 



1/2 X 3.1416 = 5.2 seconds 



time of descent. 



^,^^/8 2X 108-824. ^- 



3.1416 



i/64 X 108.824 = 83.4 ft. per second = velocity 



at bottom. 



■Partials. 



