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BPO 2024, Теория — Задача 2

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BPO6 Problems 20-24 June 2024. Montenegro · 20 юни 2024 г. · 25 т.

Условие

A track made of a vertical segment and a quarter-circle arc of radius R joining a horizontal floor. The ball at point A on the vertical part is attached to a coiled spring whose other end is fixed on the floor to the left; the floor distance from the spring anchor to point C is 2R. Radii from the arc centre to A (angle α above the horizontal) and to B (angle β below the horizontal) are drawn; C is the end of the arc on the floor.

A quarter of a circle connects the vertical and the horizontal parts of a track. The radius of circular arc is $R = 20\ \mathrm{cm}$ . A ball of unknown mass $m$ slides on the track with no friction. The ball is connected (through a slit along the track) with a stretched spring, as is shown in the figure. The length of the unstretched spring is $L_0 = 0.20\ \mathrm{m}$ , and the spring constant is $k = 100\ \mathrm{N/m}$ .

The sliding ball starts with zero initial speed from a point that is $\alpha = 45^{\circ}$ above the horizontal line when viewed from the centre of the arc, i.e. starts from point A, as is shown in the figure. The ball reaches the maximum speed at the angle $\beta = 34^{\circ}$ below the horizontal line, at point B, as is shown in the figure ( $g = 9.81\ \mathrm{m/s^2}$ ). Derive the mathematical expression and calculate the numerical value for:

a. the length of the spring when the ball is at the starting position, at point A; [3 т.]

b. the elongation of the spring when the ball is at the starting position at point A; [1 т.]

c. the length of the spring when the ball is at the position of maximal speed, at point B; [4 т.]

d. the elongation of the spring when the ball is at the of maximal speed, at point B; [1 т.]

e. the magnitude of the spring force when the ball has maximal speed, at point B; [2 т.]

f. the acceleration component of the ball, tangential to the track at point B; [2 т.]

g. the mass of the ball; [5 т.]

h. the maximum speed of the ball (at point B); [4 т.]

i. the speed of the ball at point C. [3 т.]

Отговори

Решение

The track of Problem 2 redrawn with construction lines: ball at A on the vertical part, spring anchored on the floor at the left, radii at angles α (to A) and β (to B) from the arc centre, vertical distance x_A from the centre level up to A, vertical distance x_B from the floor up to B, horizontal distance y_B from B to C, the tangent at B with angles δ and γ between the spring line, the tangent and the vertical, radius R, floor length 2R and point C.

Calculate the length of the spring when the ball is at the starting position at point A. [3 points]

$L_A^2 = R^2 + (R + x_A)^2$ [1 point]

$tg\alpha = \dfrac{x_A}{R}, \quad \alpha = 45^{0} \quad \Rightarrow \quad tg\alpha = 1 \quad \Rightarrow \quad x_A = R$ [1 point]

$\Rightarrow L_A^2 = R^2 + (2R)^2 = 5R^2$

$\Rightarrow L_A = R\sqrt{5} = 44.7cm \simeq 45cm.$ [0.5 + 0.5 point]

Calculate the elongation of the spring when the ball is at the starting position at point A. [1 points]

$\Delta L_A = L_A - L_0 = 24.7cm \simeq 25cm.$ [0.5 + 0.5 point]

Calculate the length of the spring when the ball is at the position of maximal velocity at point B. [4 points]

$L_B^2 = (2R - y_B)^2 + x_B^2$ [1.5 point]

$x_B = R - R\sin\beta = R(1 - \sin\beta)$ [0.5 point]

$y_B = R\cos\beta$ [0.5 point]

$\beta = 34^{0} \quad \Rightarrow \quad \sin\beta = 0.559, \quad \cos\beta = 0.829$

$\Rightarrow x_B = 8.82cm$

$\Rightarrow y_B = 16.6cm$

\begin{aligned} \Rightarrow L_B^2 &= (2R - R\cos\beta)^2 + R^2(1 - \sin\beta)^2 = \\ &= R^2\left[(2 - \cos\beta)^2 + (1 - \sin\beta)^2\right] = \\ &= R^2(6 - 4\cos\beta - 2\sin\beta) = 1.566R^2 \end{aligned}

[1 point]

$\Rightarrow L_B = R\sqrt{1.566} = 25.02cm \simeq 25cm.$ [0.5 point]

Calculate the elongation of the spring when the ball is at the of maximal velocity at point B. [1 points]

$\Delta L_B = L_B - L_0 = 5.02cm \simeq 5.0cm.$ [0.5 + 0.5 point]

Calculate the magnitude of the spring force when the ball has maximal velocity at point B. [2 points]

$F_k = k\Delta L_B = 5.02N \simeq 5.0N.$ [1.5 + 0.5 point]

Calculate the acceleration of the ball tangential to the track at point B. [2 points]

$v_B = v_{MAX} \quad \Rightarrow \quad a_{Bt} = 0.$

Calculate the mass of the ball. [5 points]

At point B, using $a_{Bt} = 0$ we can write

$F_k \cos\delta = mg\cos\beta$ [2.5 points]

$\Rightarrow m = \dfrac{F_k\cos\delta}{g\cos\beta}$ [1 point]

$\delta = 180^{0} - (\gamma + \beta)$ [0.5 points]

$\cos\gamma = \dfrac{x_B}{L_B} = \dfrac{8.82}{25.02} = 0.352 \quad \Rightarrow \quad \gamma = 69.3^{0}$

$\Rightarrow \delta = 180^{0} - (69.4^{0} + 34^{0}) = 76.7^{0}$ [0.5 points]

$\Rightarrow \cos\delta = 0.231^{0}$

$\Rightarrow m = \frac{5.02 \cdot 0.231}{9.81 \cdot 0.829} kg = 0.142kg \simeq 0.14kg.$ [0.5 points]

Calculate the maximum speed of the ball (at point B). [4 points]

We use work-kinetic energy theorem for positions A and B:

$mgh_A + \dfrac{1}{2}k\Delta L_A^2 = mgh_B + \dfrac{1}{2}k\Delta L_B^2 + \dfrac{1}{2}mv_{MAX}^2$ [2 points]

$\Rightarrow v_{MAX}^2 = 2g(h_A - h_B) + \dfrac{k}{m}(\Delta L_A^2 - \Delta L_B^2)$ [0.5 points]

$h_A = 2R, \quad h_B = x_B = R(1 - \sin\beta)$ [0.5 + 0.5 points]

$\Rightarrow v_{MAX}^2 = \ldots = 47.01\dfrac{m^2}{s^2}$

$\Rightarrow v_{MAX} = 6.88\frac{m}{s} \simeq 6.9\frac{m}{s}$ [0.5 points]

Calculate the speed of the ball at point C. [3 points]

We use again work-kinetic energy theorem for positions A and C:

$mgh_A + \dfrac{1}{2}k\Delta L_A^2 = mgh_C + \dfrac{1}{2}k\Delta L_C^2 + \dfrac{1}{2}mv_C^2$ [0.5 points]

$\Rightarrow v_C^2 = 2g(h_A - h_C) + \dfrac{k}{m}(\Delta L_A^2 - \Delta L_C^2)$ [0.5 points]

$h_A = 2R, \quad h_C = 0$ [0.5 + 0.5 points]

$\Delta L_C = L_C - L_0 = 2R - L_0 = 20cm$ [0.5 points]

$\Rightarrow v_C^2 = \ldots = 22.54\dfrac{m^2}{s^2}$

$\Rightarrow v_C = 4.76\frac{m}{s} \simeq 4.8\frac{m}{s}.$ [0.5 points]

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