Съдържание

IPhO 2011, Експеримент — Задача 2. Mechanical Blackbox: a cylinder with a ball inside

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Всички материали от IPhO 2011

Experimental Competition: 14 July 2011 — Problem 2 · 14 юли 2011 г.

Условие

A small massive particle (ball) of mass $m$ is fixed at distance $z$ below the top of a long hollow cylinder of mass $M$ . A series of holes are drilled perpendicularly to the central axis of the cylinder. These holes are for pivoting so that the cylinder will hang in a vertical plane.

Students are required to perform necessary nondestructive measurements to determine the numerical values of the following with their error estimates:

i. position of centre of mass of cylinder with ball inside.

Also provide a schematic drawing of the experimental set-up for measuring the centre of mass. [1 т.]

ii. distance $z$ [3,5 т.]

iii. ratio $\dfrac{M}{m}$ . [3,5 т.]

iv. the acceleration due to gravity, $g$ . [2 т.]

Equipment: a cylinder with holes plus a ball inside, a base plate with a thin pin, a pin cap, a ruler, a stopwatch, thread, a pencil and adhesive tape.

Sketch of a rectangular base plate labelled 'Base plate to be clamped to a table top' with a thin pin sticking out of its front face, labelled 'Thin pin for pivoting'.

$x_{CM}$ is the distance from the top of the cylinder to the centre of mass.

Vertical hollow cylinder of length L with a pivot hole O near the top, the centre of mass CM marked by a dot, and a ball of mass m (blue circle) below it; the cylinder is labelled M. Arrows mark z (top to ball), x_CM (top to CM), R (pivot O to CM) and L (full length); the pivot is labelled 'pivot'.

$R$ is the distance from the pivoting point to the centre of mass.

Photograph of the equipment on a green background with labels: Cylinder with holes plus ball inside, Adhesive tape, Pin cap, Stopwatch, Thread (for balancing), Base plate, Ruler.

Caution: The thin pin is sharp. When it is not in use, it should be protected with a pin cap for safety.

Useful information:

For such a physical pendulum, $\left\{(M+m)R^2+I_{CM}\right\}\dfrac{d^2\theta}{dt^2} \approx -g(M+m)R\theta$ , where $I_{CM}$ is the moment of inertia of the cylinder with a ball about the centre of mass and $\theta$ is the angular displacement.
For a long hollow cylinder of length $L$ and mass $M$ , the moment of inertia about the centre of mass with the rotational axis perpendicular to the cylinder can be approximated by $\dfrac{1}{3}M\left(\dfrac{L}{2}\right)^2$ .
The parallel axis theorem: $I = I_{\text{centre of mass}} + \mathfrak{M}x^2$ , where $x$ is the distance from the rotation point to the centre of mass, and $\mathfrak{M}$ is the total mass of the object.
The ball can be treated as a point mass and it is located on the central axis of the cylinder.
Assume that the cylinder is uniform and the mass of the end-caps is negligible.

Отговори

Решение

Solution: 2 . Mechanical Blackbox: a cylinder with a ball inside

Same cylinder diagram as in the problem: vertical cylinder with pivot O, centre of mass CM, an unlabelled blue ball, label M; arrows for z, x_CM, R and L.

In order to be able to calculate the required values in i, ii, iii, we need to know:

a. the position of the centre of mass of the tubing plus particle (object) which depends on $z, m, M$

b. the moment of inertia of the above.

The position of the CM may be found by balancing. The $I_{CM}$ can be calculated from the period of oscillation of the tubing plus object.

Analytical steps to select parameters for plotting

x_{CM} = \frac{mz + M\dfrac{L}{2}}{m+M} \qquad (1)

$L$ is readily obtainable with a ruler.

$x_{CM}$ is determined by balancing the tubing and object.

II. For small-amplitude oscillation about any point O the period $T$ is given by considering the equation:

\left\{(M+m)R^2 + I_{CM}\right\}\ddot{\theta} = -g(M+m)R\sin\theta \approx -g(M+m)R\theta \qquad (2)

T = 2\pi\sqrt{\frac{I_{CM} + (M+m)R^2}{g(M+m)R}} \qquad (3)

where

I_{CM} = \frac{1}{3}M\left(\frac{L}{2}\right)^2 + M\left(x_{CM} - \frac{L}{2}\right)^2 + m\left(z - x_{CM}\right)^2

= \frac{1}{3}ML^2 + Mx_{CM}^2 - MLx_{CM} + m\left(z - x_{CM}\right)^2 \qquad (4)

Note that

T^2\frac{g(M+m)}{4\pi^2} = \frac{I_{CM}}{R} + (M+m)R \qquad (5)

Method (a): (linear graph method)

The equation (5) may be put in the form:

T^2R = \left(\frac{4\pi^2}{g}\right)R^2 + \frac{4\pi^2 I_{CM}}{(M+m)g} \qquad (6)

