Problem 1C-04*: Elliptical motion
The position vector of a particle, r(t) = (x(t), y(t)) is described as
(a) Describe the trajectory of the motion of the particle.
(b) Calculate the velocity v(t) and the angular velocity.
(c) Calculate the acceleration vector a(t) and describe its geometric configuration.
(d) Explain why the condition A, B › 0 does not reduce the generality of the problem.
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(a) Eqs.(1) can be rewritten as
The above two equations can be considered as a set of simultaneous equations for sin(ωt) and cos(ωt), and can be solved to yield
Using the identity sin2(ωt)+cos2(ωt) = 1, we get
If α-β = nπ, where n is an integer, the term with xy vanishes. Eq.(3) then represents an elliptical trajectory if A ≠ B, or a circle if A = B. If α-β = (n+1/2)π, Eq.(3) reduces to
where + sign is for even n's, and - sign is for odd n's. Eq.(4) represents a straight-line trajectory that passes through the origin of the x-y coordinate system.
For general α and β, Eq.(3) represents an elliptical trajectory with its major axis tilted against the x-axis. To find out how much the major axis is tilted, we call the major axis of the ellipse X-axis, and its minor axis as Y-axis. The angle between the X-axis and x-axis is denoted as Φ. The whole orientation of the situation is shown in the figure at right. For an arbitrary point P = (x, y), let us denote its X-Y axis representation as P = (X, Y). Then from the figure, we have
Substituting these x and y into Eq.(3), we get
In the above equation we can group all the terms containing XY together; the coefficient of this term should be zero since in X-Y representation the major axis of the ellipse always coincides with the X-axis. Therefore, we have
(b) Using Eqs.(1), the definition of the velocity vector v(t) = (vx(t), vy(t)) leads us to
If A = B, |v(t)| = ωA and the problem reduces to Problem 1C-03 and we know already that the angular velocity is ω. For the case A ≠ B, the contact line to the ellipse at point P = (x(t), y(t)) has a slope dy/dx, and it can be written asThe right hand side of the above equation is also the slope of the velocity vector v(t), so v(t) lies on the contact line as shown in the picture.
The angular velocity isas derived in part (c) of Problem 1C-03. Then from Eqs.(1) and (5) we have
(c) The acceleration vector a(t) = (ax(t), ay(t)) is obtained from Eqs.(5) and then from Eqs.(1) asThis implies that the acceleration vector a(t) is anti-parallel to the positon vector r(t) as shown in the figure. Note that v(t)·r(t) ≠ 0, so the velocity vector v(t) is not perpendicular to the position vector r(t) in the case of an elliptical trajectory, though it is so in a circular trajectory as we have seen already in Problem 1C-03.
(d) The constraints A › 0 and B › 0 does not limit the generality of the problem. Suppose we have A ‹ 0. We can define A' = -A and x' = -x, then all the earlier discussions in this problem apply toWe can rename A' as A and x' as x. Thus no generality is lost by the above mentioned constraints on the constants A and B.
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