Exercício de Dinamica Aplicada (99)

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260 
 
PROBLEM 11.180* 
For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis. 
PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = 
(Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. 
(The space curve described by the particle is a conic helix.) 
 
SOLUTION 
First note that the vectors v and a lie in the osculating plane. 
Now ( cos ) ( sin )n nRt t ct Rt tω ω= + +r i j k 
Then (cos sin ) (sin cos )n n n n n n
dr
R t t t c R t t t
dt
ω ω ω ω ω ω= = − + + +v i j k 
and 
( )
( )
2
2
sin sin cos
cos cos sin
[ (2sin cos ) (2cos sin ) ]
n n n n n n
n n n n n n
n n n n n n n
dv
dt
R t t t t
R t t t t
R t t t t t t
ω ω ω ω ω ω
ω ω ω ω ω ω
ω ω ω ω ω ω ω
=
= − − −
+ + −
= − + + −
a
i
k
i k
 
It then follows that the vector ( )×v a is perpendicular to the osculating plane. 
 
 ( ) (cos sin ) (sin cos )
(2sin cos ) 0 (2cos sin )
n n n n n n n
n n n n n n
R R t t t c R t t t
t t t t t t
ω ω ω ω ω ω ω
ω ω ω ω ω ω
× = − +
− + −
i j k
v a 
 
( )2 2
{ (2cos sin ) [ (sin cos )(2sin cos )
(cos sin )(2cos sin )] (2sin cos )
(2cos sin ) 2 (2sin cos )
n n n n n n n n n n
n n n n n n n n n
n n n n n n n n
R c t t t R t t t t t t
t t t t t t c t t t
R c t t t R t c t t t
ω ω ω ω ω ω ω ω ω ω
ω ω ω ω ω ω ω ω ω
ω ω ω ω ω ω ω ω
= − + − + +
− − − + +
 = − − + + + 
i
j k
i j k