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IPhO 1988, Теория и експеримент — Задача 2. THEORY 2 Maxwell´s Wheel

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Всички материали от IPhO 1988

19th International Physics Olympiad - 1988 Bad Ischl / Austria

Условие

Introduction

A cylindrical wheel of uniform density, having the mass $M = 0{,}40$ kg, the radius $R = 0{,}060$ m and the thickness $d = 0{,}010$ m is suspended by means of two light strings of the same length from the ceiling. Each string is wound around the axle of the wheel. Like the strings, the mass of the axle is negligible. When the wheel is turned manually, the strings are wound up until the centre of mass is raised 1,0 m above the floor. If the wheel is allowed to move downward vertically under the pulling force of the gravity, the strings are unwound to the full length of the strings and the wheel reaches the lowest point. The strings then begin to wound in the opposite sense resulting in the wheel being raised upwards.

Analyze and answer the following questions, assuming that the strings are in vertical position and the points where the strings touch the axle are directly below their respective suspending points (see fig. 19.5).

Left: perspective drawing of the wheel (a shaded disc) on an axle with the two strings wound on the axle ends. Right: side view of the wheel hanging from suspension point P by a vertical string touching the axle at A; the centre is S, the axle radius r and the wheel radius R are marked; the wheel is shown again at its lowest position, with the fall distance s and the height h indicated by arrows. — Fig. 19.5

2.1 Determine the angular speed of the wheel when the centre of mass of the wheel covers the vertical distance s.

2.2 Determine the kinetic energy of the linear motion of the centre of mass $E_r$ after the wheel travels a distance s = 0,50 m, and calculate the ratio between $E_r$ and the energy in any other form in this problem up to this point.

Radius of the axle = 0,0030 m

2.3 Determine the tension in the string while the wheel is moving downward.

2.4 Calculate the angular speed $\omega'$ as a function of the angle $\Phi$ when the strings begin to unwind themselves in opposite sense as depicted in fig. 19.6.

Sketch a graph of variables which describe the motion (in cartesian system which suits the problem) and also the speed of the centre of mass as a function of $\Phi$ .

Left: the wheel at its lowest point, the string from the ceiling passing the axle at A and winding round the axle about the centre S in the opposite sense. Right: small side view of the wheel hanging from P with centre S, contact point A, axle radius r and wheel radius R marked. — Fig. 19.6

2.5 If the string can withstand a maximum tension $T_m = 10$ N, find the maximum length of the string which may be unwound without breaking by the wheel.

Отговори

Решение

2.1

conservation of energy: $M\cdot g\cdot s = \frac{1}{2}\cdot I_A\cdot\omega^2$ (1)

where $\omega$ is the angular speed of the wheel and $I_A$ is the moment of inertia about the axis through A.

Note: If we would take the moment of inertia about S instead of A we would have

$M\cdot g\cdot s = \frac{1}{2}\cdot I_S\cdot\omega^2 + \frac{1}{2}\cdot m\cdot v^2$

where v is the speed of the centre of mass along the vertical.

This equation is the same as the above one in meanings since

$I_A = I_S + M\cdot r^2$ and $I_S = M\cdot R^2$

From (1) we get $\omega = \sqrt{\frac{2\cdot M\cdot g\cdot s}{I_A}}$

substitute $I_A = \frac{1}{2}\cdot M\cdot r^2 + M\cdot R^2$

\omega = \sqrt{\frac{2\cdot g\cdot s}{r^2+\frac{R^2}{2}}}

Putting in numbers we get

\boxed{\omega = \sqrt{\frac{2\cdot 9{,}81\cdot 0{,}50}{9\cdot 10^{-6}+\frac{1}{2}\cdot 36\cdot 10^{-4}}} \approx 72{,}4\ \frac{\mathrm{rad}}{\mathrm{s}}}

2.2

Kinetic energy of linear motion of the centre of mass of the wheel is

E_T = \frac{1}{2}\cdot M\cdot v^2 = \frac{1}{2}\cdot M\cdot\omega^2\cdot r^2 = \frac{1}{2}\cdot 0{,}40\cdot 72{,}4^2\cdot 9\cdot 10^{-6} = 9{,}76\cdot 10^{-3}\ \mathrm{J}

Potential energy of the wheel

E_P = M\cdot g\cdot s = 0{,}40\cdot 9{,}81\cdot 0{,}50 = 1{,}962\ \mathrm{J}

Rotational kinetic energy of the wheel

E_R = \frac{1}{2}\cdot I_S\cdot\omega^2 = \frac{1}{2}\cdot 0{,}40\cdot 1{,}81\cdot 10^{-3}\cdot 72{,}4^2 = 1{,}899\ \mathrm{J}

\boxed{\frac{E_T}{E_R} = \frac{9{,}76\cdot 10^{-3}}{1{,}899} = 5{,}13\cdot 10^{-3}}

2.3

Let $\frac{T}{2}$ be the tension in each string.

