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Условие Part A (4,3 т.)Part B (4,3 т.)Part C (3,4 т.)Отговори Решение

IPhO 2003, Теория — Задача 1. A Swing with a Falling Weight

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Theoretical Question 1

Внимание: Бележка към темата

В отпечатаната схема за оценяване на (j) е дадено $\alpha_{\mathrm c}=1+2M/(3m)$ , а в уравнение (C18) на пълното решение е дадено $\alpha_{\mathrm c}=(1+2M/(3m))^{-1}$ . Двете различаващи се формули са запазени както в оригинала.

Условие

A rigid cylindrical rod of radius $R$ is held horizontal above the ground. With a string of negligible mass and length $L$ ( $L > 2\pi R$ ), a pendulum bob of mass $m$ is suspended from point $A$ at the top of the rod as shown in Figure 1a. The bob is raised until it is level with $A$ and then released from rest when the string is taut. Neglect any stretching of the string. Assume the pendulum bob may be treated as a mass point and swings only in a plane perpendicular to the axis of the rod. Accordingly, the pendulum bob is also referred to as the particle. The acceleration of gravity is $\vec{g}$ .

A circle of radius R (the rod cross-section) with centre O and a dashed vertical x-axis through O and the top point A. A horizontal string of length L runs from A to the bob m on the left. A dashed string segment s runs from the tangent point Q on the circle down-left to the particle P; the angle θ is marked from the x-axis to OQ, and unit vectors r̂ and t̂ are drawn at Q. A downward arrow labelled g⃗ is on the left.

Figure 1a

Let $O$ be the origin of the coordinate system. When the particle is at point $P$ , the string is tangential to the cylindrical surface at $Q$ . The length of the line segment $QP$ is called $s$ . The unit tangent vector and the unit radial vector at $Q$ are given by $\hat{t}$ and $\hat{r}$ , respectively. The angular displacement $\theta$ of the radius $OQ$ , as measured counterclockwise from the vertical $x$ -axis along $OA$ , is taken to be positive.

When $\theta = 0$ , the length $s$ is equal to $L$ and the gravitational potential energy $U$ of the particle is zero. As the particle moves, the instantaneous time rates of change of $\theta$ and $s$ are given by $\dot{\theta}$ and $\dot{s}$ , respectively.

Unless otherwise stated, all the speeds and velocities are relative to the fixed point $O$ .

Part A (4,3 т.)

In Part A, the string is taut as the particle moves. In terms of the quantities introduced above (i.e., $s$ , $\theta, \dot{s}$ , $\dot{\theta}$ , $R$ , $L$ , $g$ , $\hat{t}$ and $\hat{r}$ ), find:

(a) The relation between $\dot{\theta}$ and $\dot{s}$ . [0,5 т.]

(b) The velocity $\vec{v}_Q$ of the moving point $Q$ relative to $O$ . [0,5 т.]

(c) The particle’s velocity $\vec{v}^{\prime}$ relative to the moving point $Q$ when it is at $P$ [0,7 т.]

(d) The particle’s velocity $\vec{v}$ relative to $O$ when it is at $P$ . [0,7 т.]

(e) The $\hat{t}$ -component of the particle’s acceleration relative to $O$ when it is at $P$ . [0,7 т.]

(f) The particle’s gravitational potential energy $U$ when it is at $P$ . [0,5 т.]

(g) The speed $v_m$ of the particle at the lowest point of its trajectory. [0,7 т.]

Part B (4,3 т.)

In Part B, the ratio $L$ to $R$ has the following value:

\frac{L}{R} = \frac{9\pi}{8} + \frac{2}{3}\cot\frac{\pi}{16} = 3.534 + 3.352 = 6.886

(h) What is the speed $v_s$ of the particle when the string segment from $Q$ to $P$ is both straight and shortest in length? (in terms of $g$ and $R$ ) [2,4 т.]

(i) What is the speed $v_H$ of the particle at its highest point $H$ when it has swung to the other side of the rod? (in terms of $g$ and $R$ ) [1,9 т.]

Part C (3,4 т.)

