Circle, point, gravity and time!

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Circle, point, gravity and time!

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Source: South Africa 1997

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Arne

#1 h Oct 8, 2005, 5:29 PM • 1 Y

Y by Adventure10

A circle and a point $P$ higher than the circle lie in the same vertical plane. A particle moves along a straight line under gravity from $P$ to a point $Q$ on the circle. Given that the distance travelled from $P$ in time $t$ is equal to $\dfrac{1}{2}gt^2 \sin{\alpha}$ , where $\alpha$ is the angle of inclination of the line $PQ$ to the horizontal, give a geometrical characterization of the point $Q$ for which the time taken from $P$ to $Q$ is a minimum.

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yetti

#2 h Oct 9, 2005, 10:48 AM • 1 Y

Y by Adventure10

Let O be the center and r the radius of the given circle (O) and let the line PQ intersect the circle (O) at another point R different from Q. Denote $d_1 = PQ, d_2 = PR$ . The time t is given by $t = \sqrt{\frac{2d_1}{g \sin \alpha}}$ and it is minimum when $\sqrt{\frac{d_1}{\sin \alpha}}$ is minimum, i.e., when $\frac{d_1}{\sin \alpha}$ is minimum (because the square root is an increasing function). The power p of the point P to the circle (O) is constant for any angle $\alpha$ and equal to

$p = d_1d_2 = (OP - r)(OP + r) = OP^2 - r^2$

Hence, $\frac{d_1}{\sin \alpha} = \frac{p}{d_2 \sin \alpha}$ and this expression is minimum, when $\frac{1}{d_2 \sin \alpha}$ is minimum or when $d_2 \sin \alpha$ is maximum (because the reciprocal value is a decreasing function). Let the vertical line through P intersect the horizontal line through R at a point S. The distance PS is equal to $PS = d_2 \sin \alpha$ . It goes without saying anything more that this distance is maximum when the point R is identical the most bottom point B of the circle (O), i.e., with the bottom intersection of the vertical line through the center O with the circle (O). The desired point Q is the other intersection of the circle (O) with the line PB.

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