A particle moves on a spiral path. The polar coordinates ( r ,phi) as a function
ID: 2134691 • Letter: A
Question
A particle moves on a spiral path. The polar coordinates (r,phi) as a function of time t are
r = b + b (phi)/(2(pi)) and phi = wt.
(A) Calculate the acceleration vector at the point C. Give the Cartesian components ax and ay.
(B) Draw by hand a reasonably accurate graph of the spiral trajectory, and draw on the graph the acceleration vectory at C (with reasonable accuracy). [Use graph paper! And remember, you are being graded on the clarity of your answers.]
Explanation / Answer
phi = wt
==> tan(phi) = tan(wt) = y/x
==> y = x tan(w t)
r = b + b (phi)/(2 pi)
sqrt(x^2 + y^2) = b + b (wt)/(2 pi)
sqrt(x^2 + (x tan(w t))^2) = b + b (wt)/(2 pi)
x sqrt(1 + (tan(w t))^2) = b + b (wt)/(2 pi)
x/cos(wt) = b + b (wt)/(2 pi)
x = cos(wt) * (b + b (wt)/(2 pi))
==> d^2(x)/dt^2 =
at point C ===> phi = wt =pi
==> d^2x/dt^2 = (-b w^2/(2pi)) * (2*0 + (pi + 2 pi) * (-1)) = 3 b w^2/2
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y = x tan(w t) = (cos(wt) * (b + b (wt)/(2 pi))) * tan(w t)
==> dy^2/dt^2 =
==> at point C ===> phi = wt =pi
==> dy^2/dt^2 = (-b w^2/(2pi)) * ((pi + 2pi)*sin(pi) - 2 cos(pi)) = (-b w^2/(2pi)) * (-2) = b w^2/(pi)
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ax = 3 b w^2/2
ay = b w^2/pi
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