In Part C, instead of being suspended from $A$ , the pendulum bob of mass $m$ is connected by a string over the top of the rod to a heavier weight of mass $M$ , as shown in Figure 1b. The weight can also be treated as a particle.

Initially, the bob is held stationary at the same level as $A$ so that, with the weight hanging below $O$ , the string is taut with a horizontal section of length $L$ . The bob is then released from rest and the weight starts falling. Assume that the bob remains in a vertical plane and can swing past the falling weight without any interruption.

The kinetic friction between the string and the rod surface is negligible. But the static friction is assumed to be large enough so that the weight will remain stationary once it has come to a stop (i.e. zero velocity).

The bob m level with the top point A of a circle of radius R and centre O; a dashed double arrow marks the horizontal length L from m to the vertical x-axis through A. The angle θ is marked at O. The string passes over the top of the rod and down its right side to a hatched weight M hanging below the level of O.

Figure 1b

(j) Assume that the weight indeed comes to a stop after falling a distance $D$ and that $\left(L - D\right) \gg R$ . If the particle can then swing around the rod to $\theta = 2\pi$ while both segments of the string free from the rod remain straight, the ratio $\alpha = D/L$ must not be smaller than a critical value $\alpha_c$ . Neglecting terms of the order $R/L$ or higher, obtain an estimate on $\alpha_c$ in terms of $M/m$ . [3,4 т.]

Отговори

Решение

Part A

(a) Since the length of the string $L = s + R\theta$ is constant, its rate of change must be zero. Hence we have

\dot{s} + R\dot{\theta} = 0 \qquad (A1)

(b) Relative to $O$ , $Q$ moves on a circle of radius $R$ with angular velocity $\dot{\theta}$ , so

\vec{v}_Q = R\dot{\theta}\,\hat{t} = -\dot{s}\,\hat{t} \qquad (A2)

(c) Refer to Fig. A1. Relative to $Q$ , the displacement of $P$ in a time interval $\Delta t$ is $\Delta\vec{r}^{\prime} = (s\Delta\theta)(-\hat{r}) + (\Delta s)\hat{t} = [(s\dot{\theta})(-\hat{r}) + \dot{s}\,\hat{t}]\Delta t$ . It follows

\vec{v}^{\prime} = -s\dot{\theta}\,\hat{r} + \dot{s}\,\hat{t} \qquad (A3)

Small-displacement diagram at Q: two lines from Q to P and to the displaced particle with the angle Δθ between them; dashed arcs labelled s, s+Δs; the displacement Δr⃗′ from P; the arc length sΔθ; unit vectors t̂ and −r̂ at the bottom.

Figure A1

(d) The velocity of the particle relative to $O$ is the sum of the two relative velocities given in Eqs. (A2) and (A3) so that

\vec{v} = \vec{v}^{\prime} + \vec{v}_Q = (-s\dot{\theta}\,\hat{r} + \dot{s}\,\hat{t}) + R\dot{\theta}\,\hat{t} = -s\dot{\theta}\,\hat{r} \qquad (A4)

(e) Refer to Fig. A2. The ( $-\hat{t}$ )-component of the velocity change $\Delta\vec{v}$ is given by $(-\hat{t})\cdot\Delta\vec{v} = v\Delta\theta = v\dot{\theta}\,\Delta t$ . Therefore, the $\hat{t}$ -component of the acceleration $\vec{a} = \Delta\vec{v}/\Delta t$ is given by $\hat{t}\cdot\hat{a} = -v\dot{\theta}$ . Since the speed $v$ of the particle is $s\dot{\theta}$ according to Eq. (A4), we see that the $\hat{t}$ -component of the particle’s acceleration at $P$ is given by

\vec{a}\cdot\hat{t} = -v\dot{\theta} = -(s\dot{\theta})\dot{\theta} = -s\dot{\theta}^{2} \qquad (A5)

Note that, from Fig. A2, the radial component of the acceleration may also be obtained as $\vec{a}\cdot\hat{r} = -dv/dt = -d(s\dot{\theta})/dt$ .

