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IZhO 2022, Експеримент — Задача 2. Experiment 2: Rolling 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.
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Условие

In reality, even in the absence of slipping between the bodies, there are frictional forces called rolling friction forces. In this experiment, the following setup is used to study the rolling friction. A small rod is rigidly attached to the side surface of a massive solid cylinder, whose imaginary prolongation crosses the axis of the cylinder. The cylinder is located on the two horizontal plates so that it can roll over them without slipping. In this case, the rod lies always in the gap between the plates.

Let us use the following notation: the radius of the cylinder $R$ , the mass of the cylinder $M$ , the mass of the attached rod $m$ , which can be considered significantly less than the mass of the cylinder, the distance from the center of mass of the rod to the axis of the cylinder $l$ .

A solid cylinder lying on two horizontal parallel plates with a small rod rigidly attached to its side surface; a double arrow above indicates the horizontal rolling direction.

Usually, the formula for the rolling friction force is written as

F = \frac{\kappa}{R}N\,,

where $N$ is the normal reaction force, $R$ stands for the radius of the rolling body, and $\kappa$ refers to the rolling friction coefficient. In this part of the problem, it is necessary to determine the dimensionless value $\mu_r = \dfrac{\kappa}{R}$ , which for brevity is called the coefficient of rolling friction. Typically, the coefficient of rolling friction is much less in magnitude than the coefficient of sliding friction studied above.

Diagram of damped rolling oscillations: successive stoppage points x0, x2, x3, x1 shown as dashed circles rolling to the left towards the equilibrium position x = 0 where the rod is vertical downwards; the solid cylinder at initial position x0 has the rod up; an arrow indicates the motion direction.

During the experiment, the cylinder is placed into its initial position in such a way that the rod is directed vertically upwards, after which the cylinder is released without any significant push. The cylinder starts to roll on the plates, performing damped oscillations due to the action of the rolling friction force. In this case, the coordinates of successive extreme positions of the cylinder (stoppage points) are written down: $x_0$ – initial position, $x_1, x_2, x_3\dots$ – coordinates of successive stoppage points. The origin of coordinates $x=0$ corresponds to the point where the rod is directed vertically downwards. Table 2 shows the experimental values of the coordinates of the stoppage points, whose measurement error is found as $\Delta x = 0{,}2\,sm$ .

Table 2. Coordinates of the stoppage points.

$k$	$x_k$ , sm
0	15,8
1	-11,6
2	10,1
3	-9,0
4	7,7
5	-7,0
6	5,7
7	-5,3
8	4,7
9	-3,9
10	2,8

To determine the setup parameters, the period of small oscillations of the cylinder near the equilibrium position (point $x=0$ ) is measured. For this, the times of five oscillations of the cylinder with the rod are measured several times. The results of these measurements are shown in Table 3. The instrumental error of the time measurement is $\Delta t = 0{,}02s$

Table 3. The times of five oscillations

$k$	$t_5$ , s
1	7,39
2	7,21
3	7,26
4	7,47
5	7,44

Front view of the cylinder on the plates near equilibrium with the rod vertical downwards and dashed outlines showing small oscillations to either side.

2.1 Derive the formula for the period of the small fluctuations described right above. Express the oscillation period in terms of the setup parameters $M, m, R, l$ and the free fall acceleration $g$ .

2.2 Derive an equation relating the coordinates of two successive cylinder stoppage points $x_k, x_{k-1}$ in the described cylinder rolling experiment. This equation, in addition to the coordinates, may include the setup parameters $M, m, R, l$ , the free fall acceleration $g$ and the coefficient of rolling friction $\mu_r$ .

2.3 Express the coordinate of the $k$ 'th cylinder stoppage point in terms of the initial coordinate $x_0$ and the coordinates of all previous stoppage points $x_1, x_2, \dots x_{k-1}$ . This equation, in addition to the coordinates of the stoppage points, should include only the free fall acceleration g and the period of small oscillations $T$ .

2.4 Using the measurement results given in Table 3, calculate the period $T$ of small oscillations of the cylinder. Estimate the measurement error $\Delta T$ of this quantity.

2.5 Propose such values $Y(x_k)$ and $X(x_0, x_1, \dots x_k)$ such that the dependence $Y(X)$ turns linear and allows you to calculate the coefficient of rolling friction $\mu_r$ . Draw a graph of the linearized relationship $Y(X)$ .

2.6 Using the linearized relationship $Y(X)$ , calculate the coefficient of rolling friction $\mu_r$ . Estimate the error $\Delta\mu_r$ of the obtained value.

Решение

Experiment 2: Rolling friction
Theoretical part

2.1 When the axis of the cylinder is displaced by the distance $x$ , the axis of the rod deviates by the angle

\varphi = \frac{x}{R}\,. \qquad (22)

To calculate the period of small oscillations, we write the equation for the law of conservation of energy

\frac{3}{4}MV^2 + mgl(1-\cos\varphi) = mgl(1-\cos\varphi_0)\,. \qquad (23)

Hereinafter, we neglect the kinetic energy of the rod motion, since its mass is small.

The cylinder on the plate with the attached rod tilted; the horizontal displacement x of the cylinder axis from the vertical dash-dotted line is marked with an arrow, and the rotation angle φ of the rod from the vertical is marked with an arc.

On the other hand, the potential energy of the massive cylinder remains constant, so the change in the potential energy of the system is the change in the potential energy of the rod. For small oscillations, the approximate formula for the cosine of a small angle $\cos\varphi \approx 1 - \dfrac{\varphi^2}{2}$ should be used, which leads to a simplification of equation (23) to:

\frac{3}{4}MV^2 + mgl\frac{\varphi^2}{2} = mgl\frac{\varphi_0^2}{2}\,.

Using relation (22), we obtain the equation

\frac{3}{4}MV^2 + \frac{mgl}{R^2}\frac{x^2}{2} = \frac{mgl}{R^2}\frac{x_0^2}{2}\,, \qquad (24)

which is the equation of harmonic oscillations with the period of

T = 2\pi\sqrt{\frac{3MR^2}{2mgl}}\,. \qquad (25)

2.2 At the stoppage points, the kinetic energy of the cylinder with the rod is zero, so the change in the potential energy when moving from one stoppage point to the next is equal to the work of the friction force. Therefore, the following relation is valid:

mgl(1-\cos\varphi_k) = mgl(1-\cos\varphi_{k-1}) - \mu_r Mg\left|x_k - x_{k-1}\right|. \qquad (26)

Applying relation (22) between the angle of rotation and the displacement of the cylinder, we obtain the required relation

\left(1-\cos\frac{x_k}{R}\right) = \left(1-\cos\frac{x_{k-1}}{R}\right) - \mu_r\frac{M}{ml}\left|x_k - x_{k-1}\right|. \qquad (27)

2.3 Let us sum up relations, similar to (27), for all intervals of motion from the initial position to the $k$ 'th stoppage point, which finally yields

\left(1-\cos\frac{x_k}{R}\right) = \left(1-\cos\frac{x_0}{R}\right) - \mu_r\frac{M}{ml}S_k\,, \qquad (28)

where $S_k$ is the path the cylinder passes to the $k$ 'th stoppage point

S_k = \left|x_0 - x_1\right| + \left|x_2 - x_1\right| + \ldots + \left|x_k - x_{k-1}\right| = \sum_{j=1}^{k}\left|x_j - x_{j-1}\right|. \qquad (29)

In the initial position, the rod is directed vertically, i.e. $\varphi_0 = \pi$ , therefore $x_0 = \pi R$ , and one can write down

R = \frac{x_0}{\pi}\,. \qquad (30)

From the formula for the oscillation period, one can also express:

\frac{M}{ml} = \frac{T^2}{4\pi^2}\frac{2g}{3R^2} = \frac{T^2 g}{6\pi^2 R^2} = \frac{T^2 g}{6x_0^2}\,.

We substitute these values into equation (28), which yields

1-\cos\left(\pi\frac{x_k}{x_0}\right) = 2 - \mu_r\frac{T^2 g}{6x_0^2}S_k\,. \qquad (31)

Thus, the value $Y_k = 1-\cos\left(\pi\dfrac{x_k}{x_0}\right)$ linearly depends on the path $S_k$ . The slope of dependence coefficient contains the rolling friction coefficient sought as well as other known values. The unity in expression (31), of course, can be omitted. But the value $Y$ , up to a constant factor, is equal to the potential energy, therefore, in the accepted definition, the dependence $Y(S)$ is more preferable.

Processing of measurement data
2.4 The oscillation period is calculated in the traditional way.
We calculate the average value of the time of 5 oscillations: $\left\langle t_5 \right\rangle = 7.354\ s$ ;
We also calculate the random error of this value: $\Delta t_5 = 2\sqrt{\dfrac{\sum_i\left(t_i - \left\langle t \right\rangle\right)^2}{n(n-1)}} = 0.11\ s$ .
Then the oscillation period is obtained as

T = (1.47 \pm 0.02)\ s\,. \qquad (32)

2.5 The results of calculations of the quantities, appearing in formula (31), are shown in Table 4.

Table 4. Linearization.

$k$	$x_k$ , cm	$\Delta S_k = \left\vert x_k - x_{k-1}\right\vert$ , m	$S_k$ , m	$Y_k = 1-\cos\left(\pi\dfrac{x_k}{x_0}\right)$
0	15.8		0	2.000
1	-11.6	0.274	0.274	1.671
2	10.1	0.217	0.491	1.424
3	-9.0	0.191	0.682	1.217
4	7.7	0.167	0.849	0.960
5	-7.0	0.147	0.996	0.822
6	5.7	0.127	1.123	0.576
7	-5.3	0.110	1.233	0.506
8	4.7	0.100	1.333	0.406
9	-3.9	0.086	1.419	0.286
10	2.8	0.067	1.486	0.151

The dependence $Y(S)$ is shown in the following figure.

Plot of Energy, Y versus S (path), m; vertical axis 0–2.5, horizontal axis 0–1.6; eleven points descending linearly from (0, 2.0) to about (1.49, 0.15) with a fitted straight line. — Graph 4. Energy as a function of the path

2.6 The resulting dependence is linear, which confirms the theoretical model used. The slope coefficient of this graph, calculated by the least squares, is equal to

C = (1.23 \pm 0.02)\ m^{-1}\,.

The theoretical formula for this coefficient makes it possible to express the value of the rolling friction coefficient:

C = \mu_r\frac{T^2 g}{6x_0^2} \quad \Rightarrow \quad \mu_r = C\frac{6x_0^2}{T^2 g} = 8.69\cdot 10^{-3}\,. \qquad (33)

To calculate the error of this value, we make use of the formula for the error of indirect measurements:

\Delta\mu_r = \mu_r\sqrt{\left(\frac{\Delta C}{\left\langle C \right\rangle}\right)^2 + \left(2\frac{\Delta T}{\left\langle T \right\rangle}\right)^2 + \left(2\frac{\Delta x_0}{\left\langle x_0 \right\rangle}\right)^2} = \qquad (34)

= 8.69\cdot 10^{-3}\sqrt{\left(\frac{0.02}{1.23}\right)^2 + \left(2\frac{0.02}{1.47}\right)^2 + \left(2\frac{0.2}{15.8}\right)^2} = 3.8\cdot 10^{-4}

The final result is written as

\mu_r = \left(8.7 \pm 0.4\right)\cdot 10^{-3}\,. \qquad (35)

Part	Content	Points	Total
	Experiment 2: Rolling friction		10.0
	Theoretical part		2.5
2.1	Derivation of the formula for the period of small oscillations:		0.5
	initial equation: either dynamic or energetic;	0.2
	approximation of small oscillations;	0.1
	formula (25) for the period: incorrect numerical factor (-0.1);	0.2
2.2	Recursive relation:		1.0
	use of the law of conservation of energy;	0.3
	expression for the potential energy in terms of deflection angle;	0.2
	relation between the angle and coordinate;	0.1
	final form of the recursive relation: formula (27) or equivalent;	0.4
2.3	Expression for the stoppage point coordinates:		1.0
	expression for the energy change through the path traveled;	0.2
	determination of the disk radius through the initial coordinate;	0.2
	use of the formula for the period of small oscillations;	0.2
	final relation: formula (31) or equivalent;	0.4
	Processing of measurement data		7.5
2.4	Calculation of the period:		1.0
	Formula for the period;	0.2
	numerical value of the period, acceptable error ± 0,01;	0.2
	formula for the error;	0.2
	numerical value of the error;	0.2
	the dimension is indicated;	0.1
	correct rounding;	0.1
2.5	Linearization:		3.2
	choice of variables (equivalent variables are acceptable if expressed in terms of the measured known values);	1.6
	calculated for all points (graded if the choice of variables is graded); 0.1 for each point, acceptable error ±0.02;	1.1
	Plotting a graph (graded if the calculation of points is graded):
	the axes are named and ticked;	0.1
	all points are drawn;	0.2
	smoothing straight line is drawn;	0.2
2.6	Calculation of rolling friction coefficient:		3.3
	Calculation method of the friction coefficient: according to the slope coefficient – 0.6; by 1-2 points – 0.3;	0.6
	slope coefficient found: according to LSM – 0.5; graphically – 0.4; by two points – 0.2; acceptable error ± 0,05;	0.5
	error calculation method;	0.2
	numerical value of the error;	0.2
	formula for calculating the rolling friction (only through the slope factor);	0.3
	numerical value of the rolling friction coefficient (graded if the formula is graded); the error propagation rule does not apply to numerical values!	0.5
	formula for the error, the errors of all quantities included in the formula are taken into account;	0.6
	numerical value of the error;	0.2
	correct rounding;	0.2
	TOTAL		20.0

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