Complete COSMOS: Online Solutions Manual Organization System
Chapter 11, Solution 182.
From problem 11.97, the position vector is
r = ( Rt cos nt ) i + ctj + ( Rt sin nt ) k.
Differentiating to obtain v and a, dr v= = R ( cos nt nt sin nt ) i + cj + R ( sin nt + nt cos nt ) k dt dv 2 2 = R n sin nt n sin nt n t cos nt i + R n cos nt + n cos nt n t sin nt k a= dt
(
)
(
)
2 2 = R 2 n sin nt n t cos nt i + 2 n cos nt n t sin nt k
(
)(
)
)
ijk v a = vx v y vz = v y a z vz a y i + ( vz a x vx az ) j + v x a y v y a x k ax a y az
(
(
)
2 2 = cR 2 n cos nt n t sin nt i + R 2 ( sin nt + nt cos nt ) 2 n sin nt n t cos nt 2 R 2 ( cos nt nt sin nt ) 2 n cos nt n t sin nt j 2 + cR 2 n sin nt n t cos nt k
2 = cR n ( 2 cos nt nt sin nt ) i R 2 n 2 + n t 2 j + cR n ( 2sin nt + nt cos ) nt k
2 2 2 2 2 | v a | = c 2 R 2 n 4 + n 2 + R 4 n 2 + n t 2
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1/2
The binormal unit vector eb is given by Let be the angle between the y-axis and the binormal.
cos = eb j =
eb =
va | va |
( v a) j =
|va |
c 2 R 2 2 4 + 2t 2 + R 4 2 2 + 2t 2 n n n n
(
R 2 n ( 2 + nt )
)
(
)
12
Let
2 A = R 2 n 2 + n t 2 ,
(
)
A as C shown in the sketch. The angle that the osculating plane makes with the y-axis is the angle .
2 B = cR n 4 + n t 2 12
(
)
,
C=
A2 + B 2 so that cos =
2 R 2 + n t 2 A tan = = B c 4 + 2t 2 1 2 n
(
(
)
)
= tan
1
2 R 2 + n t 2 2 c 4 + n t 2
(
(
)
)
12
!
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.