Torque $\tau$ which causes the rotation is given by $\tau = M\cdot g\cdot r = I_A\cdot\alpha$

where $\alpha$ is the angular acceleration $\alpha = \frac{M\cdot g\cdot r}{I_A}$

The equation of the motion of the wheel is $M.g - T = M.a$

Substituting $a = \alpha\ .\ r$ and $I_A = \frac{1}{2}\cdot M\cdot r^2 + M\cdot R^2$ we get

T = M\cdot g + \frac{M\cdot g\cdot r^2}{\frac{1}{2}\cdot M\cdot R^2 + M\cdot r^2} = M\cdot g\cdot\left(1+\frac{2\cdot r^2}{R^2+2\cdot r^2}\right)

Thus for the tension $\frac{T}{2}$ in each string we get

\frac{T}{2} = \frac{M\cdot g}{2}\cdot\left(1+\frac{2\cdot r^2}{R^2+2\cdot r^2}\right) = \frac{0{,}40\cdot 9{,}81}{2}\cdot\left(1+\frac{2\cdot 9\cdot 10^{-6}}{3{,}6\cdot 10^{-3}+2\cdot 9\cdot 10^{-6}}\right) = 1{,}96\ \mathrm{N}

\boxed{\frac{T}{2} = 1{,}96\ \mathrm{N}}

2.4

The wheel turning about the contact point A after the string is fully unwound: two positions of the wheel circle, the string running up from A, a horizontal line through A, the radius from A to the centre S and the angle φ between them; arrows show the string moving down on one side and up on the other. — Fig 19.7

After the whole length of the strings is completely unwound, the wheel continues to rotate about A (which is at rest for some interval to be discussed). Let $\dot{\Phi}$ be the angular speed of the centre of mass about the axis through A. The equation of the rotational motion of the wheel about A may be written as $|\tau| = I_A\cdot\ddot{\Phi}$ ,

where $\tau$ is the torque about A, $I_A$ is the moment of inertia about the axis A and $\ddot{\Phi}$ is the angular acceleration about the axis through A.

Hence $M\cdot g\cdot r\cdot\cos\Phi = I_A\cdot\ddot{\phi}$

and $\ddot{\phi} = \frac{M\cdot g\cdot r\cdot\cos\Phi}{I_A}$

Multiplied with $\dot{\Phi}$ gives:

$\dot{\Phi}\cdot\ddot{\Phi} = \frac{M\cdot g\cdot r\cdot\cos\Phi\cdot\dot{\Phi}}{I_A}$ or $\frac{1}{2}\cdot\frac{d\left(\dot{\Phi}\right)^2}{dt} = \frac{M\cdot g\cdot r\cdot\cos\Phi}{I_A}\cdot\frac{d\Phi}{dt}$

this gives

$\left(\dot{\Phi}\right)^2 = \frac{2\cdot M\cdot g\cdot r\cdot\sin\Phi}{I_A} + C$ [C = arbitrary constant]

If $\Phi = 0$ [s = H] than is $\dot{\Phi} = \omega$

That gives $\omega = \frac{2\cdot M\cdot g\cdot H}{I_A}$ and therefore $C = \frac{2\cdot M\cdot g\cdot H}{I_A}$

Putting these results into the equation above one gets

Graph of ω versus φ (ticks 90°, 180°, 270°): a sinusoid oscillating about the horizontal line ω = [2MgH/I_A]^(1/2) between dashed lines ω_MAX = [2MgH/I_A (1 + r/H)]^(1/2) and ω_MIN = [2MgH/I_A (1 − r/H)]^(1/2); the vertical axis has a break mark. — Fig.19.8

\dot{\Phi} = \omega = \sqrt{\frac{2\cdot M\cdot g\cdot H\cdot\sin\Phi}{I_A}\cdot\left(1+\frac{r}{H}\right)}

For $\frac{r}{H} \lll 1$ we get:

$\omega = \omega'_{\mathrm{MAX}} = \sqrt{\frac{2\cdot M\cdot g\cdot H}{I_A}}$

and

$v = r\cdot\omega'_{\mathrm{MAX}} = r\cdot\sqrt{\frac{2\cdot M\cdot g\cdot H}{I_A}}$

Component of the displacement

along x-axis is x = r.sin $\Phi$ -r

along y-axis is y = r.cos $\Phi$ -r

2.5

Maximum tension in each string occurs $\dot{\Phi} = \omega'_{\mathrm{MAX}}$

The equation of the motion is $T_{\mathrm{MAX}} - M\cdot g = M\cdot\left(\omega'_{\mathrm{MAX}}\right)^2\cdot r$

Putting in T = 20 N and $\omega'_{\mathrm{MAX}} = \sqrt{\frac{2\cdot M\cdot g\cdot s}{I_A}}$ (where s is the maximum length of the strings supporting the wheel without breaking) and $I_A = M\cdot\left(\frac{R^2}{2}+r^2\right)$ the numbers one gets:

20 = 0{,}40\cdot 9{,}81\cdot\left(1+\frac{4\cdot 3\cdot 10^{-3}\cdot s}{36\cdot 10^{-4}+2\cdot 9\cdot 10^{-6}}\right)

This gives: s = 1,24 m

The maximum length of the strings which support maximum tension without breaking is

$\boxed{1{,}24\ \mathrm{m}}$ .

Three graphs against the angle with ticks 90° and 180°: (a) v is constant (horizontal line); (b) x rises from a minimum at 0 to a maximum at 90° and falls back to a minimum at 180°; (c) y falls from 0 to a minimum at 90° and returns to 0 at 180°. — Fig.19.9

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