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Условие APPARATUS PART 1: DETERMINATION of g USING a ROTATING LIQUID PART 2: OPTICAL SYSTEM PART 3: REFRACTIVE INDEX Отговори Решение

IPhO 2001, Експеримент — Задача 1. ROTATING LIQUID

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Experimental Competition · 30 юни 2001 г.

Please read this first:

Saturday, June 30th, 2001

The time available is 5 hours for the experimental competition.
Use only the pen provided.
Use only the front side of the paper.
Begin each part of the problem on a separate sheet.
For each question, in addition to the blank sheets where you may write, there is an answer form where you must summarize the results you have obtained. Numerical results should be written with as many digits as are appropriate to the given data.
Write on the blank sheets of paper the results of all your measurements and whatever else you consider is required for the solution of the question. Please use as little text as possible; express yourself primarily in equations, numbers, figures and plots.
Fill in the boxes at the top of each sheet of paper used by writing your Country no and Country code, your student number (Student No.), the number of the question (Question No.), the progressive number of each sheet (Page No.) and the total number of blank sheets used for each question (Total No. of pages). Write the question number and the section label of the part you are answering at the beginning of each sheet of writing paper. If you use some blank sheets of paper for notes that you do not wish to be marked, put a large X across the entire sheet and do not include it in your numbering.
At the end of the exam, arrange all sheets in the following order;
- answer form
- used sheets in order
- the sheets you do not wish to be marked
- unused sheets and the printed question

Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any sheets of paper and any material used in the experiment out of the room.

IMPORTANT NOTES

DO NOT LOOK DIRECTLY INTO THE LASER BEAM. BE AWARE THAT LASER LIGHT CAN ALSO BE DANGEROUS WHEN REFLECTED OFF A MIRROR-LIKE SURFACE. FOR YOUR OWN SAFETY USE THE GIVEN GLASSES.
Throughout the whole experiment carefully handle the cup containing glycerin.
The turntable has already been previously adjusted to be horizontal. Use bubble level only for horizontal alignment of the screen.
Throughout the entire experiment you will observe several spots on the screen produced by the reflected and/or refracted beams at the various interfaces between the air, the liquid, the screen, and the cup. Be sure to make your measurements on the correct beam.
In rotating the liquid change the speed of rotation gradually and wait for long enough times for the liquid to come into equilibrium before making any measurements.

ANSWER FORM — identity fields

Country no: ______

Country code: ______

Student No: ______

Question No: ______

Page No: ______

Total No. of pages: ______

The original form repeats these blank identity boxes on each of its three sheets. All response tables and prompts appear in the corresponding parts below. The variation-of-size cell for $\omega=0$ is printed filled black, both on the supplied answer form and in the key; it is intentionally unavailable, and its empty Markdown cell is not a missing answer.

Условие

This experiment consists of three basic parts:

investigation of the profile of the rotating liquid’s surface and the determination of the acceleration due to gravity,
investigation of the rotating liquid as an optical system,
determination of the refractive index of the liquid.

When a cylindrical container filled with a liquid rotates about the vertical axis passing through its center with a uniform angular velocity $\omega$ , the liquid’s surface becomes parabolic (see Figure 1). At equilibrium, the tangent to the surface at the point P( $x$ , $y$ ) makes an angle $\theta$ with the horizontal such that

\tan\theta = \frac{\omega^2 x}{g} \qquad \text{for } |x| \le R \qquad (1)

where $R$ is the radius of the container and $g$ is the acceleration due to gravity.

It can further be shown that for $\omega<\omega_{\max}$ (where $\omega_{\max}$ is the angular speed at which the center of the rotating liquid touches the bottom of the container)

\text{at } x=x_0=\frac{R}{\sqrt{2}},\ y(x_0)=h_0 \qquad (2)

that is; the height of the rotating liquid is the same as if it were not rotating.

The profile of the rotating liquid’s surface is a parabola defined by the equation

y = y_0 + \frac{x^2}{4C} \qquad (3)

where the vertex is at V(0, $y_0$ ) and the focus is at F(0, $y_0+C$ ). When optical rays parallel to the axis of symmetry (optical axis) reflect at the parabolic surface, they all focus at the point F (see Fig.1).

