Exercício Dinâmica - Engenharia Mecânica (1868)

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648 
PROBLEM 13.95 (Continued) 
 
 2 2 2 2
2 2
1 1 1
(0.12422)(6) (0.12422)[( ) (3) ] (0.04658)( ) 0.5
2 2 2r rv v= + + + 
 2 2
2 22.236 0.06211( ) 0.559 0.02329( ) 0.5r rv v= + + + 
 2
20.0854( ) 1.177rv = 
 2 2
2( ) 13.78 ft /srv = 2( ) 3.71 ft/srv =  
 2( ) 3.00 ft/svθ =  
 4.77 ft/sv = 
(b) Let position 3 be when collar B comes to rest. 
 3 3 3 30.33333, ( ) 0, 0ry r v y= − = = 
 Conservation of angular momentum of collar A. 
 
1 1 3 3
1 1
3
3 3 3
( ) ( )
( ) (0.33333)(6) 2
( )
A Am r v m r v
r v
v
r r r
θ θ
θ
θ
=
= = =
 
 Conservation of energy: 1 1 3 3T V T V+ = + 
 2 2 2 2 2 2
1 1 1 3 3 3 3
1 1 1 1
[( ) ( ) ] [( ) ( ) ]
2 2 2 2A r B A r B Bm v v m y m v v m y w yθ θ+ + = + + +  
 
2
2
3
3
1 1 2
(0.12422)[0 (6) ] 0 (0.12422) 0 0 (1.5)( 0.33333)
2 2
r
r
   + + = + + + − 
   
 
 32
3
0.24844
2.236 1.5 0.5r
r
= + − 
 3 2
3 31.5 2.736 0.24844 0r r− + = 
 Solving the cubic equation for r3, 
 3 1.7712 ft, 0.2805 ft, 0.33333 ftr = − 
 Since 3 1 0.33333 ft, the required root isr r> = 
 3 1.7712 ftr = 
 Corresponding velocity of collar A: 
 3( ) 0rv =  
 3
3
2 2
( )
1.7712
v
rθ = = 3( ) 1.129 ft/svθ = 
   3 1.129 ft/sv = 