Hence the plot of $T^2R$ v.s. $R^2$ will yield the straight line whose

\text{Slope}\quad \alpha = \frac{4\pi^2}{g} \qquad (7)

\text{and y-intercept}\quad \beta = \frac{4\pi^2 I_{CM}}{(M+m)g} \qquad (8)

\text{Hence,}\quad I_{CM} = (M+m)\frac{\beta}{\alpha} \qquad (9)

The value of $g$ is from equation (7):

g = \frac{4\pi^2}{\alpha} \qquad (10)

Method (b): minimum point curve method

The equation (5) implies that $T$ has a minimum value at

R = R_{\min} \equiv \sqrt{\frac{I_{CM}}{M+m}} \qquad (11)

Hence $R_{\min}$ can be obtained from the graph $T$ v.s. $R$ .

And therefore

I_{CM} = (M+m)R_{\min}^2 \qquad (12)

This equation (12) together with equation (1) will allow us to calculate the required values $z$ and $\dfrac{M}{m}$ .

At the value $R = R_{\min}$ equation (5) becomes $T_{\min}^2\dfrac{g(M+m)}{4\pi^2} = (M+m)R_{\min} + (M+m)R_{\min}$

g = \frac{2R_{\min}}{T_{\min}^2}\times 4\pi^2 = \frac{8\pi^2 R_{\min}}{T_{\min}^2} \qquad (13)

from which $g$ can be calculated.

Results

L = 30.0\ \mathrm{cm} \pm 0.1\ \mathrm{cm}

x_{CM} = 17.8\ \mathrm{cm} \pm 0.1\ \mathrm{cm}\ \text{(from top)}

$x_{CM} - R$ (cm)	time (s) for 20 cycles			$T$ (s)	$R$ (cm)	$R^2$ (cm $^2$ )	$T^2R$ (s $^2$ cm)
1.1	18.59	18.78	18.59	0.933	16.7	278.9	14.53
2.1	18.44	18.25	18.53	0.920	15.7	246.5	13.29
3.1	18.10	18.09	18.15	0.906	14.7	216.1	12.06
4.1	17.88	17.78	17.81	0.891	13.7	187.7	10.88
5.1	17.69	17.50	17.65	0.881	12.7	161.3	9.85
6.1	17.47	17.38	17.28	0.869	11.7	136.9	8.83
7.1	17.06	17.06	17.22	0.856	10.7	114.5	7.83
8.1	17.06	17.00	17.06	0.852	9.7	94.1	7.04
9.1	16.97	16.91	16.96	0.847	8.7	75.7	6.25
10.1	17.00	17.03	17.06	0.852	7.7	59.3	5.58
11.1	17.22	17.37	17.38	0.866	6.7	44.9	5.03
12.1	17.78	17.72	17.75	0.888	5.7	32.5	4.49
13.1	18.57	18.59	18.47	0.927	4.7	22.1	4.04
14.1	19.78	19.90	19.75	0.991	3.7	13.7	3.69
15.1	11.16	11.13	11.13	1.114	2.7	7.3	3.34
16.1	13.25	13.40	13.50	1.338	1.7	2.9	3.04

Notes: at $x_{CM} - R = 15.1, 16.1\ \mathrm{cm}$ , times for 10 cycles.

Method (a)

Graph of T^2R (s^2cm), 0 to 16, against R^2 (cm^2), 0 to 300: sixteen data points lying on a straight line from about 3.0 at R^2 near 3 to about 14.5 at R^2 near 279, with the fitted line drawn. — Method (a)

Calculation from straight line graph: slope $\alpha = 0.04108 \pm 0.0007\ \mathrm{s^2/cm}$ , y-intercept $\beta = 3.10 \pm 0.05\ \mathrm{s^2cm}$

g = \frac{4\pi^2}{\alpha}\ \text{giving}\ g = (961 \pm 20)\ \mathrm{cm/s^2}

\frac{\beta}{\alpha} = \frac{3.10}{0.04108} = 75.46\ \mathrm{cm^2}\ \left(\pm 2.5\ \mathrm{cm^2}\right)

I_{CM} = (M+m)\frac{\beta}{\alpha} = (75.46)(M+m)

From equation (4):

I_{CM} = \frac{1}{3}M\left(\frac{L}{2}\right)^2 + M\left(x_{CM} - \frac{L}{2}\right)^2 + m\left(z - x_{CM}\right)^2

Then

(75.46)(M+m) = 75.0M + 7.84M + m(z - 17.8)^2

-7.38\frac{M}{m} + 75.46 = (z - 17.8)^2 \qquad (14)

The centre of mass position gives:

17.8(M+m) = 15.0M + mz

\frac{M}{m} = \frac{z - 17.8}{2.8} \qquad (15)

From equations (14) and (15):

-\frac{7.38}{2.8}(z - 17.8) + 75.46 = (z - 17.8)^2

(z - 17.8) = 7.47

And

z = 25.27 = 25.3 \pm 0.1\ \mathrm{cm}

\frac{M}{m} = 2.68 = 2.7

Error Estimation

Find error for $g$ :

From (10), $g = \dfrac{4\pi^2}{\alpha}$

\Delta g = \frac{\Delta\alpha}{\alpha}g = 16.3\ \mathrm{cm/s^2} \approx 20\ \mathrm{cm/s^2}

i) Find error for $z$ :

First, find error for $r = \dfrac{\beta}{\alpha} = \dfrac{3.10}{0.04108} = 75.46\ \mathrm{cm^2}$ .

