77 



.^_f*dfidr df^dui^ fdf^dr df,du}2 fdfadr dfadui^ 

 jNow A — ] ^^^ dt "^ du dt I "^ i dr dt "^ du dt i "*"l dr dt "^ du dt J 



in which t represents the time and v the velocity. Therefore, 



f dr . dui 2 r dr , dui ^ , f b dul ^ 



(6) v^ = icosu^— rsmu^j + I smu^^ + rcosu^j + [^ -^,1 



IdtJ ^l ^47t2J idtj 



From (5) and (6). 



(^) B'^^ 





TT '^ IdtJ ^l ^4~2j [dtj 



This is the differential equation of the motion. 

 Its integral furnishes solutions of the following : 



1. What is the time of descent? 



2. What is the equation of the curs-e of quickest descent? 



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



4. What is the velocity at any instant? 



5. What is the normal pressure on the surface? 



Problem: A wheelman rides down a helix surface along tlie line of 

 pitcli 30°, keeping his wheel at a constant radial distance of 80 feet. Find 

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

 making one complete turn. 



Since r is constant and equal to r^,, we have: 

 (8) r = ro = 30. 



b = 2 - r tan 300 = 3, 1416 X 60 + i ,/'"3 = 108.824. 

 g = 32. 

 Equation (7) now becomes, 



Substituting from (8) 



du _ r 82 X 10 8.82 i I ,— ,— 



dF" '13. 1416X1199.982 J ^^i— -Sevu. 



2 /— l27r 200 / 



■ ' ~ "cjfi ^^ "96" ^ "^^ 3. 1416 = 5.2 seconds = time of descent. 



From equation (4). 



/ H2 W 1 08 824 



y = \— o i.;r. ' 1^ u ^ ^64 X 108.824 = 83.4 ft. per second = velocity 



at bottom. 



■■'Partials. 