Velocity-change diagram: string from P to Q and a rotated string by Δθ; at Q the directions −r̂ and −t̂; the velocity v⃗ and v⃗+Δv⃗ with the change Δv⃗, the angle Δθ and the arc vΔθ, a line from Q towards O.

Figure A2

(f) Refer to Fig. A3. The gravitational potential energy of the particle is given by $U = -mgh$ . It may be expressed in terms of $s$ and $\theta$ as

U(\theta) = -mg[R(1 - \cos\theta) + s\sin\theta] \qquad (A6)

Circle with centre O, top point A on the dashed x-axis, tangent point Q at angle θ, radius R; straight string s from Q down to P. Dashed levels mark the height h of A above P, split into R − R cosθ, R cosθ and s sinθ.

Figure A3

(g) At the lowest point of its trajectory, the particle’s gravitational potential energy $U$ must assume its minimum value $U_m$ . By differentiating Eq. (A6) with respect to $\theta$ and using Eq. (A1), the angle $\theta_m$ corresponding to the minimum gravitational energy can be obtained.

\begin{aligned} \frac{dU}{d\theta} &= -mg\left(R\sin\theta + \frac{ds}{d\theta}\sin\theta + s\cos\theta\right) \\ &= -mg\left[R\sin\theta + (-R)\sin\theta + s\cos\theta\right] \\ &= -mgs\cos\theta \end{aligned}

At $\theta = \theta_m$ , $\left.\dfrac{dU}{d\theta}\right|_{\theta_m} = 0$ . We have $\theta_m = \dfrac{\pi}{2}$ . The lowest point of the particle’s trajectory is shown in Fig. A4 where the length of the string segment of QP is $s = L - \pi R/2$ .

Circle with centre O and top point A on the dashed x-axis; Q at angle π/2 on the left end of the horizontal radius R; the string s hangs vertically from Q down to P.

Figure A4

From Fig. A4 or Eq. (A6), the minimum potential energy is then

U_m = U\left(\pi/2\right) = -mg[R + L - (\pi R/2)] \qquad (A7)

Initially, the total mechanical energy $E$ is 0. Since $E$ is conserved, the speed $v_m$ of the particle at the lowest point of its trajectory must satisfy

E = 0 = \frac{1}{2}mv_m^{2} + U_m \qquad (A8)

From Eqs. (A7) and (A8), we obtain

v_m = \sqrt{-2U_m/m} = \sqrt{2g[R + (L - \pi R/2)]} \qquad (A9)

Part B

(h) From Eq. (A6), the total mechanical energy of the particle may be written as

E = 0 = \frac{1}{2}mv^{2} + U(\theta) = \frac{1}{2}mv^{2} - mg[R(1 - \cos\theta) + s\sin\theta] \qquad (B1)

From Eq. (A4), the speed $v$ is equal to $s\dot{\theta}$ . Therefore, Eq. (B1) implies

v^{2} = (s\dot{\theta})^{2} = 2g[R(1 - \cos\theta) + s\sin\theta] \qquad (B2)

Let $T$ be the tension in the string. Then, as Fig. B1 shows, the $\hat{t}$ -component of the net force on the particle is $-T + mg\sin\theta$ . From Eq. (A5), the tangential acceleration of the particle is $(-s\dot{\theta}^{2})$ . Thus, by Newton’s second law, we have

m(-s\dot{\theta}^{2}) = -T + mg\sin\theta \qquad (B3)

Circle with centre O, top point A on the dashed x-axis, tangent point Q at angle θ, radius R; the string s from Q to P with the tension T along it; at P the weight mg vertically down and its component mg sinθ along the string, the angle θ between them.

Figure B1

According to the last two equations, the tension may be expressed as

\begin{aligned} T &= m(s\dot{\theta}^{2} + g\sin\theta) = \frac{mg}{s}[2R(1 - \cos\theta) + 3s\sin\theta] \\ &= \frac{2mgR}{s}\left[\tan\frac{\theta}{2} - \frac{3}{2}\left(\theta - \frac{L}{R}\right)\right](\sin\theta) \\ &= \frac{2mgR}{s}(y_1 - y_2)(\sin\theta) \end{aligned} \qquad (B4)

The functions $y_1 = \tan(\theta/2)$ and $y_2 = 3(\theta - L/R)/2$ are plotted in Fig B2.

