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BPO 2022, Теория — Задача 4. Galileo Galilei – “The minimum descent time”

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Problem 1: Resistor circuit · Problem 2: The descent of a skier · Problem 3: “The bead on the spring” · Problem 4: Galileo Galilei – “The minimum descent time” · 25 т.

Условие

This year the world of Physics commemorates 380 years since Galileo Galilei (1564-1642), one of the founders of modern science, passed away. During the final years of his life, Galileo approached an interesting problem dedicated to the ”minimum descent time”, or, what is today called the ”Galilei’s problem”.

A vertical circle with point A at its top; several chords (ramps) go from A down to the circle at different angles, one of them making angle α_k with the vertical diameter and carrying a small black ball; an arrow g points vertically down to the right of the circle.

A. (5 points) From a point A, with zero initial speed, a big number of identical material points move in a vertical plane, in a uniform gravitational field (the gravitational acceleration, g, is known), sliding without friction along ramps, inclined at different angles to the vertical (see the adjacent figure). If the point A lies at the highest point of a vertical circle of radius R, determine:

A1. on which of the ramps the material points reach back to the circle in the shortest time; [2 т.]

A2. the geometric locus of the points occupied by all the material points at any given time during their motion on the ramps (i.e., the geometrical figure described by the moving material points, at any given moment of time, if a snapshot is taken). [3 т.]

B. (9 points) An inclined plane makes the angle $\alpha = 60^{\circ}$ with the horizontal plane (see the figure below). A point A is situated above the inclined plane, at the height $H = \mathrm{AC} = 29{,}43\ \mathrm{m}$ , measured along the vertical direction, above the inclined plane.

B1. Find the position of a point B on the inclined plane (i.e., the mathematical expression for the length of the segment BC), such that a material point, leaving A with zero initial speed, and sliding without friction along the rectilinear ramp AB, to arrive in B in the shortest time. The gravitational acceleration is $g = 9{,}81\ \mathrm{m/s^2}$ and has the same value in all the points. Calculate the numerical value for the length of the segment BC.

A right-triangle inclined plane falling to the right with angle α at its lower right corner; point A above the plane, a vertical dashed segment H from A down to point C on the plane; a thick straight ramp from A to point B on the plane, higher up the slope than C; an arrow g points vertically down.

B2. Determine the mathematical expression for the minimum descent time on the ramp AB and calculate its numerical value.

C. (11 points) In his last book¹, in which motion is treated mathematically for the first time in history, Galileo analyzes the problem of the fastest descent in uniform gravitational field without friction and concludes that the arc of a circle is faster than any number of its chords. Although it was later proved that the circle is not the path of fastest descent, Galilei admits that any possible error will be removed after future advances in Mathematics. The problem of fastest descent (also called the brachistochrone problem) – formulated in 1696 by Johann Bernoulli and correctly solved by his older brother Jakob Bernoulli – actually shows that the curve is different from a circle and is called cycloid, a name given by Galilei himself.

Points A and B at the same height a distance L apart (double arrow); a symmetric path of three straight ramps: from A down at angle α to a horizontal middle ramp and up at angle α to B (solid lines); a second path drawn dashed with steeper ramps at angle α′ and a lower horizontal ramp; an arrow g points down in the middle.

Since the brachistochrone problem is complicated, it can be simply modeled by three connected rectilinear ramps, the descent time surpassing the cycloidal one with just a few percent. To further simplify the model, a symmetric arrangement will be considered, as illustrated in the figure above. A material point is released in A, with zero initial speed and slides on the ramps without friction from A to B. The middle ramp is horizontal and the transition from one ramp to the next is smooth. The horizontal distance between the fixed points A and B is $L = 5{,}66\ \mathrm{m}$ .

Note: By modifying the angle $\alpha$ , while keeping the points A and B fixed, the lengths of all the ramps will change. For an angle $\alpha' > \alpha$ , the new configuration is represented with dashed lines on the above figure.

¹ Galileo Galilei, Discourses and Mathematical Demonstrations Relating to Two New Sciences, 1638, Leiden, South Holland.

