From Advanced Mathematics by Richard Brown: "A piston rod PQ, 4 units long, is connected to the rim of a wheel at point P, and to a piston at point Q. As P moves counter-clockwise around the wheel at 1 radian per second, Q slides left and right in the piston. What are the coordinates of P and Q in terms of time t in seconds? Assume that P is at (1,0) at t = 0."
I believe that the coordinates of P (on the rim of the unit circle wheel) will be P = (cos pi t, sin pi t). at t=0, this gives p = (1,0). At t =.5 seconds this gives p = (0,1). At one second, it will have turned pi radians, and cos pi = -1, sin pi = 0, (-1,0) so point P is on the opposite side of the wheel. Finally, at t=1.5 (3/4 of a turn), P should be at the bottom of the unit circle, and (cos 1.5 pi, sin 1.5 pi) is in fact (0,-1)
Please correct me if I am wrong.
As for Q, I know that it oscillates from x=5 (at t=0) to x=3 at t=1 (4+1, 4-1), as the rod itself is length 4. At t =.5 and t=1.5, I assume that x = sq. rt. 15 at both times via Pythagorean theorem, as the rod, the hypotenuse, is always 4. The Y value of Q is always 0.
I do not know how to create an equation for this in terms of t, that is, if my assumptions are correct, which might not be the case. I'm studying on my own; any help is greatly appreciated.
Thank You,
Victor Jaroslaw.
Given that P moves 1 radian/s, its parametrization is
$$x(t)=\cos t, \>\>\> y(t)=\sin t$$
For Q, let its x-axis coordinate be a(t), which s given by
$$a(t)= x(t)+ \sqrt{4^2- y^2(t)} $$
or,
$$a(t)= \cos t + \sqrt{16- \sin^2 t} $$