Cross-section of a cylindrical container rotating with angular speed ω about the vertical y axis. The liquid surface is a parabola with vertex V(0,y0) on the axis and a point P(x,y) on it with a tangent line making angle θ with the horizontal; the focus F(0,y0+C) lies on the axis above. A dashed horizontal line marks the initial liquid height h0 above the bottom; R marks the radius; the x axis runs along the bottom. — Figure 1. Definitions of the bank angle θ at point P(x,y), the vertex V and the focus F for the parabolic surface produced by rotating the liquid, of initial height h0 and radius R, at a constant angular speed ω around the y-axis.

APPARATUS

A cylindrical rigid plastic cup containing liquid glycerin. Millimetric scales are attached to the bottom and the sidewall of this cup.
A turntable driven by a small dc electric motor powered by a variable voltage supply, which controls the angular velocity.
A transparent horizontal screen on which you can put transparent or semi-transparent millimetric scales. The location of the screen can be adjusted along the vertical and horizontal directions.
A laser pointer mounted on a stand. The position of the pointer can be adjusted. The head of the pointer can be changed.
Additional head for the laser pointer.
A ruler.
A highlighter pen.
A stopwatch. Push the left button to reset, the middle button to select the mode, and the right button to start and stop the timing.
Transmission gratings with 500 or 1000 lines/mm.
Bubble level.
Glasses.

PART 1: DETERMINATION of g USING a ROTATING LIQUID

Derive Equation 1.
Measure the height $h_0$ of the liquid in the container and the inner diameter $2R$ of the container.

Perspective drawing of the setup: a laser pointer (1) on a horizontal stand above a transparent screen (2) marked with point +P on the vertical y axis; below, a cylindrical container (7) of liquid of height h0 and radius R on a turntable (5) rotating with ω about the axis (6); a motor (3) and motor controller (4) connected to 220V AC; the distance H from the screen to the turntable is marked, with arrows showing the screen and pointer can be moved. — Figure 2 Experimental setup for parts 1 and 2. 1. Laser pointer on a stand, 2. Transparent screen, 3. Motor, 4. Motor controller, 5. Turntable, 6. Axis of rotation, 7. Cylindrical container.

Insert the screen between the light source and the container. Measure the distance $H$ between the screen and the turntable (see Figure 2).
Align the laser pointer such that the beam points vertically downward and hits the surface of the liquid at a distance $x_0=\frac{R}{\sqrt{2}}$ from the center of the container.
Rotate the turntable slowly. Be sure that the center of the rotating liquid is not touching the bottom of the container.
It is known that at $x_0=\frac{R}{\sqrt{2}}$ the height of the liquid remains the same as the original height $h_0$ , regardless of the angular speed $\omega$ . Using this fact and measurements of the angle $\theta$ of the surface at $x_0$ for various values of $\omega$ , perform an experiment to determine the gravitational acceleration $g$ .
Prepare tables of measured and calculated quantities for each $\omega$ .
Produce the necessary graph to calculate $g$ .
Calculate the value of $g$ and the experimental error in it
Copy the values $2R$ , $x_0$ , $h_0$ , $H$ and the experimental value of $g$ and its error onto the answer form.

ANSWER FORM

1) Determination of g using a rotating liquid

$2R$	$x_0$	$h_0$	$H$

Experimental value of $g$ : [7,5 т.]

PART 2: OPTICAL SYSTEM

In this part of the experiment the rotating liquid will be treated as an image forming optical system. Since the curvature of the surface varies with the angular speed of rotation, the focal distance of this optical system depends on $\omega$ .