Graph of y against θ from 0 to 3π (ticks π/2, π, 3π/2, 2π, 5π/2, 3π; y from −30 to 30): the curves y1 = tan(θ/2) with asymptotes at π and 3π, and the straight line y2 = (3/2)(θ − L/R); their crossing θs just after π is marked with an arrow, and the region between the θ-axis and the crossing level is hatched.

Figure B2

From Eq. (B4) and Fig. B2, we obtain the result shown in Table B1. The angle at which . $y_2 = y_1$ is called $\theta_s$ ( $\pi < \theta_s < 2\pi$ ) and is given by

\frac{3}{2}\left(\theta_s - \frac{L}{R}\right) = \tan\frac{\theta_s}{2} \qquad (B5)

or, equivalently, by

\frac{L}{R} = \theta_s - \frac{2}{3}\tan\frac{\theta_s}{2} \qquad (B6)

Since the ratio $L/R$ is known to be given by

\frac{L}{R} = \frac{9\pi}{8} + \frac{2}{3}\cot\frac{\pi}{16} = \left(\pi + \frac{\pi}{8}\right) - \frac{2}{3}\tan\frac{1}{2}\left(\pi + \frac{\pi}{8}\right) \qquad (B7)

one can readily see from the last two equations that $\theta_s = 9\pi/8$ .

Table B1

	$(y_1 - y_2)$	$\sin\theta$	tension $T$
$0 < \theta < \pi$	positive	positive	positive
$\theta = \pi$	$+\infty$	0	positive
$\pi < \theta < \theta_s$	negative	negative	positive
$\theta = \theta_s$	zero	negative	zero
$\theta_s < \theta < 2\pi$	positive	negative	negative

Table B1 shows that the tension $T$ must be positive (or the string must be taut and straight) in the angular range $0 < \theta < \theta_s$ . Once $\theta$ reaches $\theta_s$ , the tension $T$ becomes zero and the part of the string not in contact with the rod will not be straight afterwards. The shortest possible value $s_{\min}$ for the length $s$ of the line segment $QP$ therefore occurs at $\theta = \theta_s$ and is given by

s_{\min} = L - R\theta_s = R\left(\frac{9\pi}{8} + \frac{2}{3}\cot\frac{\pi}{16} - \frac{9\pi}{8}\right) = \frac{2R}{3}\cot\frac{\pi}{16} = 3.352R \qquad (B8)

When $\theta = \theta_s$ , we have $T = 0$ and Eqs. (B2) and (B3) then leads to $v_s^{2} = -gs_{\min}\sin\theta_s$ . Hence the speed $v_s$ is

\begin{aligned} v_s &= \sqrt{-gs_{\min}\sin\theta_s} = \sqrt{\frac{2gR}{3}\cot\frac{\pi}{16}\sin\frac{\pi}{8}} = \sqrt{\frac{4gR}{3}}\cos\frac{\pi}{16} \\ &= 1.133\sqrt{gR} \end{aligned} \qquad (B9)

(i) When $\theta \ge \theta_s$ , the particle moves like a projectile under gravity. As shown in Fig. B3, it is projected with an initial speed $v_s$ from the position $P = (x_s, y_s)$ in a direction making an angle $\phi = (3\pi/2 - \theta_s)$ with the $y$ -axis.

Circle with centre O, dashed x-axis upwards and y-axis to the left; the bob m level with the top at distance L; the dashed string s to the earlier position; angles θ, θs and φ at O; tangent point Q at the bottom right with the straight segment smin to P = (xs, ys); at P the velocity vs at angle θs − π to the vertical, the parabolic path to the highest point with horizontal speed vH, and the horizontal distance H.