C1. Determine the angle $\alpha$ of the inclined ramps such that the transition time $\tau$ from A to B to be minimum. [8 т.]

C2. Derive the expression for the length of the entire path of the material point, $L_{tot}$ , when the transition time is minimal and calculate its numerical value. [2 т.]

C3. Derive the mathematical expression for the minimum transition time $\tau_{min}$ and calculate its numerical value. [1 т.]

Отговори

Решение

A1.

Solution: For any given ramp, inclined with an angle $\alpha_k$ with respect to the vertical diameter of the circle, the acceleration of the material point is

a_k = gcos\alpha_k, \qquad \textbf{0.5 p}

which is constant.

The length of that ramp is

L_k = 2Rcos\alpha_k. \qquad \textbf{0.5 p}

On the other hand,

L_k = \frac{a_k t_k^2}{2}, \qquad \textbf{0.5 p}

t_k = \sqrt{\frac{2L_k}{a_k}},

t_k = 2\sqrt{\frac{R}{g}}, \qquad \textbf{0.5 p}

which means that all the material points will reach the circle at the same time.

A2.

Solution: After a time $t$ , the material point falling vertically will reach the point $\mathrm{B_0}$ , the length of the segment $\mathrm{AB_0}$ being

\mathrm{AB_0} = \frac{gt^2}{2}, \qquad \textbf{0.5 p}

while the material point sliding along the ramp inclined with an angle $\alpha_k$ with respect to the vertical diameter, will reach the point $\mathrm{B_k}$ , the length $\mathrm{AB_k}$ being

\mathrm{AB}_k = \frac{a_k t^2}{2} = \frac{gt^2}{2}cos\alpha_k. \qquad \textbf{0.5 p}

drawing: 0.5 p

Drawing for A2: a vertical circle with A at the top and B0 at the bottom joined by the vertical diameter; a chord from A at angle α_k to the vertical ends at Bk on the circle and carries a small black ball; a dashed segment joins Bk and B0; an arrow g points down.

Hence

\mathrm{AB}_k = AB_0 cos\alpha_k, \qquad \textbf{0.5 p}

which means that, for any angle $\alpha_k$ ,

m\left(\widehat{AB_kB_0}\right) = 90^{\circ}. \qquad \textbf{0.5 p}

This means that, at any instant $t$ , all the material points are on the vertical circle passing through A (the uppermost point of the circle), having the radius

r = \frac{\mathrm{AB_0}}{2} = \frac{gt^2}{4}. \qquad \textbf{0.5 p}

B1.

Solution 1 (geometrical): 4.5p

We saw above that, for any inclination of the ramp AB, after a time $t$ , the material point is on the circle having the radius

r = \frac{gt^2}{4},

which means that the sliding time on the ramp AB is minimum when the circle radius is minimum. 0.25 p

For the material point to reach the inclined plane, the circle should intersect the plane (the circle can be tangent or secant to the inclined plane). The circle with the minimal value of its radius, satisfying this condition, is the circle tangent to the plane, so B is the tangency point of the circle with the inclined plane (see the figure below). 0.5 p

Drawing for Solution 1 of B1: an inclined plane falling to the right with angle α; a dashed circle with centre O and top point A tangent to the plane at B; the vertical dashed segment H from A to C on the plane and the dashed ramp AB making angle β with the vertical; a right-angle mark at B; an arrow g points down.

Drawing: 0.75 p

The problem is solved if we can derive the expression of the angle $\beta$ which gives the inclination of the ramp AB with respect to the vertical.

