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IZhO 2022, Експеримент — Задача 1. Dry friction — Experiment 1: Sliding friction

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XVIII International Zhautykov Olimpiad/Experimental Competition · 17 февруари 2022 г.

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The duration of the experimental competition is 4 hours. There are two problems.
You can use your own calculator for numerical calculations.
You are provided with Writing sheet and additional white sheets of paper. You can use the additional sheets of paper for drafts of your solutions, but these sheets will not be graded. Your final solutions should be written on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures, and plots.
Use only the front side of Writing sheets. Write only inside the boxed areas.
Start putting down your solution to each problem on a new Writing sheet.
Graphs must be drawn on Writing sheets with the graph paper. All drawings must be done with a pen, not a pencil!
Fill in the boxes at the top of each Writing sheet with your country (Country), your student code (Student Code), problem number (Question Number), the progressive number of each Writing sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be graded, put a large X across the entire sheet and do not include it in your numbering.

Условие

To measure the coefficient of sliding friction, the rolling of a solid homogeneous cylinder on an inclined plane is studied at various angles of inclination of the plane to the horizon. We assume that the force of sliding friction is described by the well-known Coulomb-Amonton law:

F = \mu_s N\,,\qquad (1)

where $N$ stands for the normal reaction force and $\mu_s$ denotes the coefficient of sliding friction.

Consider the experimental setup to be a wide flat steel plate, whose angle of inclination to the horizon can be arbitrarily varied. The error in setting the angle of inclination of the plane to the horizon is $\Delta\alpha = 0.2^{\circ}$ .

A solid cylinder on an inclined steel plate with three optical sensors labelled 0, 1, 2 mounted on stands along the plate; distances S1 and S2 measured from sensor 0 along the plate; the angle of inclination α is marked at the lower end of the plate.

The solid homogeneous steel cylinder can roll down the plate with its axis remaining horizontal all the time. To measure the acceleration, three optical sensors (0, 1, 2) are fixed on the plate. Each sensor consists of a light source and a photodetector, which are mounted on the identical stands. The solid homogeneous cylinder is placed on the plate and released. The cylinder moves down between these stands and blocks the light beam so that at the moment of light interruption, an electrical impulse is generated that controls an electronic stopwatch (not shown in the figure). When the cylinder passes sensor 0, a stopwatch is started, when the cylinder passes sensors 1 and 2, the times of these passages are recorded.

Thus, the following time intervals are measured in the experiment:

$t_1$ – cylinder motion time interval from sensor 0 to sensor 1;

$t_2$ – cylinder motion time interval from sensor 0 to sensor 2.

These time intervals are recorded with 4 significant digits. The instrumental error in measuring time intervals $t_1$ and $t_2$ is equal $\Delta t = 2\cdot10^{-4}s$ .

The distances between the sensors are measured as:

from sensor 0 to sensor 1 $S_1 = (50.0\pm0.2)\,sm$ ;

from sensor 0 to sensor 2 $S_2 = (100.0\pm0.2)\,sm$ .

When making numerical calculations, assume that the free fall acceleration is equal $g = 9.81\,\mathrm{m/s^2}$ .

The results of measuring time intervals $t_1$ and $t_2$ are given in Table 1 for different angles of inclination of the plate to the horizon.

Table 1. Time intervals of the cylinder motion.

$\alpha^{\circ}$	$t_1$ , s	$t_2$ , s
20	0,4546	0,7187
25	0,3936	0,6290
30	0,3462	0,5589
35	0,3229	0,5211
40	0,3358	0,5283
45	0,3084	0,4911
50	0,2682	0,4347
55	0,2816	0,4432
60	0,2600	0,4113
65	0,2461	0,3908
70	0,2308	0,3675
75	0,2218	0,3542

1.1 Derive formulas for the acceleration of the cylinder axis in two cases: A) the motion of the cylinder occurs without slipping – acceleration $a_1$ ; B) when moving along the plate, the cylinder slips – acceleration $a_2$ . Express your answers in terms of $\alpha, g, \mu_s$ .

1.2 Express the maximum angle $\alpha = \alpha_{cr}$ of inclination of the plate, at which the motion of the cylinder still occurs without slipping, in terms of the coefficient of friction $\mu_s$ .