Align the laser pointer such that the laser beam is directed vertically downward at the center of the container. Mark the point $P$ where the beam strikes the screen. Thus the line joining this point to the center of the cup is the optical axis of this system (see Figure 2).
Since the surface of the liquid behaves like a parabolic mirror, any incident beam parallel to the optical axis will pass through the focal point F on the optical axis after reflection.
Adjust the speed of rotation to locate the focal point on the screen. Measure the angular speed of rotation $\omega$ and the distance $H$ between the screen and the turntable.
Repeat the above steps for different $H$ values.
Copy the measured values of $2R$ and $h_0$ and the value of $\omega$ at each $H$ onto the answer form.
With the help of an appropriate graph of your data, find the relationship between the focal length and the angular speed. Copy your result onto the answer form.

2a) Investigation of the focal distance

2R	$h_0$

H	$\omega$

Relation between focal length and $\omega$ : [5,5 т.]

In this part of the experiment the properties of the “image” produced by this optical system will be analyzed. To do so, follow the steps given below.

Remove the head of the laser pointer by turning it counterclockwise.
Mount the new head (provided in an envelope) by turning it clockwise. Now your laser produces a well defined shape rather than a narrow beam.
Adjust the position of the laser pointer so that the beam strikes at about the center of the cup almost normally.
Put a semitransparent sheet of paper on the horizontal screen, which is placed close to the cup, such that the laser beam does not pass through the paper, but the reflected beam does.
Observe the size and the orientation of the “image” produced by the source beam and the beam reflected from the liquid when it is not rotating.
Start the liquid rotating, and increase the speed of rotation gradually up to the maximum attainable speed while watching the screen. As $\omega$ increases you might observe different frequency ranges over which the properties of the “image” are drastically different. To describe these observations complete the table on the answer form by adding a row to this table for each such frequency range and fill it in by using the appropriate notations explained on that page.

2b) Analysis of the “image”

Use the appropriate notations explained below to describe what you see on the screen due to reflected beam

$\omega$ range: For the frequency ranges only approximate values are required.

Orientation (in comparison with the object beam as seen on the transparent screen):

Inverted : INV

Erect : ER

Variation of the size with increasing $\omega$ :

Increases : I

Decreases : D

No change : NC

For the frequency ranges you have found above:

Write “R” if the screen is above the focal point.

Write “V” if the screen is below the focal point.

$\omega$ Range	Orientation	Variation of the size	“image”
$\omega$ =0

[3,5 т.]

PART 3: REFRACTIVE INDEX

In this part of the experiment the refractive index of the given liquid will be determined using a grating. When monochromatic light of wavelength $\lambda$ is incident normally on a diffraction grating, the maxima of the diffraction pattern are observed at angles $\alpha_m$ given by the equation

m\lambda = d\sin\alpha_m \qquad (4)

where, $m$ is the order of diffraction and $d$ is the distance between the rulings of the grating. In this part of the experiment a diffraction grating will be used to determine the wavelength of the laser light and the refractive index of the liquid (see Figure 3).

Top view of a circular container (4) filled with liquid of index n and radius R, with a grating on a holder (2) along a vertical diameter at the center. A laser pointer (3) on the right sends a beam along the horizontal diameter onto the grating; on the left, the scaled sidewall (1) receives the zeroth-order beam and a diffracted beam at angle αm. — Figure 3 Top view of the grating in a liquid experiment. 1. Scaled sidewall, 2. Grating on a holder, 3. Laser pointer, 4. Cylindrical container.

Use the grating to determine the wavelength of the laser pointer. Copy your result onto the answer form.
Immerse the grating perpendicularly into the liquid at the center of the cup.
Align the laser beam such that it enters the liquid from the sidewall of the cup and strikes the grating normally.
Observe the diffraction pattern produced on the millimetric scale attached to the cup on the opposite side. Make any necessary distance measurements.
Calculate the refractive index $n$ of the liquid by using your measurements. (Ignore the effect of the plastic cup on the path of the light.)
Copy the result of your experiment onto the answer form.

3) Refractive index

Wavelength =

Experimental value for $n$ = [3,5 т.]

Отговори

Решение

Part 1

Theory:

Consider a small mass m of the liquid at the surface (Figure 4). At dynamic equilibrium

N\cos\theta = mg

and

N\sin\theta = mw^2x

Therefore:

\tan\theta = \frac{w^2x}{g}.