In $\Delta\mathrm{OBC}$ , the angle $\widehat{BOC}$ has the measure

m\left(\widehat{BOC}\right) = 2\beta = \alpha, \qquad \textbf{2 x 0.25 p}

\boxed{\beta = \frac{\alpha}{2}}. \qquad \textbf{0.25 p}

The length of the segment BC is

\mathrm{BC} = rtan2\beta = rtan\alpha. \qquad \textbf{0.25 p}

The radius of the circle follows from

cos2\beta = \frac{\mathrm{OB}}{\mathrm{OC}} = \frac{r}{\mathrm{AC} - \mathrm{AO}} = \frac{r}{H - r}, \qquad \textbf{3 x 0.25 p}

hence

r = H\frac{cos\alpha}{1 + cos\alpha}. \qquad \textbf{0.25 p}

So,

\boxed{\mathrm{BC} = H\frac{sin\alpha}{1 + cos\alpha} = Htan\frac{\alpha}{2}}. \qquad \textbf{0.5 p}

The numerical value for the length of this segment is

\boxed{\mathrm{BC} = 9.81\sqrt{3}\ \mathrm{m} = 16.99\ \mathrm{m}}. \qquad \textbf{0.5 p}

Solution 2 (algebraical): 7.5p

Drawing for Solution 2 of B1: an inclined plane falling to the right with angle α; from A a dashed perpendicular AD to the plane (right-angle mark at D), the dashed ramp AB at angle β to the vertical and the vertical dashed segment H from A to C; the angle α is marked at A between AD and the vertical; an arrow g points down.

Drawing: 0.75 p

The length of the ramp AB is

\mathrm{AB} = \frac{at^2}{2} \qquad \textbf{0.5 p}

and the acceleration of the material point is

a = gcos\beta, \qquad \textbf{0.5 p}

t = \sqrt{\frac{2\mathrm{AB}}{gcos\beta}}. \qquad \textbf{0.5 p}

This means that $t$ is minimum when $\frac{\mathrm{AB}}{cos\beta} = min$ . But 0.25 p

\mathrm{AB} = \frac{\mathrm{AD}}{cos(\alpha - \beta)}, \qquad \textbf{0.5 p}

while

\mathrm{AD} = \mathrm{AC}cos\alpha = Hcos\alpha. \qquad \textbf{0.5 p}

Hence

\frac{\mathrm{AB}}{cos\beta} = \frac{Hcos\alpha}{cos\beta cos(\alpha - \beta)}. \qquad \textbf{0.5 p}

Since $Hcos\alpha = const.$ , $t = min$ when the function

f(\beta) = cos\beta cos(\alpha - \beta) = max. \qquad \textbf{0.5 p}

This function can be rewritten as

\begin{aligned} f(\beta) &= cos^2\beta cos\alpha + sin\beta cos\beta sin\alpha = \frac{1}{2}(cos2\beta + 1)cos\alpha + \frac{1}{2}sin2\beta sin\alpha = \\ &= \frac{1}{2}cos\alpha + \frac{1}{2}(cos2\beta cos\alpha + sin2\beta sin\alpha) = \\ &= \frac{1}{2}cos\alpha + \frac{1}{2}cos(2\beta - \alpha). \end{aligned} \qquad \textbf{0.5 p}

This function is maximum when $cos(2\beta - \alpha) = 1$ , which means that 0.25 p

\boxed{\beta = \frac{\alpha}{2}}, \qquad \textbf{0.5 p}

meaning that AB is the bisector of the angle $\widehat{DAC}$ .

Then

\begin{aligned} \mathrm{BC} &= \mathrm{DC} - \mathrm{DB} = Hsin\alpha - \mathrm{AD}tan\frac{\alpha}{2} = Hsin\alpha - Hcos\alpha tan\frac{\alpha}{2} = \\ &= H\left(tan\alpha - tan\frac{\alpha}{2}\right)cos\alpha. \end{aligned} \qquad \textbf{0.5 p}

Since

\mathrm{tan}\alpha = \frac{sin\alpha}{cos\alpha} = \frac{2sin\frac{\alpha}{2}cos\frac{\alpha}{2}}{cos^2\frac{\alpha}{2} - sin^2\frac{\alpha}{2}} = \frac{2tan\frac{\alpha}{2}}{1 - tan^2\frac{\alpha}{2}}, \qquad \textbf{0.25 p}

then

\mathrm{BC} = H\ tan\frac{\alpha}{2}\frac{1 + tan^2\frac{\alpha}{2}}{1 - tan^2\frac{\alpha}{2}}cos\alpha.