Figure B3

The speed $v_H$ of the particle at the highest point of its parabolic trajectory is equal to the $y$ -component of its initial velocity when projected. Thus,

v_H = v_s\sin(\theta_s - \pi) = \sqrt{\frac{4gR}{3}}\cos\frac{\pi}{16}\sin\frac{\pi}{8} = 0.4334\sqrt{gR} \qquad (B10)

The horizontal distance $H$ traveled by the particle from point $P$ to the point of maximum height is

H = \frac{v_s^{2}\sin 2(\theta_s - \pi)}{2g} = \frac{v_s^{2}}{2g}\sin\frac{9\pi}{4} = 0.4535R \qquad (B11)

The coordinates of the particle when $\theta = \theta_s$ are given by

x_s = R\cos\theta_s - s_{\min}\sin\theta_s = -R\cos\frac{\pi}{8} + s_{\min}\sin\frac{\pi}{8} = 0.358R \qquad (B12)

y_s = R\sin\theta_s + s_{\min}\cos\theta_s = -R\sin\frac{\pi}{8} - s_{\min}\cos\frac{\pi}{8} = -3.478R \qquad (B13)

Evidently, we have $|y_s| > (R + H)$ . Therefore the particle can indeed reach its maximum height without striking the surface of the rod.

Part C

(j) Assume the weight is initially lower than $O$ by $h$ as shown in Fig. C1.

The bob m level with the top point A of a circle of radius R and centre O; a dashed double arrow marks the length L; angle θ at O; the string passes over the rod to the hatched weight M, which hangs a distance h below the level of O (dashed double arrow h).

Figure C1

When the weight has fallen a distance $D$ and stopped, the law of conservation of total mechanical energy as applied to the particle-weight pair as a system leads to

-Mgh = E' - Mg(h + D) \qquad (C1)

where $E'$ is the total mechanical energy of the particle when the weight has stopped. It follows

E' = MgD \qquad (C2)

Let $\Lambda$ be the total length of the string. Then, its value at $\theta = 0$ must be the same as at any other angular displacement $\theta$ . Thus we must have

\Lambda = L + \frac{\pi}{2}R + h = s + R\left(\theta + \frac{\pi}{2}\right) + (h + D) \qquad (C3)

Noting that $D = \alpha L$ and introducing $\ell = L - D$ , we may write

\ell = L - D = (1 - \alpha)L \qquad (C4)

From the last two equations, we obtain

s = L - D - R\theta = \ell - R\theta \qquad (C5)

After the weight has stopped, the total mechanical energy of the particle must be conserved. According to Eq. (C2), we now have, instead of Eq. (B1), the following equation:

E' = MgD = \frac{1}{2}mv^{2} - mg\left[R(1 - \cos\theta) + s\sin\theta\right] \qquad (C6)

The square of the particle’s speed is accordingly given by

v^{2} = (s\dot{\theta})^{2} = \frac{2MgD}{m} + 2gR\left[(1 - \cos\theta) + \frac{s}{R}\sin\theta\right] \qquad (C7)

Since Eq. (B3) stills applies, the tension $T$ of the string is given by

-T + mg\sin\theta = m(-s\dot{\theta}^{2}) \qquad (C8)

From the last two equations, it follows

\begin{aligned} T &= m(s\dot{\theta}^{2} + g\sin\theta) \\ &= \frac{mg}{s}\left[\frac{2M}{m}D + 2R(1 - \cos\theta) + 3s\sin\theta\right] \\ &= \frac{2mgR}{s}\left[\frac{MD}{mR} + (1 - \cos\theta) + \frac{3}{2}\left(\frac{\ell}{R} - \theta\right)\sin\theta\right] \end{aligned} \qquad (C9)

where Eq. (C5) has been used to obtain the last equality.

We now introduce the function

f(\theta) = 1 - \cos\theta + \frac{3}{2}\left(\frac{\ell}{R} - \theta\right)\sin\theta \qquad (C10)

From the fact $\ell = (L - D) \gg R$ , we may write

f(\theta) \approx 1 + \frac{3}{2}\frac{\ell}{R}\sin\theta - \cos\theta = 1 + A\sin(\theta - \phi) \qquad (C11)

where we have introduced

A = \sqrt{1 + \left(\frac{3}{2}\frac{\ell}{R}\right)^{2}} \quad , \quad \phi = \tan^{-1}\left(\frac{2R}{3\ell}\right) \qquad (C12)

