Съдържание

APhO 2003, Експеримент — Задача 2. II. Cylindrical Bore

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Всички материали от APhO 2003

4th Asian Physics Olympiad — Experimental Competition — 25 April 2003 · 25 април 2003 г. · 10 т.

Условие

Background

There are many techniques to study the object with a bore inside. Mechanical oscillation method is one of the non-destructive techniques. In this problem, you are given a brass cube of uniform density with cylindrical bore inside. You are required to perform non-destructive mechanical measurements and use these data to plot the appropriate graph to find the ratio of the radius of the bore to the side of the cube.

The cube of sides $a$ has a cylindrical bore of radius $b$ along the axis of symmetry as shown in Fig. 2.1. This bore is covered by very thin discs of the same material. A, B and C represent small holes at the corners of the cube. These holes can be used for suspending the cube in two configurations. Fig. 2.2(a) shows the suspension using B and C; the other suspension is by using A and B as shown in Fig 2.2(b).

Perspective drawing of a cube of side a (arrow along the bottom front edge) with a dotted vertical cylindrical bore of diameter 2b along its axis; small holes marked by dots at corners C (top front left), B (top back right) and A (bottom back right, directly below B). — Fig. 2.1 Geometry of cube with cylindrical bore

Two bifilar suspensions hanging from the top of a stand, with gravity g pointing down between them. 2.2(a): the cube hangs by two vertical threads attached at corners C and B, bore vertical, rotation about the vertical axis marked I1; corner A lies below B. 2.2(b): the cube hangs by threads at corners B and A, tilted so the bore axis is horizontal, rotation about the vertical axis marked I2; corner C at the bottom. The label “Top of stand” points to both stand tops. — Fig. 2.2 Two configurations of cube’s suspension

Students may use the following in their derivation of necessary formulae:

For a solid cube of side $a$

Cube with a vertical axis Y and a horizontal axis X, both passing through the centre of mass marked c.m.

I = \frac{1}{6}Ma^{2}\ \text{about both axes}

c.m. = centre of mass

For a solid cylinder of radius $b$ length $a$

Solid cylinder with its symmetry axis Y vertical and an axis X perpendicular to it through its centre.

I_{Y} = \frac{1}{2}mb^{2}

I_{X} = \frac{1}{12}ma^{2} + \frac{1}{4}mb^{2}

Materials and apparatus

brass cube
stop watch
stand
thread
ruler/ centimeter stick
linear graph papers

Experiment

a) Choose only one of the two bifilar suspensions as shown in Fig. 2.2, and derive the expressions for the moment of inertia and the period of oscillation about the vertical axis through the centre of mass in terms of $\ell, d, b, a$ and $g$ where $\ell$ is the length of each thread and $d$ is the separation between threads. (2 points)

b) Perform necessary non-destructive mechanical measurements and use these data to plot an appropriate graph and then find the value of $\dfrac{b}{a}$ . (8 points)

The value of $g$ for Bangkok = $9.78\ \mathrm{m/s^{2}}$ .

Отговори

Решение

Solution and Marking Scheme

Experiment

II. Cylindrical Bore

a) Derivation of moment of inertia $I$ (0.5 points)

Configuration Fig. 1.2(a)

I_1 = \frac{1}{6}Ma^{2} - \frac{1}{2}mb^{2} = \frac{1}{6}\left(\rho a^{3}\right)a^{2} - \frac{1}{2}\left(\rho\pi b^{2}a\right)b^{2}

= \frac{1}{6}\rho a^{5} - \frac{1}{2}\rho\pi ab^{4}

Configuration Fig. 1.2(b)

I_2 = \frac{1}{6}Ma^{2} - \frac{1}{12}ma^{2} - \frac{1}{4}mb^{2} = \frac{1}{6}\left(\rho a^{3}\right)a^{2} - \frac{1}{12}\left(\rho\pi b^{2}a\right)a^{2} - \frac{1}{4}\left(\rho\pi b^{2}a\right)b^{2}

= \frac{1}{6}\rho a^{5} - \frac{1}{12}\rho\pi a^{3}b^{2} - \frac{1}{4}\rho\pi ab^{4}

Derivation of period of oscillation $T$

For both configurations:

the restoring torque $\quad \tau = Fd$

where $\quad F = \frac{1}{2}m_0 g\frac{\delta s}{\ell} \quad$ and $\quad \frac{\delta s}{d/2} \approx \theta$

F \approx \frac{1}{2}m_0 g\,\frac{d}{2\ell}\theta \qquad \text{(0.5 points)}

net mass $\quad m_0 = \rho a^{3}\left(1 - \pi\frac{b^{2}}{a^{2}}\right) = \rho a^{3}\left(1 - \pi x^{2}\right) \quad$ where $\quad x \equiv \frac{b}{a}$

since $\quad \tau = I\alpha, \quad \alpha = \dfrac{\frac{1}{4}m_0 g\frac{d^{2}}{\ell}\theta}{I}$ (0.5 points)