1.3 Using the measurement results given in Table 1, calculate the acceleration with which the cylinder axis moved for each angle of inclination of the plate. Put down the formula for the acceleration, according to which the calculations are carried out. Draw a graph of the acceleration versus the angle and with its aid find an approximate value of the critical $\alpha_{cr}$ .

1.4 Carry out the linearization of the obtained dependence, i.e. find such values $X(\alpha, a)$ and $Y(\alpha, a)$ so that the dependence $Y(X)$ turns linear for the both cases when the cylinder moves without slipping and with slipping. Draw a linearized dependence $Y(X)$ for all experimental points.

1.5 Using the linearized relationship $Y(X)$ , calculate the coefficient of friction $\mu_s$ between the cylinder and the plate. Estimate the error $\Delta\mu_s$ of the obtained value. Put down the formulas used in your calculations.

1.6 Using the linearized relationship $Y(X)$ , calculate the value of the critical angle $\alpha_{cr}$ , at which the cylinder starts to slip. Estimate the error of the obtained value. Put down the formulas used in your calculations.

Решение

SOLUTION
Dry friction
Experiment 1: Sliding friction
Theoretical part

1.1 The formulas for the accelerations of the cylinder axis can be obtained in various ways, but they are all based on the use of Newton's second law. When a cylinder rolls down an inclined plate, it is subject to the following forces: $m\vec{g}$ – gravity due to the Earth, $\vec{N}$ – normal reaction force, $\vec{F}$ – friction force.

A sphere (cylinder) on an inclined plane; arrows labelled N (normal reaction, pointing up-left from the cylinder axis), F (friction, along the plane to the upper left), a (acceleration, down the slope), mg (gravity, vertically down); the angle α is marked at the lower right of the plate.

A) In this case, the friction force does not perform any work, so we can write down the equations of the law for the conservation of mechanical energy as follows

\frac{3}{4}mR^{2}V_{_c}^{2}=mgS\sin\alpha\ , \qquad (1)

which takes into account the energy of rotation of the cylinder around its own axis.
To determine the acceleration, we calculate the time derivative of this equation to obtain

\frac{3}{4}mR^{2}\cdot 2V_{_C}a=mg\sin\alpha\cdot V_{_C}. \qquad (2)

Here $V_{_C}=\dfrac{dS}{dt}$ stands for the speed of the cylinder axis, and $a=\dfrac{dV_{_C}}{dt}$ refers to its acceleration.
It follows from equation (2) that when rolling without slipping, the acceleration of the cylinder axis is described by the formula

a_{_1}=\frac{2}{3}g\sin\alpha\,. \qquad (3)

An alternative way to derive this formula is to use the equation for the rotational motion of the cylinder.
B) If the cylinder slips during its downward motion, then the friction force is determined by the formula

F=\mu_{_s}N=\mu_{_s}mg\cos\alpha\,, \qquad (4)

then the equation of Newton's second law in projection onto an inclined plane has the form

ma=mg\sin\alpha-\mu_{_s}mg\cos\alpha\,, \qquad (5)

It follows from this equation that the acceleration of the cylinder in this case is equal to

a_{_2}=g\left(\sin\alpha-\mu_{_s}\cos\alpha\right), \qquad (6)

1.2 It is obvious that the motion without slipping occur at angles of inclination of the plate that are less than some critical value $\alpha_{cr}$ , whose magnitude can be found in various ways. For example, it can be found by equating the accelerations described by formulas (3) and (6).
Here is another method for obtaining the critical angle. From equation (2) of Newton's law, we obtain

ma=mg\sin\alpha-F\,.

Let us express the value of the static friction force, taking into account the found acceleration, as

F=mg\sin\alpha-ma=\frac{1}{3}mg\sin\alpha

and assume that this force does not exceed the force of sliding friction

F<\mu_{_s}N=\mu_{_s}mg\cos\alpha\,.

It is derived from the last inequality that

\frac{1}{3}mg\sin\alpha<\mu_{_s}mg\cos\alpha\ \ \Rightarrow\ \ tg\,\alpha<3\mu_{_s}. \qquad (7)

Thus, the value of the critical angle is determined by the formula

\alpha_{cr}=arctg\left(3\mu_{_s}\right). \qquad (8)

Processing of measurement data

1.3 To calculate accelerations according to the measurement results given in the problem formulation, we write expressions for the distances traveled by the balls

\begin{cases} S_{_1}=V_{_0}t_{_1}+\dfrac{at_{_1}^{2}}{2} \\ S_{_2}=V_{_0}t_{_2}+\dfrac{at_{_2}^{2}}{2} \end{cases}\ \Rightarrow\ \ \frac{S_{_2}}{t_{_2}}-\frac{S_{_1}}{t_{_1}}=\frac{a}{2}\left(t_{_2}-t_{_1}\right). \qquad (9)

The formula for calculating accelerations is then found as:

a=\frac{2\left(\dfrac{S_{_2}}{t_{_2}}-\dfrac{S_{_1}}{t_{_1}}\right)}{t_{_2}-t_{_1}}. \qquad (10)

The results of acceleration calculations according to formula (10) are shown in Table 1.

Table 1. Calculation of accelerations.

α°	t₁, s	t₂, s	a, m/s²
20	0.4546	0.7187	2.208
25	0.3936	0.6290	2.715
30	0.3462	0.5589	3.244
35	0.3229	0.5211	3.739
40	0.3358	0.5283	4.196
45	0.3084	0.4911	4.543
50	0.2682	0.4347	5.239
55	0.2816	0.4432	5.950
60	0.2600	0.4113	6.718
65	0.2461	0.3908	7.286
70	0.2308	0.3675	8.116
75	0.2218	0.3542	8.595

Based on these data, Graph 1 of the dependence of acceleration on the angle of inclination of the plate is drawn.

Scatter plot with fitted two-branch curve; x-axis: alfa (degrees) 0–90, y-axis: acceleration, m/s2 0–10. Data points from 20° to 75°; the lower branch (green points, rolling without slipping) bends over to a steeper straight branch (red points, with slipping) above about 45–50 degrees; the kink is near alfa ≈ 46°. — Graph 1. Acceleration dependence on the inclination angle

It is clearly seen in the graph that it consists of two different branches: at small angles, the motion occurs without slipping (acceleration is described by formula (3)), whereas at large angles, the cylinder slips, so the acceleration is described by formula (6). The abscissa of the intersection point of these graphs is the critical angle, which is seen to be approximately equal to

\alpha_{cr}\approx 46^{\circ}\,.

Note: In principle, it is not needed to specify the free fall acceleration, which can be determined from the behavior of the acceleration at small angles. However, this is not required in this problem.
The linearization of the obtained dependence can be carried out in two alternative ways, which are approximately equivalent. Let us consider these methods in detail.

Solution 1

1.4 As a variable $Y$ we take the combination

Y=\frac{a}{g\cos\alpha}, \qquad (11)

then it follows from the formulas for accelerations:

\begin{cases} a_{_1}=\dfrac{2}{3}g\sin\alpha \\ a_{_2}=g\left(\sin\alpha-\mu_{_s}\cos\alpha\right) \end{cases}\ \Rightarrow\ \ \begin{cases} \dfrac{a_{_1}}{g\cos\alpha}=\dfrac{2}{3}tg\,\alpha \\ \dfrac{a_{_2}}{g\cos\alpha}=tg\,\alpha-\mu_{_s} \end{cases}\ ,

i.e. the new variable $Y$ is a linear function of $X=tg\,\alpha$ , such that

\begin{cases} Y=\dfrac{2}{3}X, & \alpha<\alpha_{cr} \\ Y=X-\mu_{_s}, & \alpha>\alpha_{cr} \end{cases}\,. \qquad (12)

Calculations of the chosen variables are provided in Table 2. In the table the highlighted values correspond to the motion without slipping.

Table 2. Linearization 1.

α°	a, m/s²	X = tg α	Y = a/(g cos α)
20	2.208	0.3640	0.2395
25	2.715	0.4663	0.3053
30	3.244	0.5774	0.3818
35	3.739	0.7002	0.4653
40	4.196	0.8391	0.5584
45	4.543	1.0000	0.6549
50	5.239	1.1918	0.8308
55	5.950	1.4281	1.0574
60	6.718	1.7321	1.3697
65	7.286	2.1445	1.7575
70	8.116	2.7475	2.4188
75	8.595	3.7321	3.3851

Graph 2 shows the linearized dependencies.

Plot of a/(g cos(alfa)) versus tg(alfa) 0–4; two straight lines through the data: a line of slope ≈ 2/3 through the green (no-slip) points and a unit-slope line with negative intercept −μs through the red (slipping) points; the lines intersect near tg(alfa) ≈ 1.4. — Graph 2. Linearization 1.

1.5 Indeed, both dependences turn out to be linear. We represent these dependencies in the form

Y=cX+b\,. \qquad (13)

The calculation of the coefficients of these dependencies by the least squares gives the following results:
At the motion without slipping: $c_{_1}=0.659\pm0.008$ , $b_{_1}=0.0004\pm0.005$ , with slipping: $c_{_2}=1.011\pm0.009$ , $b_{_2}=-0.38\pm0.02$ .
It follows from formula (12) that the coefficient of sliding friction is equal to

\mu_{_s}=-b_{_2}=0.38\pm0.02\,. \qquad (14)

Note that the calculated random error (of the order of 10%) significantly exceeds the relative errors of direct measurements of both distances between sensors and motion times. Therefore, the latter can be ignored.

1.6 The value of the critical angle can be found by equating the two linear relations above:

\frac{2}{3}X=X-\mu_{_s}\ \ \Rightarrow\ \ X_{cr}=tg\,\alpha_{cr}=3\mu_{_s}\,, \qquad (15)

\alpha_{cr}=arctg\left(3\mu_{_s}\right)=0.855=49^{\circ}

The error of this value is equal to

\Delta a_{cr}=\frac{3\Delta\mu_{_S}}{\left(3\mu_{_S}\right)^{2}+1}=0.03\,,

and we finally get

\alpha_{cr}=0.86\pm0.03=49^{\circ}\pm2^{\circ}\,. \qquad (16)

Solution 2

1.4 As a variable $Y$ we take the combination

Y=\frac{a}{g\sin\alpha}, \qquad (17)

then it follows from the formulas for accelerations:

\begin{cases} a_{_1}=\dfrac{2}{3}g\sin\alpha \\ a_{_2}=g\left(\sin\alpha-\mu_{_s}\cos\alpha\right) \end{cases}\ \Rightarrow\ \ \begin{cases} \dfrac{a_{_1}}{g\sin\alpha}=\dfrac{2}{3} \\ \dfrac{a_{_2}}{g\sin\alpha}=1-\mu_{_s}ctg\,\alpha \end{cases}\ ,

i.e. the new variable $Y$ is a linear function of $X=ctg\,\alpha$ , such that

\begin{cases} Y=\dfrac{2}{3}, & \alpha<\alpha_{cr} \\ Y=1-\mu_{_s}X, & \alpha>\alpha_{cr} \end{cases}\,. \qquad (18)

Calculations of the chosen variables are provided in Table 3. In the table the highlighted values correspond to the motion without slipping.

Table 3. Linearization 2.

α°	a, m/s²	X = ctg α	Y = a/(g sin α)
20	2.208	2.7475	0.6580
25	2.715	2.1445	0.6548
30	3.244	1.7321	0.6613
35	3.739	1.4281	0.6645
40	4.196	1.1918	0.6655
45	4.543	1.0000	0.6549
50	5.239	0.8391	0.6972
55	5.950	0.7002	0.7404
60	6.718	0.5774	0.7908
65	7.286	0.4663	0.8195
70	8.116	0.3640	0.8804
75	8.595	0.2679	0.9070

Graph 3 shows the linearized dependencies.

Plot of a/(g sin(alfa)) versus ctg(alfa) 0–3; a horizontal line at ≈ 2/3 fits the green (no-slip) points, while a straight line of slope −μs ≈ −0.38 and intercept ≈ 1 fits the red (slipping) points, intersecting the horizontal line near ctg(alfa) ≈ 1.0. — Graph 3. Linearization 2.

1.5 Indeed, both dependences turn out to be linear. We represent these dependencies in the form

Y=cX+b\,. \qquad (19)

The calculation of the coefficients of these dependencies by the least squares gives the following results:
At the motion without slipping: $c_{_1}=-0.002\pm0.003$ , $b_{_1}=0.663\pm0.006$ , with slipping: $c_{_2}=-0.38\pm0.02$ , $b_{_2}=1.01\pm0.01$ .
It follows from formula (18) that the coefficient of sliding friction is equal to

\mu_{_s}=-c_{_2}=0.38\pm0.02\,. \qquad (20)

In this case, the errors of direct measurements can also be ignored.

1.6 The calculation of the critical angle is carried out similarly and we obtain

\alpha_{cr}=arctg\left(3\mu_{_s}\right)=0.84=48^{\circ}

The error of this value is equal to

\Delta a_{cr}=\frac{3\Delta\mu_{_S}}{\left(3\mu_{_S}\right)^{2}+1}=0.02\,,

and we finally get

\alpha_{cr}=0{,}84\pm0{,}01=48^{\circ}\pm1^{\circ}\,. \qquad (21)

Part	Content	Points	Total
	Experiment 1: Sliding friction		10.0
	Theoretical part		1.5
1.1	Motion without slipping:		1.0
	equation of motion;	0.2
	formula (3);	0.3
	Motion with slipping:
	equation of motion;	0.2
	formula (6);	0.3
1.2	Critical angle value:		0.5
	the condition for the start of slippinge: the boundary value of the static friction force; equality of accelerations (3) and (6);	0.2
	relation between the critical angle and the friction coefficient: formula (7) or (8);	0.3
	Processing of measurement data		8.5
1.3	Calculation of accelerations:		2.5
	formula (10) for acceleration;	0.2
	all accelerations are calculated (graded if the formula for calculating acceleration is graded); 0.1 for each point, acceptable error of acceleration calculation is ±0.01;	1.2
	Plotting a graph (graded if the results of acceleration calculations are graded):
	the axes are named and ticked;	0.1
	all points are drawn;	0.1
	two smoothing curved lines are drawn;	0.2
	two dependences are obtained: two different intersecting curves are visible in the graph;	0.3
	Critical angle value;
	determination method: the point of intersection of two curves;	0.2
	numerical value: acceptable error ±1°;	0.2
1.4	Linearization of the dependence:		2.8
	New variables: any reasonable: $\left(\dfrac{a}{\sin\alpha},\ ctg\alpha\right)$ , $\left(\dfrac{a}{\cos\alpha},\ tg\alpha\right)$ , and equivalent, leading to linear dependencies;	1.0
	Numerical calculations: graded if the choice of new variables $X,Y$ is graded; 0.1 for each correctly calculated point, acceptable error ±0.02;	1.2
	Plotting a graph (graded if the results of the calculations of the variables $X,Y$ are graded):
	the axes are named and ticked;	0.1
	all points are drawn;	0.1
	two straight lines are visible;	0.2
	two straight lines are drawn;	0.2
1.5	Calculation of the coefficient of sliding friction:		1.9
	coefficients of the linear dependences are calculated: 0.2 for each coefficient (graphically for all points, LSM); 0.1 – by two points; the use of theoretical values is not graded;	0.8
	formula for the coefficient of friction through the coefficients of linear dependencies;	0.2
	numerical value of the coefficient of sliding friction: graded if the formula for the coefficient of friction is graded, the acceptable error ±0.05;	0.4
	formula for the error: only through the coefficients of the linear dependencies;	0.2
	numerical value of the error: graded if the formula is graded;	0.2
	correct rounding;	0.1
1.6	Critical angle value:		1.3
	calculation method: analytical calculation of the intersection point of two lines, the use of numerical theoretical values of the coefficients is acceptable;	0.5
	numerical value of the critical angle: graded if the method of determination is graded;	0.2
	formula for the error: through the errors of the coefficients of the straight lines;	0.3
	numerical value of the error: graded if the formula is graded;	0.2
	correct rounding;	0.1

Оригинал в Архива: IZhO-2022-Exp_eng.pdf · официални решения: IZhO-2022-Exp_eng_sol.pdf