Because

\frac{1 + tan^2\frac{\alpha}{2}}{1 - tan^2\frac{\alpha}{2}}cos\alpha = \frac{cos^2\frac{\alpha}{2} + sin^2\frac{\alpha}{2}}{cos^2\frac{\alpha}{2} - sin^2\frac{\alpha}{2}}cos\alpha = \frac{1}{cos\alpha}cos\alpha = 1,

then

\boxed{\mathrm{BC} = H\ tan\frac{\alpha}{2}}. \qquad \textbf{0.5 p}

The numerical value for the length of this segment is

\boxed{\mathrm{BC} = 9.81\sqrt{3}\ \mathrm{m} = 16.99\ \mathrm{m}}. \qquad \textbf{0.5 p}

B2.

(For solution 1): 4.5p

The length of the ramp AB is

\mathrm{AB} = \frac{at^2}{2} \qquad \textbf{0.5 p}

and the acceleration of the material point is

a = gcos\beta, \qquad \textbf{0.5 p}

t = \sqrt{\frac{2\mathrm{AB}}{gcos\beta}}, \qquad \textbf{0.5 p}

where

\mathrm{AB} = \frac{\mathrm{AD}}{cos(\alpha - \beta)}, \qquad \textbf{0.5 p}

while

\mathrm{AD} = \mathrm{AC}cos\alpha = Hcos\alpha. \qquad \textbf{0.5 p}

Hence

\frac{\mathrm{AB}}{cos\beta} = \frac{Hcos\alpha}{cos\beta cos(\alpha - \beta)}, \qquad \textbf{0.5 p}

t = \sqrt{\frac{2Hcos\alpha}{gcos\beta cos(\alpha - \beta)}}. \qquad \textbf{0.5 p}

The minimum descent time is ( $\beta = \alpha/2$ )

t_{min} = \sqrt{\frac{2Hcos\alpha}{gcos^2\frac{\alpha}{2}}},

which gives

\boxed{t_{min} = 2\sqrt{\frac{H}{g}\frac{cos\alpha}{1 + cos\alpha}}} \qquad \textbf{0.5 p}

The numerical value of the minimum descent time is

\boxed{t_{min} = 2.000\ \mathrm{s}}. \qquad \textbf{0.5 p}

(For Solution 2): 1.5p

Since

t = \sqrt{\frac{2Hcos\alpha}{gcos\beta cos(\alpha - \beta)}}. \qquad \textbf{0.5 p}

The minimum descent time is ( $\beta = \alpha/2$ )

t_{min} = \sqrt{\frac{2Hcos\alpha}{gcos^2\frac{\alpha}{2}}},

which gives

\boxed{t_{min} = 2\sqrt{\frac{H}{g}\frac{cos\alpha}{1 + cos\alpha}}} \qquad \textbf{0.5 p}

The numerical value of the minimum descent time is

\boxed{t_{min} = 2.000\ \mathrm{s}}. \qquad \textbf{0.5 p}

C1.

Solution: Due to the symmetry of the path, the total transition time is

\tau = t_1 + t_2 + t_1 = 2t_1 + t_2. \qquad \textbf{0.5 p}

If the length of one of the inclined ramps is denoted by $L_1$ , the descending time on the first inclined ramp is

t_1 = \sqrt{\frac{2L_1}{a}}, \qquad \textbf{0.5 p}

where

a = gsin\alpha. \qquad \textbf{0.5 p}

The transition time on the horizontal ramp is

t_2 = \frac{L - 2L_1cos\alpha}{v}, \qquad \textbf{2 x 0.5 p}

where the speed of the material point on the horizontal path is 2 x 0.5 p

v = \sqrt{2gh} = \sqrt{2gL_1sin\alpha}.