\Delta r = \left(\frac{\Delta\alpha}{\alpha} + \frac{\Delta\beta}{\beta}\right)r = 2.5\ \mathrm{cm^2}

Since error from $r$ contributes most ( $\dfrac{\Delta r}{r} \sim 0.03$ while $\dfrac{\Delta L}{L}, \dfrac{\Delta x_{cm}}{x_{cm}} \sim 0.005$ ), we estimate error propagation from $r$ only to simplify the analysis by substituting the min and max values into equation (4).

Now, we use $r_{\max} = r + \Delta r = 75.46 + 2.5 = 77.96$ . The corresponding quadratic equation is $(z - 17.8)^2 + 1.743(z - 17.8) - 77.96 = 0$ The corresponding solution is $(z - 17.8)_{\max} = 7.55\ \mathrm{cm}$

If we use $r_{\min} = r - \Delta r = 75.46 - 2.5 = 72.96$ , the corresponding quadratic equation is

$(z - 17.8)^2 + 3.529(z - 17.8) - 72.96 = 0$

The corresponding solution is $(z - 17.8)_{\min} = 6.96\ \mathrm{cm}$

So $\Delta(z - 17.8) = \dfrac{7.55 - 6.96}{2} = 0.3\ \mathrm{cm}$

Note that $\dfrac{\Delta(z - 17.8)}{z - 17.8} \sim 0.04$ . So, we still ignore the error propagation due to $\Delta L, \Delta x_{cm}$

The error $\Delta z$ can be estimated from $\Delta z \approx \Delta(z - 17.8) = 0.3\ \mathrm{cm}$

ii) Find error for $\dfrac{M}{m}$ :

We know that $\dfrac{M}{m} = \dfrac{z - 17.8}{2.8}$

\Delta\left(\frac{M}{m}\right) = \frac{\Delta(z - 17.8)}{2.8} = 0.11

Method (b)

Calculation from T-R plot:

Graph of T (s), 0.8 to 1.4, against R (cm), 0 to 18: sixteen data points on a smooth curve falling steeply from 1.34 s at R = 1.7 cm to a minimum of about 0.85 s near R = 8.7 cm, then rising slowly to 0.93 s at R = 16.7 cm.

Using the minimum position: $T = T_{\min}$ at $I_{CM} = (M+m)R_{\min}^2$ and $g = \dfrac{8\pi^2 R_{\min}}{T_{\min}^2}$

From graph: $R_{\min} = 8.9 \pm 0.2\ \mathrm{cm}$ and $T_{\min} = 0.846 \pm 0.005\ \mathrm{s}$

\therefore\quad g = 982 \pm 40\ \mathrm{cm/s^2}

I_{CM} = (M+m)(8.9)^2 = (79.21)(M+m) \qquad (16)

From equations (14), (15), (16):

(79.21)(M+m) = 75.0M + 7.84M + m(z - 17.8)^2

-3.63M + 79.21m = m(z - 17.8)^2

(x - 17.8)^2 + \frac{3.63}{2.8}(x - 17.8) - 79.21 = 0

(z - 17.8) = 8.28

And

z = 26.08 = 26.1 \pm 0.7\ \mathrm{cm}

\frac{M}{m} = 2.95 = 3.0 \pm 0.3

Error estimation

i) Find error for $g$ :

Using the minimum position: $g = \dfrac{8\pi^2 R_{\min}}{T_{\min}^2}$ , we have

\Delta g = \left(\frac{\Delta R_{\min}}{R_{\min}} + 2\frac{\Delta T_{\min}}{T_{\min}}\right)g = 34 \approx 30\ \mathrm{cm/s^2}

ii) Find error for $z$ :

First, find error for $r = R_{\min}^2 = 79.21\ \mathrm{cm^2}$ .

\Delta r = 2R_{\min}\Delta R_{\min} = 3.56\ \mathrm{cm^2}

This $r$ is equivalent to $r$ in part 1. So, one can follow the same error analysis. As a result, we have

z = 26.08 \approx 26.1\ \mathrm{cm}

\Delta z = 0.8\ \mathrm{cm}

i) Find error for $\dfrac{M}{m}$ :

Following the same analysis as in part I, we found that

\frac{M}{m} = 2.96;\ \Delta\left(\frac{M}{m}\right) = 0.15

NOTE: This minimum curve method is not as accurate as the usual straight line graph.

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