The profile of the liquid surface can be found as follows:

\tan\theta = \frac{dy}{dx}, \qquad \frac{dy}{dx} = \frac{w^2x}{g}

so that

y = \frac{w^2x^2}{2g} + y_0

where $y_0$ is the height at $x = 0$ .

At a certain point $x = x_0$ , height of the liquid $h_0$ would be the same as if it not rotating. In this case,

h_0 = y_0 + \frac{w^2x_0^2}{2g} \qquad (1)

and,

x_0^2 = \frac{2g(h_0-y_0)}{w^2}.

Since the volume of the liquid is constant,

\pi R^2h_0 = \int y(2\pi x\,dx) = 2\pi\int\left(y_0+\frac{w^2x^2}{2g}\right)x\,dx,

y_0 = h_0 - \frac{w^2R^2}{4g} \qquad (2)

From Eq.1 and Eq.2 one obtains

x_0 = \frac{R}{\sqrt{2}}.

Force diagram: a small mass m on the curved (parabolic) liquid surface, with the rotation axis arrow labelled w, the weight mg pointing down, the normal force N perpendicular to the surface, and the tangent line making an angle with the dotted horizontal. — Figure 1.

Experiment:

2R(mm)	$x_0$ (mm)	$h_0$ (mm)	H(mm)
145	51	30	160

$H-h_0$ =130 mm

Measure 10T at small speeds and measure 15T-20T at high speeds.

Use $\tan(2\theta) = \frac{x}{H-h_0}$ and $w = \frac{2\pi}{T}$ .

2R(mm)	$x_0$ (mm)
145	51

$h_0$ (mm)	H(mm)	$H-h_0$ (mm)
30	160	130

x(mm)	10T(s)	w(rad/s)	tan(2θ)	θ(rad)	θ(deg)	tan(θ)	$w^2$ (rad/s) $^2$
11	21.34	2.94	0.08	0.04	2.4	0.04	8.67
20	15.80	3.98	0.15	0.08	4.4	0.08	15.81
26	14.22	4.42	0.20	0.10	5.7	0.10	19.52
30	12.99	4.84	0.23	0.11	6.5	0.11	23.40
40	11.74	5.35	0.31	0.15	8.6	0.15	28.64
51	10.45	6.01	0.39	0.19	10.7	0.19	36.15
56	9.90	6.35	0.43	0.20	11.7	0.21	40.28
65	9.40	6.68	0.50	0.23	13.3	0.24	44.68
70	9.08	6.92	0.54	0.25	14.2	0.25	47.88
85	8.39	7.49	0.65	0.29	16.6	0.30	56.08
100	7.71	8.15	0.77	0.33	18.8	0.34	66.41
112	7.43	8.46	0.86	0.36	20.4	0.37	71.51
132	7.00	8.98	1.02	0.40	22.7	0.42	80.57
61.4	11.19	6.20	0.47	0.21	11.98	0.21	41.51	Ave.

The last line is for error calculation only.

The slope of the Figure 5 is 0.0052 (s/rad) $^2$ which gives

g = \frac{x_0}{\mathit{slope}} = \frac{5.1}{0.0052} = 980\ \mathrm{cm/s^2}.

Graph titled 'tanθ vs w² at x0': tanθ (0,00 to 0,45) against (w(rad/s))² (0,00 to 100,00) with data points on a straight line through the origin; fit y = 0,0052x, R² = 0,9987. — Figure 2.