\omega^{2} = \frac{4\pi^{2}}{T^{2}} = \frac{\frac{1}{4}m_0 g\frac{d^{2}}{\ell}}{I}

T^{2} = \frac{4\pi^{2}I\ell}{\frac{1}{4}m_0 gd^{2}} = \left(\frac{16\pi^{2}I}{m_0 gd^{2}}\right)\ell

For configuration in Fig. 2.2(a),

T_1^{2} = \left(\frac{16\pi^{2}}{gd^{2}}\,\frac{\frac{1}{6}\rho a^{5} - \frac{1}{2}\rho\pi ab^{4}}{\rho a^{3}\left(1 - \pi x^{2}\right)}\right)\ell

= \frac{8\pi^{2}}{3g}\left(\frac{a}{d}\right)^{2}\left(\frac{1 - 3\pi x^{4}}{1 - \pi x^{2}}\right)\ell \qquad \text{(0.5 points)}

For configuration in Fig. 2.2(b),

T_2^{2} = \left(\frac{16\pi^{2}\frac{1}{6}\rho a^{5}\left(1 - \frac{\pi}{2}\frac{b^{2}}{a^{2}} - \frac{3\pi}{2}\frac{b^{4}}{a^{4}}\right)}{\rho a^{3}(1 - \pi x^{2})gd^{2}}\right)\ell

= \frac{8\pi^{2}}{3g}\left(\frac{a}{d}\right)^{2}\left(\frac{1 - \frac{\pi x^{2}}{2} - \frac{3\pi}{2}x^{4}}{1 - \pi x^{2}}\right)\ell

b) For configuration in Fig. 2.2(a), $\quad d = 7.0\ \mathrm{cm}$

$\ell$ (cm)	$T_1$ for 40 oscillations(s)			$T_1$ (s)	$(T_1)^2$ (s $^2$ )
16.5	20.60	20.50	20.70	0.515	0.265
17.9	21.35	21.35	21.30	0.533	0.284
22.6	24.05	24.00	24.00	0.601	0.362
27.4	26.55	26.45	26.55	0.663	0.440
29.0	27.40	27.40	27.40	0.685	0.469
34.2	29.75	29.70	29.65	0.743	0.551
36.1	30.60	30.60	30.50	0.764	0.584
43.0	33.40	33.35	33.50	0.835	0.698

(3 points):

3 sets of $n$ oscillations (1 point) [2 sets –0.3, 1 set –0.7]

$n \ge 20$ (1 point) [ $\ge 15$ , -0.3, $\ge 10$ , -0.7, $<10$ , -1.0]

number of lengths, $\ell$ , $\ge 5$ (1 point) [4, -0.3, 3, -0.5, 1 or 2, -1.0]

Graph of T1^2 (s^2) against ℓ (cm) for configuration 2.2(a): eight data points from ℓ = 16.5 to 43.0 cm on a straight line through the origin; vertical axis 0.00 to 0.80, horizontal axis 0.0 to 50.0.

slope of graph: $\quad s_1 = \dfrac{0.698 - 0.265}{\left(43.0 - 16.5\right)\times 10^{-2}} = \dfrac{0.433}{26.5\times 10^{-2}} = 1.634\ \mathrm{s^{2}/m}$

x = \frac{b}{a} = 0.24 \pm 0.02

For configuration in Fig. 2.2(b), $\quad d = 4.9\ \mathrm{cm}$

$\ell$ (cm)	$T_2$ for 40 oscillations (s)			$T_2$ (s)	$(T_2)^2$ (s $^2$ )
43.8	46.95	46.90	46.80	1.172	1.374
36.0	42.70	42.45	42.50	1.064	1.132
30.9	39.60	39.40	39.35	0.986	0.973
26.5	36.40	36.30	36.45	0.909	0.827
19.5	30.80	30.85	30.75	0.776	0.593

Graph of T2^2 (s^2) against ℓ (cm) for configuration 2.2(b): five data points from ℓ = 19.5 to 43.8 cm on a straight line through the origin; vertical axis 0.00 to 1.60, horizontal axis 0.0 to 50.0.

slope of graph: $\quad s_2 = \dfrac{1.374 - 0.827}{\left(43.8 - 26.5\right)\times 10^{-2}} = \dfrac{0.547}{17.3\times 10^{-2}} = 3.14\ \mathrm{s^{2}/m}$

x = \frac{b}{a} = 0.25 \pm 0.03

graph (3.0 points):

good graph (1.5 points)
slope (1.0 point)
error of experimental points (0.5 point)

calculation for $\dfrac{b}{a}$ (1.0 point) $\qquad$ error estimation (1.0 point)

Оригинал в Архива: exp2.pdf · официални решения: exp2_sol.pdf