Kinetics – Constrained Motion of a Particle in a Plane (continued)
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Step 4A: Geometric
constraints. Use of the
"chain rule" for taking the derivative when deriving
the geometric constraints. i.e. dy/dt
= (dy/dx)(dx/dt) Suppose y(x) = A sin bx, Here
dy/dx = A b cos bx. then dy/dt = (Abcos bx) dx/dt (1) and d2y/dt2 = ˗
Ab2 sin bx) (dx/dt)2
+ Abcos bx (d2x/dt2)
(2) |
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Step 4B: Kinematic constraints involving speed: Speed
= v(t) = √ [ (dx/dt)2 +
(dy/dt)2 ] (3) Or
v2 = (dx/dt)2 +
(dy/dt)2 (4) |
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Suppose
the kinematic constraint, speed v is prescribed, say v = Vo and y(x)
= A sin bx. Then
from above, dy/dt = (Abcos bx) dx/dt and
by (2) v2 = Vo 2 = (dx/dt)2 +
(Abcos bx)2 (dx/dt)2 and
from (3) Vo 2 = (dx/dt)2[1 + (Abcos bx)2] so
(dx/dt)2 = Vo 2 / [1 + (Abcos
bx)2] and dx/dt = ± Vo
/√ 1 + (Abcos bx)2]
(5) Note: For a particle moving in the + x-direction,
use the + sign. If in the ˗x-direction,
use the ˗ sign. |
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Step 4C: Kinematic constraints involving change of speed: Suppose
the kinematic constraint involving change of speed, dv/dt is prescribed, say
dv/dt = C Note:
C could be positive, negative,
or zero. Use
either (3) or (4) to evaluate dv/dt .
i.e. From
(4) 2v dv/dt = 2
(dx/dt)(d2x/dt2)
+ 2(dy/dt)(d2y/dt2) so v dv/dt =
(dx/dt)(d2x/dt2)
+ (dy/dt)(d2y/dt2) So d2x/dt2 = [
v dv/dt ˗ (dy/dt)(d2y/dt2)
] / (dx/dt) (6) Put
(6) into (2) to find d2y/dt2 . |
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Click
here to continue with discussion. |