Error Calculation (possible methods):

g = \frac{w^2x_0}{\tan\theta}

\frac{\Delta g}{g} = \sqrt{4\left(\frac{\Delta w}{w}\right)^2 + \left(\frac{\Delta x_0}{x_0}\right)^2 + \left(\frac{\Delta(\tan\theta)}{\tan\theta}\right)^2} \qquad \frac{\Delta w}{w} = \frac{\Delta T}{T}

\frac{\Delta(\tan\theta)}{\tan\theta} \approx \frac{\Delta\theta}{\theta}

(since from the table $\tan\theta \cong \theta$ )

\theta \approx \frac{x}{H-h_0}, \qquad \frac{\Delta\theta}{\theta} = \sqrt{\left(\frac{\Delta x}{x}\right)^2 + \left(\frac{\Delta H+\Delta h_0}{H-h_0}\right)^2}

\frac{\Delta g}{g} = \sqrt{4\left(\frac{\Delta T}{T}\right)^2 + \left(\frac{\Delta x_0}{x_0}\right)^2 + \left(\frac{\Delta x}{x}\right)^2 + \left(\frac{\Delta H+\Delta h_0}{H-h_0}\right)^2}

Using the values H=160 mm, $\Delta H$ =1mm, $h_0$ =30 mm, $\Delta h_0$ =1mm, $x_{av}$ =61.4 mm, $\Delta x_{av}$ =1mm, $T_{av}$ =1.1s, $\Delta T$ =0.01 s, $x_0$ =51 mm, $\Delta x_0$ =1mm one obtains

g = 980±34 cm/s $^2$

Note that from the method of least squares one obtains the following results:

g = 982 cm/s $^2$ with a standard deviation of $\sigma$ = 33 cm/s $^2$
From the linear regression of the data slope $\tan\theta$ vs $w^2$ is found to be 0.052 with a standard error of 5.14x10 $^{-5}$ , therefore:

\frac{\Delta g}{g} = \sqrt{\left(\frac{\Delta(\mathit{slope})}{\mathit{slope}}\right)^2 + \left(\frac{\Delta x_0}{x_0}\right)^2} = 0.02

g = 980±20 cm/s $^2$

Grading Scheme for Experimental Competition

Part 1	7.5 pts
Derivation of Equation 1	1.0 pts
Calculation of $\omega$ using period measurements	1.0 pts
At low speeds 10T is OK
At high speeds 20T is expected	-0.2 pts
Missing units	-0.2 pts
Calculation of tan2 $\theta$ , tan $\theta$ at each $\omega$	1.0 pts
Calculation of tan2 $\theta$	0.5 pts
Calculation of tan $\theta$	0.5 pts
Plot of tan $\theta$ vs $\omega^2$	1.5 pts
Axes with labels and units	0.4 pts
Drawing best fit line	0.5 pts
At least 6 different data in a wide range of $\omega$	0.6 pts
No. of measurements 5:	-0.2 pts
No of measurements 4:	-0.4 pts
No of measurements 3 or less:	-0.6 pts
Calculations	2.0 pts
calculation of slope with unit	1.0 pts
calculation of g	1.0 pts
FULL credit for
$9.3<g<10.3$ m/s $^2$ (±5% error)
For g values credits to be subtracted from the total credit of 7.5:
$10.3<g<10.5$ m/s $^2$ , $9.1<g<9.3$ m/s $^2$	-0.5 pts
$8.8<g<9.1$ m/s $^2$ , $10.3<g<10.8$ m/s $^2$	-1.0 pts
outside the above ranges	-1.5 pts
Error Calculation	1.0 pts

Part 2a

H(mm)	10T(s)	w(rad/s)	lnw	$H-h_0$ (mm)	ln( $H-h_0$ )
158	10.31	6.09	0.784921	128	2.107
209	13.19	4.76	0.677935	179	2.253
190	11.70	5.37	0.729994	160	2.204
150	9.80	6.41	0.806954	120	2.079
129	9.21	6.82	0.83392	99	1.996
119	8.75	7.18	0.856172	89	1.949
110	8.10	7.76	0.889695	80	1.903

Graph titled 'ln(H-h0) vs lnw': ln(H-h0) (0,8 to 1,4) against lnw (0,6 to 1) with seven data points on a falling straight line; fit y = -1,7446x + 2,4608, R² = 0,9771. — Figure 3.

Thus the focal length depends on w as

f = Aw^{n},

and

n ~ -1.7.

The plot of $H-h_0$ vs. $1/w^2$ is also acceptable as a correct plot.

Part 2a	5.5 pts
Measurements of H vs $\omega$	0.6 pts
Calculation of $\omega$ using period measurements	0.4 pts
At low speeds 10T is OK
At high speeds 20T is expected	-0.2 pts
H- $\omega$ table	0.2 pts
Plot of F vs $\omega$	2.4 pts
Calculation of F=H- $h_0$	0.5 pts
Plot with axis labels	0.8 pts
Drawing best fit line	0.5 pts
At least 6 different data in a wide range of $\omega$	0.6 pts
No. of measurements 5:	-0.2 pts
No of measurements 4:	-0.4 pts
No of measurements 3 or less:	-0.6 pts
Calculations	2.5 pts
Calculation of slope with unit	1.0 pts
Dependence F $\alpha$ 1/ $\omega^2$	1.5 pts

Part 2b

$\omega$ Range(rad/s)	Orientation	Variation of the size	Image
$\omega$ =0	ER		V
$0<\omega<8.2^{}$ $0<\omega<6.3^{*}$	ER	D	V
$8.2<\omega<14.6^{}$ $6.3<\omega<14.0^{*}$	INV	I	R
$14.6<\omega<\omega_{max}^{}$ $14.0<\omega<\omega_{max}^{*}$	ER	NC	V

* for H=110 mm

** for H=240 mm

$\omega$ values depend on the initial values of H, $h_0$ , etc.

Note that measurements only at one H value are required from the students.

Part 2b	3.5 pts
Every correct item in the table	0.25 pts

Part 3

Measurement of wavelength

Both the grating and the screen are in air. Normal incidence.

Screen to grating distance	: L
Distance between the diffraction spots seen on the screen	: x
Order of diffraction	: m

L= 225 mm, $x_{av}$ =77 mm for m=±1 d=1/500 mm

\tan\alpha = \frac{x_{av}}{L} = \frac{77}{225}

\lambda = \frac{1}{500}\sin\alpha = 647\ \mathrm{nm}

L= 128 mm, $x_{av}$ =44 mm for m=±1, d=1/500 mm

\tan\alpha = \frac{44}{128}

\lambda = \frac{1}{500}\sin\alpha = 650\ \mathrm{nm}

L= 128 mm, $x_{av}$ =111 mm for m=±2, d=1/500 mm

\tan\alpha = \frac{111}{128}

\lambda = \frac{1}{2x500}\sin\alpha = 655\ \mathrm{nm}

The average value of $\lambda$ is $\lambda_{av}$ =651 nm.

Measurement of refractive index

2R=145 mm

Distance between the spots measured on the curved screen = R $\alpha$

R $\alpha_{av}$ = 17 mm for m=±1 $\alpha_{av}$ = 0.234 rad

using $n = \frac{m\lambda}{d\sin(\alpha)}$ , one obtains n=1.40

If the curvature of the screen is neglected:

\tan\alpha = \frac{17}{72.5}

\alpha = 13.20^{\circ}

n = \frac{\lambda}{d\sin(\alpha)} = \frac{651(nm)}{\frac{1}{500}(mm)x10^6\sin(\alpha)} = 1.43

Part 3	3.5 pts
(At least 3 measurements at different orders are required)
Wavelength measurement	1.2 pts
Distance measurements and calculation of angle	0.6 pts
Calculation of $\lambda$	0.6 pts
Credits to be subtracted from the total credit of 1.2:
If $\lambda$ is outside the range 600-700 nm	-0.4 pts
If less then 3 measurements	-0.4 pts
Measurement of n	2.3 pts
Distance measurements and calculation of angle	0.6 pts
Realizing $\lambda$ /n	0.8 pts
Calculation of n	0.9 pts
credits to be subtracted from the total credit of 2.3:
If n is outside range 1.3-1.6	-0.4 pts
If less then 3 measurements	-0.4 pts

Оригинал в Архива: exp.pdf · официални решения: exp_sol.pdf · допълнителни данни: IPhO2001-Exp Question(pdf).pdf · допълнителни данни: IPhO2001-Exp Solution(pdf).pdf