Hence, the total transit time is

\tau = 2\sqrt{\frac{2L_1}{gsin\alpha}} + \frac{L - 2L_1cos\alpha}{\sqrt{2gL_1sin\alpha}}. \qquad \textbf{0.25 p}

For optimal values of $L_1$ and $\alpha$ , the transit time $\tau$ will be minimal. Let’s consider that $\alpha$ already has that optimal value and that the only variable is $L_1$ . Under these circumstances, we can write

\begin{aligned} \tau &= 2\sqrt{\frac{2L_1}{gsin\alpha}} + \frac{L}{\sqrt{2gL_1sin\alpha}} - \frac{2\sqrt{L_1}cos\alpha}{\sqrt{2gsin\alpha}} = \\ &= \frac{2(2 - cos\alpha)}{\sqrt{2gsin\alpha}}\sqrt{L_1} + \frac{1}{\sqrt{L_1}}\frac{L}{\sqrt{2gsin\alpha}}. \end{aligned} \qquad \textbf{0.5 p}

This expression has the form

\tau = ax + \frac{b}{x},

where

x = \sqrt{L_1},\ a = \frac{2(2 - cos\alpha)}{\sqrt{2gsin\alpha}},\ \text{and}\ b = \frac{L}{\sqrt{2gsin\alpha}}.

It can be written as

\tau = ax + \frac{b}{x} - 2\cdot\sqrt{ax}\cdot\sqrt{\frac{b}{x}} + 2\cdot\sqrt{ax}\cdot\sqrt{\frac{b}{x}} = \left(\sqrt{ax} - \sqrt{\frac{b}{x}}\right)^2 + 2\cdot\sqrt{ab}.

From here we can see that $\tau$ is minimal when the binomial is zero, or

\sqrt{ax} = \sqrt{\frac{b}{x}},

Which gives

x = \sqrt{\frac{b}{a}}. \qquad \textbf{1 p}

In our case, the optimal value for $L_1$ is

L_1 = \frac{L}{2(2 - cos\alpha)}. \qquad \textbf{0.5 p}

For this result the transit time will be

\tau = 2\sqrt{\frac{L(2 - cos\alpha)}{gsin\alpha}}. \qquad \textbf{0.25 p}

Now, since $L_1$ is fixed, we can find the optimal value of $\alpha$ which minimizes the value of $\tau$ . This will happen when the ratio

r = \frac{2 - cos\alpha}{sin\alpha} \qquad \textbf{0.25 p}

is minimal. From here, by eliminating the denominator and squaring the equation, it follows that

r^2sin^2\alpha = 4 - 4cos\alpha + cos^2\alpha,

r^2(1 - cos^2\alpha) = 4 - 4cos\alpha + cos^2\alpha.

Rearranging the terms, we get

(1 + r^2)cos^2\alpha - 4cos\alpha + 4 - r^2 = 0. \qquad \textbf{0.5 p}

The solutions of this quadratic equation are

cos\alpha = \frac{2 \pm r\sqrt{r^2 - 3}}{1 + r^2}. \qquad \textbf{0.25 p}

The minimal value for $r$ to obtain real solutions is

r_{min} = \sqrt{3}. \qquad \textbf{0.25 p}

For this value of $r$ , the optimal value of $\alpha$ is given by

cos\alpha = \frac{1}{2}, \qquad \textbf{0.25 p}

which gives the value

\boxed{\alpha = 60^{\circ}}. \qquad \textbf{0.5 p}

C2.

Solution: The optimal value for $L_1$ is

L_1 = \frac{L}{3}. \qquad \textbf{0.5p}

Since the horizontal path has the length

L_2 = L - 2L_1cos\alpha = \frac{2L}{3}, \qquad \textbf{0.25 p}

then

L_{tot} = 2L_1 + L_{2,} \qquad \textbf{0.25 p}

having the expression

\boxed{L_{tot} = \frac{4L}{3}} \qquad \textbf{0.5 p}

and the numerical value

\boxed{L_{tot} = 7.55\ \mathrm{m}}. \qquad \textbf{0.5 p}

C3.

Solution: Since

\tau_{min} = 2\sqrt{\frac{Lr_{min}}{g}}, \qquad \textbf{0.25 p}

then

\boxed{\tau_{min} = 2\sqrt{\frac{L\sqrt{3}}{g}}}. \qquad \textbf{0.25 p}

The numerical value of this time is

\boxed{\tau_{min} = 2.00\ \mathrm{s}}. \qquad \textbf{0.5 p}

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf