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APhO 2016, Теория — Задача 1. Mechanics of a Deformable Lattice

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Asian Physics Olympiad 2016 — Theoretical Question – T1 — Mechanics of a Deformable Lattice (Total Marks : 20) · 20 т.

Условие

Here we study a deformable lattice hanging in gravity which acts as a deformable physical pendulum. It has only one degree of freedom, i.e. only one way to deform it and the configuration is fully described by an angle $\alpha$ . Such structures have been studied by famous physicist James Maxwell in 19th century, and some surprising behaviors have been discovered recently.

As shown in the figure 1, $N^2$ identical triangular plates (red triangle) are freely hinged by identical rods and form an $N \times N$ lattice ( $N > 1$ ). The joints at the vertices are denoted by small circles. The sides of the equilateral triangles and the rods have the same length $l$ . The dashed lines in the figure represent four tubes; each tube confines $N$ vertices (grey circles) on the edge and the $N$ vertices can slide in the tube, i.e. the tube is like a sliding rail.

The four tubes are connected in a diamond shape with two angles fixed at 60° and another two angles at 120° as shown in Figure 1. Each plate has a uniform density with mass $M$ , and the other parts of the system are massless. The configuration of the lattice is uniquely determined by the angle $\alpha$ , where $0^{\circ} \le \alpha \le 60^{\circ}$ (please see the examples of different angle $\alpha$ in Figure 1). The system is hung vertically like a “curtain” with the top tube fixed along the horizontal direction.

The coordinate system is shown in Figure 2. The zero level of the potential energy is defined at $y = 0$ . A triangular plate is denoted by a pair of indices ( $m$ , $n$ ), where $m, n = 0, 1, 2, \cdots, N-1$ representing the order in the $x$ and $y$ directions respectively. A( $m$ , $n$ ), B( $m$ , $n$ ) and C( $m$ , $n$ ) denote the positions of the 3 vertices of the triangle ( $m$ , $n$ ). The top-left vertex, A(0, 0) (the big black circle), is fixed.

The motion of the whole system is confined in the $x$ - $y$ plane. The moment of inertia of a uniform equilateral triangular plate about its center of mass is $I = Ml^2/12$ . The free fall acceleration is $g$ . Please use $E_\mathrm{k}$ and $E_\mathrm{p}$ to denote kinetic energy and potential energy respectively.

Four drawings of the hinged lattice of red equilateral triangles inside a diamond of dashed tubes, for α = 0°, α = 18°, α = 48° and α = 60°; the fixed top-left vertex is a big black circle, the vertices sliding in the tubes are grey circles, the other joints small white circles, and a small blue arrow marks the angle α at the first triangle. — Figure 1

Left: the 4 × 4 lattice with triangles labelled (0,0) to (3,3), the x axis to the right along the top tube and the y axis upwards from the fixed black vertex, with a downward arrow labelled g. Right: one red triangle labelled (m, n) with vertices A(m, n) at the top-left, B(m, n) on the right and C(m, n) at the bottom, the angle α between side AB and the horizontal line through A, and side AC of length l. — Figure 2

Section A: When $N$ =2 (as shown in figure 3):

The 2 × 2 lattice of four red triangles inside the dashed tubes for α = 0°, α = 18°, α = 48° and α = 60°, with the fixed black vertex at the top-left and a blue arrow marking α. — Figure 3

A1 What is the potential energy $E_\mathrm{p}$ of the system for a general angle $\alpha$ when $N = 2$ ? [2 т.]

A2 What is the equilibrium angle $\alpha_\mathrm{E}$ of the system under gravity when $N = 2$ ? [1 т.]

A3 The system follows a simple harmonic oscillation under a small perturbation from equilibrium. Calculate the kinetic energy of this system in terms of $\Delta\dot{\alpha} \equiv \mathrm{d}(\Delta\alpha)/\mathrm{d}t$ . Calculate the oscillation frequency $f_\mathrm{E}$ when $N = 2$ . [5 т.]

Section B: For arbitrary $N$ :

B1 What is the equilibrium angle $\alpha'_\mathrm{E}$ under gravity when $N$ is arbitrary? [3 т.]

B2 Consider the case when $N \to \infty$ . Under a small perturbation of angle $\alpha$ , the change of potential energy of the system is $\Delta E_\mathrm{p} \propto N^{\gamma_1}$ , the kinetic energy of the system is $E_\mathrm{k} \propto N^{\gamma_2}$ , and the oscillation frequency is $f'_\mathrm{E} \propto N^{\gamma_3}$ . Find the values of $\gamma_1$ , $\gamma_2$ and $\gamma_3$ . [3 т.]

Section C: A force is exerted on one of the $3N^2$ triangle vertices so that the system maintains at $\alpha_\mathrm{m} = 60^{\circ}$ .

Left: the 4 × 4 lattice at α = 60°, where the red triangles fill a parallelogram, with light blue x and y axes through the fixed black vertex and a downward arrow labelled g. Right: a vector F making the angle θ_F with the x axis. — Figure 4

C1 Which vertex should we choose to minimize the magnitude of this force? [1 т.]

C2 What are the direction and magnitude of this minimum force? Describe the direction in terms of the angle $\theta_F$ defined in Figure 4. [5 т.]

Отговори

Решение

(Full Mark = 20)

A1 — Marks: 2

The potential energy for $N = 2$ is:

E_\mathrm{p}(\alpha) = Mg \cdot y_{\mathrm{c\cdot m.}(0,0)} \times 4 + Mg \cdot \Delta y \times 2 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (1)}

where

y_{\mathrm{c\cdot m.}(0,0)} = -\frac{\sqrt{3}l}{3}\sin\left(\frac{\pi}{6} + \alpha\right) \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (2)}

is the $y$ coordinate of center of mass of triangle (0,0), and

\begin{aligned} \Delta y &= y_{\mathrm{A}(0,1)} - y_{\mathrm{A}(0,0)} \\ &= -l\left[\sin\left(\frac{\pi}{3} + \alpha\right) + \sin\left(\frac{\pi}{3} - \alpha\right)\right] \\ &= -\sqrt{3}l\cos\alpha \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (3)} \end{aligned}

is the translational difference of two neighbouring triangles in $y$ -direction. Solving Eqs. (1), (2) and (3), we obtain

E_\mathrm{p}(\alpha) = -\frac{2}{3}Mgl\left(4\sqrt{3}\cos\alpha + 3\sin\alpha\right) \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (4)}

Sketch for A2: a blue curve of E_p against α with a minimum at α_E and rising up to 60°, with the label “restoring torque=−dE_p/dα”; dashed vertical lines mark α_E and 60° on the α axis.

A2 — Marks: 1

At equilibrium, the potential energy reaches a minimum, which gives:

\left.\frac{dE_\mathrm{p}(\alpha)}{d\alpha}\right|_{\alpha=\alpha_\mathrm{E}} = 0 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (5)}

\sqrt{3}\sin\alpha_\mathrm{E} + 3\cos\alpha_\mathrm{E} = 0 \qquad \text{- Eq. (6)}

\alpha_\mathrm{E} = \tan^{-1}\frac{\sqrt{3}}{4} \qquad \textbf{(0.5 point)} \qquad \text{- Eq. (7)}

A3 — Marks: 5

If the total energy of the oscillation has the following form

E(\Delta\alpha, \Delta\dot{\alpha}) = E_\mathrm{p} + E_\mathrm{k} = \frac{1}{2}K(\Delta\alpha)^2 + \frac{1}{2}I(\Delta\dot{\alpha})^2 , \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (8)}

where $E_\mathrm{p}$ and $E_\mathrm{k}$ are the potential and kinetic energies of the system respectively, then the motion is a simple harmonic oscillation with angular frequency $\omega = \sqrt{K/I}$ . Here $\Delta\alpha = \alpha - \alpha_\mathrm{E}$ . Under a small perturbation, the potential energy change is:

\begin{aligned} \Delta E_\mathrm{p} &\approx \frac{1}{2}\left.\frac{d^2E_\mathrm{p}}{\mathrm{d}\alpha^2}\right|_{\alpha=\alpha_\mathrm{E}}(\Delta\alpha)^2 \\ &= \left(\frac{1}{2}\right)\left(\frac{2}{3}Mgl\right)(4\sqrt{3}\cos\alpha_\mathrm{E} + 3\sin\alpha_\mathrm{E})(\Delta\alpha)^2 \\ &= \frac{\sqrt{57}}{3}Mgl(\Delta\alpha)^2 \qquad \textbf{(1 point)} \qquad \text{- Eq. (9)} \end{aligned}

The total kinetic energy of the system includes the translational kinetic energy of every plate and the rotational kinetic energy of every plate relative to its center of mass

E_\mathrm{k} = \sum E_\mathrm{k}^{\mathrm{trans}} + \sum E_\mathrm{k}^{\mathrm{rot}} \qquad \text{- Eq. (10)}

The rotational kinetic energy is

\sum E_\mathrm{k}^{\mathrm{rot}} = 4 \times \frac{1}{2}\frac{Ml^2}{12}(\Delta\dot{\alpha})^2 = \frac{1}{6}Ml^2(\Delta\dot{\alpha})^2 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (11)}

$E_\mathrm{k}^{\mathrm{trans}}$ can be obtained by considering the motion of the center of mass of each triangle and setting $N = 2$ .

x_{\mathrm{c.m.}(m,n)} = m(2l\cos\alpha) + n(2l\cos\alpha)\cos\frac{\pi}{3} + \frac{l}{\sqrt{3}}\cos\left(\alpha + \frac{\pi}{6}\right),

y_{\mathrm{c.m.}(m,n)} = -n(2l\cos\alpha)\sin\frac{\pi}{3} - \frac{l}{\sqrt{3}}\sin\left(\alpha + \frac{\pi}{6}\right). \qquad \textbf{(0.5 point)}

Differentiating and substituting

\sin\alpha = \frac{\sqrt{3}}{\sqrt{19}}, \cos\alpha = \frac{4}{\sqrt{19}}, \sin\left(\alpha + \frac{\pi}{6}\right) = \frac{7}{2\sqrt{19}}, \cos\left(\alpha + \frac{\pi}{6}\right) = \frac{3\sqrt{3}}{2\sqrt{19}},

\dot{x}_{\mathrm{c.m.}(m,n)} = -\left(2m + n + \frac{7}{6}\right)\frac{3}{\sqrt{57}}l\Delta\dot{\alpha}, \qquad \dot{y}_{\mathrm{c.m.}(m,n)} = \frac{3(2n-1)}{2\sqrt{19}}l\Delta\dot{\alpha}.

v^2_{\mathrm{c.m.}(m,n)} = \dot{x}^2_{\mathrm{c.m.}(m,n)} + \dot{y}^2_{\mathrm{c.m.}(m,n)} = \frac{(12m+6n+7)^2+27}{228}l^2(\Delta\dot{\alpha})^2, \qquad \textbf{(1 point)}

E^{\mathrm{trans}}_{\mathrm{c.m.,k}} = \frac{M}{2}\left[v^2_{\mathrm{c.m.}(0,0)} + v^2_{\mathrm{c.m.}(0,1)} + v^2_{\mathrm{c.m.}(1,0)} + v^2_{\mathrm{c.m.}(1,1)}\right] = \frac{164}{57}Ml^2(\Delta\dot{\alpha})^2.

E^{\mathrm{trans}}_\mathrm{k} = E^{\mathrm{trans}}_{\mathrm{c.m.,k}} + E^{\mathrm{rot}}_\mathrm{k} = \frac{347}{114}Ml^2(\Delta\dot{\alpha})^2. \qquad \textbf{(1 point)}

Alternatively, another way to get $E_k^{\mathrm{trans}}$ is based on the center of mass of the whole system:

E_\mathrm{k} = \sum E^{\mathrm{trans}}_{\mathrm{c.m.,k}} + \sum E^{\mathrm{rot}}_{\mathrm{r.c.,k}} \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (12)}

where

E^{\mathrm{trans}}_{\mathrm{r.c.,k}} = \frac{M}{2}\left[v^2_{\mathrm{r.c.}(0,0)} + v^2_{\mathrm{r.c.}(1,0)} + v^2_{\mathrm{r.c.}(0,1)} + v^2_{\mathrm{r.c.}(1,1)}\right] \qquad \text{- Eq. (13)}

is the translational kinetic energy relative to the center of mass of the system and

E^{\mathrm{trans}}_{\mathrm{c.m.,k}} = \frac{4M}{2}v^2_{\mathrm{c.m.}} \qquad \text{- Eq. (14)}

is the translational kinetic energy of the center of mass of the system.

The center of mass of each of the 2×2 = 4 triangles always form diamond shape with lateral length $2l$ cos α. The center of mass of the whole system is at the center of the diamond shape. Hence

v_{\mathrm{r.c.}(0,0)} = v_{\mathrm{r.c.}(1,1)} = \left.\frac{d(\sqrt{3}l\cos\alpha)}{d\alpha}\right|_{\alpha=\alpha_\mathrm{E}}\Delta\dot{\alpha}

v_{\mathrm{r.c.}(1,0)} = v_{\mathrm{r.c.}(0,1)} = \left.\frac{d(l\cos\alpha)}{d\alpha}\right|_{\alpha=\alpha_\mathrm{E}}\Delta\dot{\alpha} \qquad \text{- Eq. (15)}

Substituting Eqs. (14) and (15) into Eq. (13), we obtain

E^{\mathrm{trans}}_{\mathrm{r.c.,k}} = 4\sin\alpha_\mathrm{E}^2 Ml^2(\Delta\dot{\alpha})^2 \qquad \text{- Eq. (16)}

For $E^{\mathrm{trans}}_{\mathrm{c.m.,k}}$ ,

v_{\mathrm{c.m.}} = \left.\sqrt{\left(\frac{\mathrm{d}x_{\mathrm{c.m.}}}{\mathrm{d}\alpha}\right)^2 + \left(\frac{\mathrm{d}y_{\mathrm{c.m.}}}{\mathrm{d}\alpha}\right)^2}\,\right|_{\alpha=\alpha_\mathrm{E}}\Delta\dot{\alpha} \qquad \text{- Eq. (17)}

is the velocity of the center-of-mass of the four triangular plates, with

\begin{aligned} x_{\mathrm{c.m.}} &= x_{\mathrm{c.m.}(0,0)} + \frac{1}{2}\left(x_{\mathrm{B}(0,0)} + x_{\mathrm{A}(1,0)}\right) \\ &= \frac{\sqrt{3}l}{3}\cos\left(\frac{\pi}{6} + \alpha\right) + \frac{3}{2}l\cos\alpha \qquad \text{- Eq. (18)} \\ y_{\mathrm{c.m.}} &= y_{\mathrm{c.m.}(0,0)} + \frac{1}{2}\Delta y \\ &= -\frac{\sqrt{3}l}{3}\sin\left(\frac{\pi}{6} + \alpha\right) - \frac{\sqrt{3}}{2}l\cos\alpha \qquad \text{- Eq. (19)} \end{aligned}

Substituting Eqs. (17), (18) and (19) and into Eq. (14), we obtain

E^{\mathrm{trans}}_{\mathrm{c.m.,k}} = \left(\frac{2}{3} + 10\sin^2\alpha_E\right)Ml^2(\Delta\dot{\alpha})^2 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (20)}

Combining Eqs. (12), (16) and (20), we obtain

\begin{aligned} E_\mathrm{k} &= E^{\mathrm{rot}}_\mathrm{k} + E^{\mathrm{trans}}_{\mathrm{r.c.,k}} + E^{\mathrm{trans}}_{\mathrm{c.m.,k}} \\ &= \left(\frac{5}{6} + 14\sin^2\alpha_E\right)Ml^2(\Delta\dot{\alpha})^2 \\ &= \frac{347}{114}Ml^2(\Delta\dot{\alpha})^2 \qquad \textbf{(1.5 points)} \qquad \text{- Eq. (21)} \end{aligned}

According to Eqs. (8), (9) and (21),

f = \frac{1}{2\pi}\sqrt{\frac{\frac{\sqrt{57}}{3}Mgl}{\frac{347}{114}Ml^2}} = \frac{1}{2\pi}\sqrt{\frac{38\sqrt{57}}{347}\frac{g}{l}} \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (22)}

[Note 1: 0.5 point should be deducted if there are numerical mistakes, but all steps are correct.

Note 2: A rough estimate of $f \sim \sqrt{\frac{g}{l}}$ can get 0.5 points out of 5 points.]

B1 — Marks: 3

For arbitrary $N$ , the total potential energy

E_\mathrm{p} = \sum_{m,n=0}^{N-1}E_\mathrm{p}(m,n) \qquad \text{- Eq. (23)}

where

E_\mathrm{p}(m,n) = \frac{1}{3}Mg\left[y_{\mathrm{A}(m,n)} + y_{\mathrm{B}(m,n)} + y_{\mathrm{C}(m,n)}\right] \qquad \text{- Eq. (24)}

(0.5 points for Eqs. (23) and (24))

and

\begin{aligned} y_{\mathrm{A}(m,n)} &= -nl\sin\left(\frac{\pi}{3} - \alpha\right) - nl\sin\left(\frac{\pi}{3} + \alpha\right) = -\sqrt{3}nl\cos\alpha \\ y_{\mathrm{B}(m,n)} &= y_{\mathrm{A}(m,n)} - l\sin\alpha = -\sqrt{3}nl\cos\alpha - l\sin\alpha \\ y_{\mathrm{C}(m,n)} &= y_{\mathrm{A}(m,n)} - l\sin\left(\frac{\pi}{3} + \alpha\right) = -\sqrt{3}nl\cos\alpha - l\sin\left(\frac{\pi}{3} + \alpha\right) \qquad \text{- Eq. (25)} \end{aligned}

(0.5 points for all three correct coordinates)

Thus,

E_\mathrm{p}(m,n) = -\frac{1}{3}Mgl\left[3\sqrt{3}n\cos\alpha + \sin\alpha + \sin\left(\frac{\pi}{3} + \alpha\right)\right] \qquad \text{- Eq. (26)}

and

\begin{aligned} E_\mathrm{p} &= \sum_{m,n=0}^{N-1}E_\mathrm{p}(m,n) \\ &= -\frac{1}{3}Mgl\sum_{m,n=0}^{N-1}\left[3\sqrt{3}n\cos\alpha + \sin\alpha + \sin\left(\frac{\pi}{3} + \alpha\right)\right] \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (27)} \end{aligned}

Using the mathematical relations

\sum_{m=0}^{N-1}1 = \sum_{n=0}^{N-1}1 = N

and

\sum_{m=0}^{N-1}m = \sum_{n=0}^{N-1}n = \frac{N(N-1)}{2} \qquad \text{- Eq. (28),}

Eq. (27) becomes

\begin{aligned} E_\mathrm{p} &= -\frac{1}{3}N^2Mgl\left[\frac{3\sqrt{3}(N-1)\cos\alpha}{2} + \sin\alpha + \sin\left(\frac{\pi}{3} + \alpha\right)\right] \\ \text{or} \quad &= -\frac{1}{3}N^2Mgl\left[\frac{\sqrt{3}(3N-2)\cos\alpha}{2} + \frac{3}{2}\sin\alpha\right] \qquad \textbf{(1 points)} \qquad \text{- Eq. (29)} \end{aligned}

At equilibrium, $\frac{\mathrm{d}E_\mathrm{p}}{\mathrm{d}\alpha} = 0$ , therefore

-\frac{3\sqrt{3}(N-1)\sin\alpha'_\mathrm{E}}{2} + \cos\alpha'_\mathrm{E} + \cos\left(\frac{\pi}{3} + \alpha'_\mathrm{E}\right) = 0 \qquad \text{- Eq. (30)}

\alpha'_\mathrm{E} = \tan^{-1}\left(\frac{\sqrt{3}}{3N-2}\right) \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (31)}

[Remark: Increasing α lowers each triangle relative to its vertex A, but globally raises the system, i.e. the bottom tube is raised higher. When $N \to \infty$ , the global displacement dominates, consequently $\alpha \to 0$ .]

B2 — Marks: 3

Under a small perturbation, the potential energy change, according to Eq. (29) is

\Delta E_\mathrm{p} \approx \frac{1}{2}\left.\frac{d^2E_\mathrm{p}}{\mathrm{d}\alpha^2}\right|_{\alpha=\alpha'_\mathrm{E}}(\Delta\alpha)^2 \sim N^3 \quad \text{or} \ \gamma_1 = 3 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (32)}

[Remark: There are $N^2$ triangles and the $y$ coordinate of the total center of mass is proportional to $N$ , hence $E_\mathrm{p} \sim N^3$ and $\gamma_1 = 3$ . Using this argument to derive the correct $\gamma_1$ can also get 0.5 points.]

The kinetic energy of a triangle includes the translational energy of its center of mass and the rotational energy about its center of mass. Hence the total kinetic energy of the $N^2$ triangles is

E_\mathrm{k} = \sum_{m,n}E_{\mathrm{c.m.}(m,n)} + \sum_{m,n}E_{\mathrm{r.c.}(m,n)} \qquad \text{- Eq. (33)}

where

E_{\mathrm{r.c.}(m,n)} = \frac{1}{2}\frac{Ml^2}{12}(\Delta\dot{\alpha})^2 = \frac{1}{24}Ml^2(\Delta\dot{\alpha})^2 \sim 1 \qquad \text{- Eq. (34)}

and

\begin{aligned} E_{\mathrm{c.m.}(m,n)} &= \frac{M}{2}v^2_{\mathrm{c.m.}(m,n)} \\ &= \frac{M(\Delta\dot{\alpha})^2}{2}\left[\left(\frac{\mathrm{d}x_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha}\right)^2 + \left(\frac{\mathrm{d}y_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha}\right)^2\right]_{\alpha=\alpha'_\mathrm{E}} \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (35)} \end{aligned}

Since

\begin{aligned} x_{\mathrm{c.m.}(m,n)} &= x_{\mathrm{A}(m,n)} + \frac{\sqrt{3}l}{3}\cos\left(\frac{\pi}{6} + \alpha\right) \\ &= (2m + n)l\cos\alpha + \frac{l}{2}\cos\alpha - \frac{\sqrt{3}l}{6}\sin\alpha \end{aligned}

and

\begin{aligned} y_{\mathrm{c.m.}(m,n)} &= y_{\mathrm{A}(m,n)} + \frac{\sqrt{3}l}{3}\sin\left(\frac{\pi}{6} + \alpha\right) \\ &= \sqrt{3}nl\cos\alpha + \frac{\sqrt{3}l}{6}\cos\alpha + \frac{l}{2}\sin\alpha \qquad \text{- Eq. (36)} \end{aligned}

(0.5 points for correct $x$ and $y$ )

\frac{\mathrm{d}x_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha} = \left[-(2m + n)\sin\alpha - \frac{1}{2}\sin\alpha - \frac{\sqrt{3}}{6}\cos\alpha\right]l

\frac{\mathrm{d}y_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha} = \left[-\sqrt{3}n\sin\alpha - \frac{\sqrt{3}}{6}\sin\alpha + \frac{1}{2}\cos\alpha\right]l

we have

E_{\mathrm{c.m.}(m,n)} = \frac{1}{2}Ml^2(\Delta\dot{\alpha})^2\begin{bmatrix} (4m^2 + 4n^2 + 4mn + 2m + 2n)\sin^2\alpha'_\mathrm{E} \\ + \frac{2\sqrt{3}}{3}(m - n)\sin\alpha'_\mathrm{E}\cos\alpha'_E + \frac{1}{3} \end{bmatrix} \qquad \text{- Eq. (37)}

Since $\alpha'_\mathrm{E} \sim \frac{1}{N}$ in Eq. (31), we have

E_{\mathrm{c.m.}(m,n)} = A \cdot N^2 \cdot \frac{1}{N^2} + B \cdot N \cdot \frac{1}{N} + C \sim 1 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (38)}

According to Eqs. (33), (34) and (38), we have

E_\mathrm{k} = \sum_{m,n}E_{\mathrm{c.m.}(m,n)} + \sum_{m,n}E_{\mathrm{r.c.}(m,n)} \sim N \times N \times 1 \sim N^2

\text{or} \ \gamma_2 = 2 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (39)}

[Remarks: $E_\mathrm{k} \sim N^2$ because there are $N^2$ triangles, each contribute $E_{\mathrm{r.c.}}(m,n) \sim 1$ (relative-to-center-of-mass kinetic energy) and $E_{\mathrm{c.m.}}(m,n) \sim 1$ (center-of-mass kinetic energy).]

Note that $E_{\mathrm{r.c.}}(m,n) \sim 1$ is true for arbitrary α while $E_{\mathrm{c.m.}}(m,n) \sim 1$ is only true for the special case of $\alpha'_\mathrm{E} \to 0$ or $N \to \infty$ .

Therefore

f'_\mathrm{E} \sim \sqrt{\frac{E_\mathrm{p}}{E_\mathrm{k}}} \sim \sqrt{N}

\text{or} \ \gamma_3 = 0.5 \qquad \textbf{(0.5 points)} \qquad \text{- Eq. (40)}

C1 — Marks: 1

The minimum force should act on the farthest triangle ( $N - 1$ , $N - 1$ ), whose motion can be decomposed into the motion of the center of mass and the rotation around the center of mass: $\vec{v} = \vec{v}_{\mathrm{c.m.}} + \vec{v}_{\mathrm{rot}}$ . As shown in the figure, $\vec{v}_{\mathrm{rot}}$ of vertex C makes the smallest angle relative to the direction of $\vec{v}_{\mathrm{c.m.}}$ near $\alpha_\mathrm{m} \equiv \pi/3$ . Hence its displacement is the largest and its corresponding force is minimum, i.e. the minimum force should act on vertex C( $N - 1$ , $N - 1$ ). (1 point)

Sketch for C1: a yellow equilateral triangle with dashed lines from its vertices to the center; arrows show V_cm from the center, and the rotational velocities V_rot_A at the top vertex (horizontal), V_rot_B at the bottom-right vertex and V_rot_C at the bottom-left vertex.

[Remarks: A rigorous calculation is given in Appendix 3.]

C2 — Marks: 5

At $\alpha = \alpha_\mathrm{m} \equiv \pi/3$ , a small change in $\alpha$ will change the potential energy by:

\begin{aligned} \Delta E_\mathrm{p}(\alpha_\mathrm{m}) &= \left.\frac{dE_\mathrm{p}}{d\alpha}\right|_{\alpha=\alpha_\mathrm{m}}\Delta\alpha \\ &= \frac{1}{3}N^2Mgl\left[\left(\frac{3\sqrt{3}N}{2} - \sqrt{3}\right)\sin\alpha_\mathrm{m} - \frac{3}{2}\cos\alpha_\mathrm{m}\right]\Delta\alpha \\ &= \frac{3}{4}(N - 1)N^2Mgl\Delta\alpha \qquad \textbf{(1 point)} \qquad \text{- Eq. (41)} \end{aligned}

The displacement of $C$ ( $m$ , $n$ ) point is

\begin{aligned} \Delta x_{\mathrm{C}(m,n)} &= -\left[(2m + n)\sin\alpha_\mathrm{m} - \sin\left(\frac{\pi}{3} + \alpha_\mathrm{m}\right)\right]l\Delta\alpha \\ &= \frac{(2m+n+1)\sqrt{3}}{2}l\Delta\alpha \qquad \textbf{(0.5 points)} \\ \Delta y_{\mathrm{C}(m,n)} &= -\left[\sqrt{3}n\sin\alpha_\mathrm{m} - \cos\left(\frac{\pi}{3} + \alpha_\mathrm{m}\right)\right]l\Delta\alpha \\ &= \frac{(3n+1)}{2}l\Delta\alpha \qquad \textbf{(0.5 points)} \end{aligned}

For C( $N$ -1, $N$ -1), $\Delta r = \sqrt{(\Delta x)^2 + (\Delta y)^2} = (3N - 2)(l\Delta\alpha)$ . (1 point)

Hence

F_{\min} = \frac{\Delta E_\mathrm{p}(\alpha_\mathrm{m})}{\Delta r_{\max}} = \frac{3(N-1)N^2}{4(3N-2)}Mg \qquad \textbf{(1 point)} \qquad \text{- Eq. (42)}

and

\begin{aligned} \theta_{F_{\min}} &= \tan^{-1}\left[\frac{\Delta y_{\mathrm{C}(N-1,N-1)}}{\Delta x_{\mathrm{C}(N-1,N-1)}}\right] + \pi \\ &= -\tan^{-1}\frac{\sqrt{3}}{3} + \pi = \frac{5\pi}{6} \qquad \textbf{(1 point)} \qquad \text{- Eq. (43)} \end{aligned}

[Remarks: This $\theta_{F_{\min}}$ is not perpendicular to the $C$ ( $N$ -1, $N$ -1)–A(0,0) direction because of the constraints of the tunes, e.g. $A$ (1,0), $A$ (2,0), $A$ (3,0), ⋯ , are also the holding points.]

Appendix 1: — Marks: N/A

(a) Calculation of the exact $E_\mathrm{p}$ , $E_\mathrm{k}$ and $f'_\mathrm{E}$ in Parts (C), (D) and € for arbitrary $N$

Under a small perturbation, the potential energy change is

\begin{aligned} \Delta E_\mathrm{p} &\approx \frac{1}{2}\left.\frac{d^2E_\mathrm{p}}{d\alpha^2}\right|_{\alpha=\alpha'_E}(\Delta\alpha)^2 \\ &= \frac{1}{3}N^2Mgl\left(\frac{3\sqrt{3}N - 2\sqrt{3}}{2}\cos\alpha'_\mathrm{E} + \frac{3}{2}\sin\alpha'_\mathrm{E}\right)\frac{(\Delta\alpha)^2}{2} \\ &= \frac{\sqrt{3(3N-2)^2+9}}{12}N^2Mgl(\Delta\alpha)^2 \qquad \text{- Eq. (44)} \end{aligned}

The kinetic energy of a triangle includes the translational energy of its center of mass and the rotational energy around its center of mass. Hence the total kinetic energy of the $N^2$ triangles is

E_\mathrm{k} = \sum_{m,n}E_{\mathrm{c.m.}(m,n)} + \sum_{m,n}E_{\mathrm{r.c.}(m,n)} \qquad \text{- Eq. (45)}

where

E_{\mathrm{r.c.}(m,n)} = \frac{1}{2}\frac{Ml^2}{12}(\Delta\dot{\alpha})^2 = \frac{1}{24}Ml^2(\Delta\dot{\alpha})^2 \qquad \text{- Eq. (46)}

and

\begin{aligned} E_{\mathrm{c.m.}(m,n)} &= \frac{M}{2}v^2_{\mathrm{c.m.}(m,n)} \\ &= \frac{M(\Delta\dot{\alpha})^2}{2}\left[\left(\frac{\mathrm{d}x_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha}\right)^2 + \left(\frac{\mathrm{d}y_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha}\right)^2\right]_{\alpha=\alpha'_\mathrm{E}} \qquad \text{- Eq. (47)} \end{aligned}

Since

\begin{aligned} x_{\mathrm{c.m.}(m,n)} &= x_{\mathrm{A}(m,n)} + \frac{\sqrt{3}l}{3}\cos\left(\frac{\pi}{6} + \alpha\right) \\ &= (2m + n)l\cos\alpha + \frac{l}{2}\cos\alpha - \frac{\sqrt{3}l}{6}\sin\alpha \end{aligned}

and

\begin{aligned} y_{\mathrm{c.m.}(m,n)} &= y_{\mathrm{A}(m,n)} - \frac{\sqrt{3}l}{3}\sin\left(\frac{\pi}{6} + \alpha\right) \\ &= -\sqrt{3}nl\cos\alpha - \frac{\sqrt{3}l}{6}\cos\alpha - \frac{l}{2}\sin\alpha \qquad \text{- Eq. (48)} \end{aligned}

Hence,

\frac{\mathrm{d}x_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha} = \left[-(2m + n)\sin\alpha - \frac{1}{2}\sin\alpha - \frac{\sqrt{3}}{6}\cos\alpha\right]l

\frac{\mathrm{d}y_{\mathrm{c.m.}(m,n)}}{\mathrm{d}\alpha} = \left[-\sqrt{3}n\sin\alpha + \frac{\sqrt{3}}{6}\sin\alpha - \frac{1}{2}\cos\alpha\right]l

We have

E_{\mathrm{c.m.}(m,n)} = \frac{1}{2}Ml^2(\Delta\dot{\alpha})^2\begin{bmatrix} (4m^2 + 4n^2 + 4mn + 2m + 2n)\sin^2\alpha'_\mathrm{E} \\ + \frac{2\sqrt{3}}{3}(m - n)\sin\alpha'_\mathrm{E}\cos\alpha'_\mathrm{E} + \frac{1}{3} \end{bmatrix} \qquad \text{- Eq. (49)}

and

\begin{aligned} E_k &= \sum_{m,n}E_{\mathrm{c.m.}(m,n)} + \sum_{m,n}E_{\mathrm{r.c.}(m,n)} \\ &= \left[\frac{1}{6}(11N - 1)(N - 1)\sin^2\alpha'_\mathrm{E} + \frac{5}{24}\right]N^2Ml^2(\Delta\dot{\alpha})^2 \\ &= \left[\frac{(11N-1)(N-1)}{2(3N-2)^2+6} + \frac{5}{24}\right]N^2Ml^2(\Delta\dot{\alpha})^2 \qquad \text{- Eq. (50)} \end{aligned}

With Eqs. (44) and (50), we have

\begin{aligned} f'_\mathrm{E} &= \frac{1}{2\pi}\sqrt{\frac{\frac{\sqrt{3(3N-2)^2+9}}{12}N^2Mgl}{\left[\frac{(11N-1)(N-1)}{2(3N-2)^2+6} + \frac{5}{24}\right]N^2Ml^2}} \\ &= \frac{1}{2\pi}\sqrt{\frac{2\sqrt{3(3N-2)^2+9}}{\left[\frac{12(11N-1)(N-1)}{(3N-2)^2+3} + 5\right]}\frac{g}{l}} \qquad \text{- Eq. (51)} \end{aligned}

(b) Center of mass movement of the whole system

According to Eq. (48), we have

\begin{aligned} x_{\mathrm{c.m.(sys.)}}(\alpha) &= \frac{\sum_{m,n}x_{\mathrm{c.m.}(m,n)}}{N^2} \\ &= \frac{\sum_{m,n}\left[(2m + n)l\cos\alpha + \frac{l}{2}\cos\alpha - \frac{\sqrt{3}l}{6}\sin\alpha\right]}{N^2} \\ &= \left(\frac{3N-2}{2}\right)l\cos\alpha - \frac{\sqrt{3}l}{6}\sin\alpha \end{aligned}

and

\begin{aligned} y_{\mathrm{c.m.}(m,n)}(\alpha) &= \frac{\sum_{m,n}y_{\mathrm{c.m.}(m,n)}}{N^2} \\ &= -\frac{\sum_{m,n}\left[\sqrt{3}nl\cos\alpha + \frac{\sqrt{3}l}{6}\cos\alpha + \frac{l}{2}\sin\alpha\right]}{N^2} \\ &= -\left(\frac{3N-2}{6}\right)\sqrt{3}l\cos\alpha - \frac{l\sin\alpha}{2} \qquad \text{- Eq. (52)} \end{aligned}

Eq. (52) is the trajectory of the center of mass for the whole system, which is not a straight line.

