Exercício Dinâmica - Engenharia Mecânica (1850)

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630 
PROBLEM 13.82* (Continued) 
 
 Thus, 2 2 1/2
2 2 1/2 00
2 2 1/2 2 1/2
( 2 )
( 2 )
( 2 ) (2 ) ( 3 2)
a a
BD
BD
OABD O A A B B D
y
U dy y a
y a
U a a a a
U U U U− − −
= = +
+
= + − = −
= + +
 
 ( 2 1) ( 3 2)a a a= + − + − 3OABDU a=  
 2 2 2 1/2
( , , ) (0,0,0)
( ) 0
ODV V a a a V
a a a
Δ = −
= − + + − 3ODV aΔ = −  
 Thus, OABD ODU V= −Δ