From Eq. (C11), the minimum value of $f(\theta)$ is seen to be given by

f_{\min} = 1 - A = 1 - \sqrt{1 + \left(\frac{3}{2}\frac{\ell}{R}\right)^{2}} \qquad (C13)

Since the tension $T$ remains nonnegative as the particle swings around the rod, we have from Eq. (C9) the inequality

\frac{MD}{mR} + f_{\min} = \frac{M(L - \ell)}{mR} + 1 - \sqrt{1 + \left(\frac{3\ell}{2R}\right)^{2}} \ge 0 \qquad (C14)

\left(\frac{ML}{mR}\right) + 1 \ \ge\ \left(\frac{M\ell}{mR}\right) + \sqrt{1 + \left(\frac{3\ell}{2R}\right)^{2}} \approx \left(\frac{M\ell}{mR}\right) + \left(\frac{3\ell}{2R}\right) \qquad (C15)

From Eq. (C4), Eq. (C15) may be written as

\left(\frac{ML}{mR}\right) + 1 \ \ge\ \left(\frac{ML}{mR} + \frac{3L}{2R}\right)(1 - \alpha) \qquad (C16)

Neglecting terms of the order ( $R/L$ ) or higher, the last inequality leads to

\alpha \ \ge\ 1 - \frac{\left(\dfrac{ML}{mR}\right) + 1}{\left(\dfrac{ML}{mR} + \dfrac{3L}{2R}\right)} = \frac{\dfrac{3L}{2R} - 1}{\dfrac{ML}{mR} + \dfrac{3L}{2R}} = \frac{1 - \dfrac{2R}{3L}}{\dfrac{2M}{3m} + 1} \approx \frac{1}{1 + \dfrac{2M}{3m}} \qquad (C17)

The critical value for the ratio $D/L$ is therefore

\alpha_c = \frac{1}{1 + \dfrac{2M}{3m}} \qquad (C18)

Marking Scheme

Theoretical Question 1 — A Swing with a Falling Weight

Total Scores	Sub Scores	Marking Scheme for Answers to the Problem
Part A	(a)	Relation between $\dot{\theta}$ and $\dot{s}$ . ( $\dot{s} = -R\dot{\theta}$ )
4.3 pts.	0.5	➢ 0.2 for $\dot{\theta} \propto \dot{s}$ .
		➢ 0.3 for proportionality constant (- $R$ ).
	(b)	Velocity of $Q$ relative to $O$ . ( $\vec{v}_Q = R\dot{\theta}\,\hat{t}$ )
	0.5	➢ 0.2 for magnitude $R\dot{\theta}$ .
		➢ 0.3 for direction $\hat{t}$ .
	(c)	Particle’s velocity at $P$ relative to $Q$ . ( $\vec{v}^{\prime} = -s\dot{\theta}\,\hat{r} + \dot{s}\,\hat{t}$ )
	0.7	➢ 0.2+0.1 for magnitude and direction of $\hat{r}$ -component.
		➢ 0.3+0.1 for magnitude and direction of $\hat{t}$ -component.
	(d) 0.7	Particle’s velocity at $P$ relative to $O$ . ( $\vec{v} = \vec{v}^{\prime} + \vec{v}_Q = -s\dot{\theta}\,\hat{r}$ )
		➢ 0.3 for vector addition of $\vec{v}^{\prime}$ and $\vec{v}_Q$ .
		➢ 0.2+0.2 for magnitude and direction of $\vec{v}$ .
	(e)	$\hat{t}$ -component of particle’s acceleration at $P$ .
	0.7	➢ 0.3 for relating $\vec{a}$ or $\vec{a}\cdot\hat{t}$ to the velocity in a way that implies $\lvert\vec{a}\cdot\hat{t}\rvert = v^{2}/s$ .
		➢ 0.4 for $\vec{a}\cdot\hat{t} = -s\dot{\theta}^{2}$ (0.1 for minus sign.)
	(f)	Potential energy $U$ .
	0.5	➢ 0.2 for formula $U = -mgh$ .
		➢ 0.3 for $h = R(1 - \cos\theta) + s\sin\theta$ or $U$ as a function of $\theta$ , $s$ , and $R$ .
	(g)	Speed at lowest point $v_m$ .
	0.7	➢ 0.2 for lowest point at $\theta = \pi/2$ or $U$ equals minimum $U_m$ .
		➢ 0.2 for total mechanical energy $E = mv_m^{2}/2 + U_m = 0$ .
		➢ 0.3 for $v_m = \sqrt{-2U_m/m} = \sqrt{2g[R + (L - \pi R/2)]}$ .
Part B	(h)	Particle’s speed $v_s$ when $\overline{QP}$ is shortest.
4.3 pts.	2.4	➢ 0.4 for tension $T$ becomes zero when $\overline{QP}$ is shortest.
		➢ 0.3 for equation of motion $-T + mg\sin\theta = m(-s\dot{\theta}^{2})$ .
		➢ 0.3 for $E = 0 = m(s\dot{\theta})^{2}/2 - mg[R(1 - \cos\theta) + s\sin\theta]$ .
		➢ 0.4 for $\frac{3}{2}\left(\theta_s - \frac{L}{R}\right) = \tan\frac{\theta_s}{2}$ .
		➢ 0.5 for $\theta_s = 9\pi/8$ .
		➢ 0.3+0.2 for $v_s = \sqrt{4gR/3}\cos\pi/16 = 1.133\sqrt{gR}$
	(i)	The speed $v_H$ of the particle at its highest point.
	1.9	➢ 0.4 for particle undergoes projectile motion when $\theta \ge \theta_s$ .
		➢ 0.3 for angle of projection $\phi = (3\pi/2 - \theta_s)$ .
		➢ 0.3 for $v_H$ is the $y$ -component of its velocity at $\theta = \theta_s$ .
		➢ 0.4 for noting particle does not strike the surface of the rod.
		➢ 0.3+0.2 for $v_H = \sqrt{4gR/3}\cos(\pi/16)\sin(\pi/8) = 0.4334\sqrt{gR}$ .
Part C	(j)	The critical value $\alpha_c$ of the ratio $D/L$ .
3.4 pts	3.4	➢ 0.4 for particle’s energy $E' = MgD$ when the weight has stopped.
		➢ 0.3 for $s = L - D - R\theta$ .
		➢ 0.3 for $E' = MgD = mv^{2}/2 - mg[R(1 - \cos\theta) + s\sin\theta]$ .
		➢ 0.3 for $-T + mg\sin\theta = m(-s\dot{\theta}^{2})$ .
		➢ 0.3 for concluding $T$ must not be negative.
		➢ 0.6 for an inequality leading to the determination of the range of $D/L$ .
		➢ 0.6 for solving the inequality to give the range of $\alpha = D/L$ .
		➢ 0.6 for $\alpha_c = (1 + 2M/3m)$ .

Answer Sheet

Answer Sheet — Theoretical Question 1 — A Swing with a Falling Weight

(a) The relation between $\dot{\theta}$ and $\dot{s}$ is

(b) The velocity of the moving point $Q$ relative to $O$ is

$\vec{v}_Q =$

$\vec{v}^{\prime} =$

(d) When at $P$ , the particle’s velocity relative to $O$ is

$\vec{v} =$

(e) When at $P$ , the $\hat{t}$ -component of the particle’s acceleration relative to $O$ is

(f) When at $P$ , the particle’s gravitational potential energy is

$U =$

(g) The particle’s speed when at the lowest point of its trajectory is

$v_m =$

(h) When line segment $QP$ is straight with the shortest length, the particle‘s speed is (Give expression and value in terms of g and R )

$v_s =$

(i) At the highest point, the particle’s speed is (Give expression and value in terms of g and R)

$v_H =$

(j) In terms of the mass ratio $M/m$ , the critical value $\alpha_{\mathrm{c}}$ of the ratio $D/L$ is

$\alpha_{\mathrm{c}} =$

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf · допълнителни данни: IPhO_2003_Theoretical Question 1.pdf · допълнителни данни: IPhO_2003_Theoretical Solution 1.pdf