Appendix 2: Calculation of the moment of inertia of a triangular plate — Marks: N/A

Appendix 2: a red equilateral triangle of side l divided into four small equilateral triangles; the centres c_1 (central triangle), c_2 (top), c_2′ (right) and c_2″ (left) are marked, and the vertical distance d between c_2 and c_1 is indicated.

An equilateral triangle with lateral length $l$ can be divided into four small equilateral triangles with lateral length $l$ /2. For the central small triangle centered at $c_1$ , its moment of inertia is

I_1 = \beta\frac{M}{4}\left(\frac{l}{2}\right)^2 \qquad \text{- Eq. (53)}

For the non-central small triangle centered at $c_2$ , $c_2'$ and $c_2''$ ,

I_2 = I_1 + \frac{M}{4}d^2 \qquad \text{- Eq. (54)}

where $d = \sqrt{3}l/6$ is the distance between the centers of triangles 1 and 2. The second term is from the parallel-axis theorem. The moment of inertia of the whole triangle is the sum of the moment of inertia of the four sub-triangles:

\beta Ml^2 = 4 \times \beta\frac{M}{4}\left(\frac{l}{2}\right)^2 + 3 \times \frac{M}{4}d^2 \qquad \text{- Eq.(55)}

Thus

\beta = \frac{1}{12} \qquad \text{- Eq. (56)}

Appendix 3: The minimum force corresponds to the maximum displacement of the exerting point of this force. — Marks: N/A

Consider the position of vertices A, B, C of a triangle ( $m$ , $n$ ) :

\begin{aligned} x_{\mathrm{A}(m,n)} &= (2m + n)\cos\alpha_\mathrm{m}\,l \\ y_{\mathrm{A}(m,n)} &= -\sqrt{3}n\cos\alpha_\mathrm{m}\,l \\ x_{\mathrm{B}(m,n)} &= (2m + n + 1)\cos\alpha_\mathrm{m}\,l \\ y_{\mathrm{B}(m,n)} &= -\left(\sqrt{3}n\cos\alpha_\mathrm{m} + \sin\alpha_\mathrm{m}\right)l \\ x_{\mathrm{C}(m,n)} &= \left[(2m + n)\cos\alpha_\mathrm{m} + \cos\left(\frac{\pi}{3} + \alpha_\mathrm{m}\right)\right]l \\ y_{\mathrm{C}(m,n)} &= -\left[\sqrt{3}n\cos\alpha_\mathrm{m} + \sin\left(\frac{\pi}{3} + \alpha_\mathrm{m}\right)\right]l \qquad \text{- Eq. (57)} \end{aligned}

Taking derivatives on α on the above coordinates we get

\begin{aligned} \Delta x_{\mathrm{A}(m,n)} &= -(2m + n)\sin\alpha_\mathrm{m}\,l\Delta\alpha = -\frac{(2m + n)\sqrt{3}}{2}l\Delta\alpha \\ \Delta y_{\mathrm{A}(m,n)} &= \sqrt{3}n\sin\alpha_\mathrm{m}\,(l\Delta\alpha) = \frac{3n}{2}l\Delta\alpha \\ \Delta x_{\mathrm{B}(m,n)} &= -(2m + n + 1)\sin\alpha_\mathrm{m}\,l\Delta\alpha = -\frac{(2m + n + 1)\sqrt{3}}{2}l\Delta\alpha \\ \Delta y_{\mathrm{B}(m,n)} &= -\left(-\sqrt{3}n\sin\alpha_\mathrm{m} + \cos\alpha_\mathrm{m}\right)l\Delta\alpha = \frac{3n - 1}{2}l\Delta\alpha \\ \Delta x_{\mathrm{C}(m,n)} &= \left[-(2m + n)\sin\alpha_\mathrm{m} - \sin\left(\frac{\pi}{3} + \alpha_\mathrm{m}\right)\right]l\Delta = -\frac{(2m + n + 1)\sqrt{3}}{2}l\Delta\alpha \\ \Delta y_{\mathrm{C}(m,n)} &= -\left[-\sqrt{3}n\sin\alpha_\mathrm{m} + \cos\left(\frac{\pi}{3} + \alpha_\mathrm{m}\right)\right]l\Delta\alpha = \frac{(3n+1)}{2}l\Delta\alpha \qquad \text{- Eq. (58)} \end{aligned}

For $\Delta r = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ , we have

\begin{aligned} \Delta r_{\mathrm{A}(m,n)} &= \sqrt{3m^2 + 3n^2 + 3mn}(l\Delta\alpha) \\ \Delta r_{\mathrm{B}(m,n)} &= \sqrt{3m^2 + 3n^2 + 3mn + 3m + 1}(l\Delta\alpha) \\ \Delta r_{\mathrm{C}(m,n)} &= \sqrt{3m^2 + 3n^2 + 3mn + 3m + 3n + 1}(l\Delta\alpha) \qquad \text{- Eq. (59)} \end{aligned}

Thus we find

\Delta r_{\mathrm{C}(m,n)} > \Delta r_{\mathrm{B}(m,n)} > \Delta r_{\mathrm{A}(m,n)} \qquad \text{- Eq. (60)}

Therefore, we should choose point C of the triangle ( $N - 1$ , $N - 1$ ) to obtain

\Delta r_{\max} = (3N - 2)l\Delta\alpha \qquad \text{- Eq. (61)}

so that the force is minimal.

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf