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IZhO 2007, Теория — Задача 1. A block with an object

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III International Zhautykov Olympiad — THEORETICAL COMPETITION, January 16, 2007 · 16 януари 2007 г.

Please read this first:

The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be 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 bordered area.
Begin each question on a separate sheet.
Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each 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 evaluated, 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 for each problem in the following order:
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Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper out of the room.

Условие

A weightless rod is attached to a massive block with a mass M. An object with a mass m is attached to the rod by a weightless string of length R. The object can rotate in a vertical plane not touching the rod. The coefficient of friction between the block and the floor depends on the direction of sliding, $k1=\mu$ , $k2 = \infty$ . The object is launched with an initial velocity $V_0$ in the direction as shown in the picture 1. $V_0$ is sufficiently big so that the object makes at least one rotation around the rod.

In this problem you are asked to examine the movement of the system during the first rotation of the object. You can assume that the block does not detach from the floor and does not turn over and that the string is stretched while the object is moving. Let’s describe the position of the object by the angle $\varphi$ (Fig 1). Assume $M \gg m$ .

A block M on a hatched floor with a vertical rod; a point mass m on a string of length R hangs from the rod top and moves on a dashed circle; the angle φ is measured from the rod; at the lowest point the initial velocity V0 points left; below the floor, arrows k1 (left) and k2 (right). — Fig.1

1. [1 point] In this part of the question assume that the parameters are such that the block does not move at all. Derive the condition that will guarantee that the change in absolute velocity of the object is considerably smaller than $V_0$ in the process of rotation. [1 т.]

From this point onwards you can assume that the block’s velocity is considerably smaller than $V_0$ , and that the change in absolute velocity of the object during first rotation is considerably smaller than $V_0$ .

2. [2 points] Find the minimum initial velocity $V_0^{\min}$ which will cause the block to slide when the object is rotating. [2 т.]

3. a) For $V_0 > V_0^{\min}$ find:

[1 point] At which angle $\varphi_0$ sliding will start; [1 т.]

3. b) [1 point] At which angle $\psi$ the velocity of block’s motion V will be largest. [1 т.]

4. a) [1 point] Find $V_{max}$ which corresponds to the angle $\psi$ from b) above; [1 т.]

4. b) [1 point] Write down the condition for angle $\theta$ , at which the block stops. [1 т.]

5. a) Numerical data:

\mu = \frac{\pi}{2} - 1 \approx 0.57,

g = 10\ m/s^2,

M = 10\ kg

V_0 = 10\ m/s,

R = 0.1\ m,

m = 0.057\ kg.

[0.5 points] Calculate $\varphi_0$ , $\psi$ и $\theta$ . [0,5 т.]

5. b) [0.5 points] Calculate $V_{max.}$ [0,5 т.]

5. c) [1 point] For the given numerical values, calculate the displacement of the block in one rotation of the object. [1 т.]

5. d) [1 point] In first non vanishing order, estimate the change in the velocity of the object at the lowest point after the first rotation in comparison to $V_0$ . [1 т.]

Note:

The following formulas might be useful for solution:

\cos\varphi + \mu\ \sin\varphi = \sqrt{1+\mu^2}\ \cos(\varphi - \alpha),

\cos\varphi - \mu\ \sin\varphi = \sqrt{1+\mu^2}\ \cos(\varphi + \alpha),

where $\alpha = \operatorname{arctg}\mu = \arccos(1/\sqrt{1+\mu^2}) = \arcsin(\mu/\sqrt{1+\mu^2})$ .

\int\cos\varphi d\varphi = \sin\varphi,\ \int\sin\varphi d\varphi = -\cos\varphi

Отговори

Решение

The same drawing as Fig.1 of the problem: block M on the floor with a vertical rod, mass m on a string of length R moving on a dashed circle, angle φ, initial velocity V0 at the lowest point pointing left, arrows k1 (left) and k2 (right) under the floor. — Fig.1

[1 point] From the energy conservation law for the object:

V_0^{\,2} - V_{top}^{\,2} = 4gR \approx 2V_0\Delta V,\quad \Delta V / V = 2gR/V_0^{\,2} \ll 1

[2 point] Maximal angle of deviation of the floor’s total reaction force (Normal force + friction force) from vertical is

\alpha = \operatorname{arctg}\mu.

Force diagram: a vertical arrow Mg from point A down to point O, the centre of a dashed circle; an arrow T from O to point B on the circle; the resultant arrow from A to B passes point C, where the line AB enters the circle; the angle α is marked at A between AO and AB. — Fig.2

Denote

T = mV_0^{\,2}/R,\qquad \omega = V_0/R.

From the picture above we can see (see Fig.2) that the maximum angle of deviation of

M\vec{g} + \vec{T}

is $\arcsin(T/Mg)$ when $T \le Mg$ ( $M\vec{g} + \vec{T}$ is tangent to the circle). If the deviation angle of $M\vec{g} + \vec{T}$ from the vertical line is greater than $\alpha$ , then the equilibrium is impossible and the block will start to slide. The condition that we are looking for is

V_0^{\min} = \sqrt{\frac{MgR\mu}{m\sqrt{1+\mu^2}}}

From Fig 2, the beginning of sliding corresponds to point B. The block accelerates on the portion of path BC, and decelerates afterwards.. Since the line intersects the rotation circle at two points, $\angle ABO < \pi/2$ , and therefore from the sine theorem for the triangle ABO :

\angle ABO = \arcsin\left(\mathrm{Mg}\frac{\sin\alpha}{m\omega^2\ R}\right) = \arcsin\left(\frac{\mu Mg}{m\omega^2\ R\sqrt{1+\mu^2}}\right)

a) [1 point] Then

\varphi_0 = \alpha + \angle ABO = \operatorname{arctg}\mu + \arcsin\left(\frac{\mu Mg}{m\omega^2\ R\sqrt{1+\mu^2}}\right).

b) [1 point] Similarly

\psi = \alpha + \pi - \angle ABO = \operatorname{arctg}\mu + \pi - \arcsin\left(\frac{\mu Mg}{m\omega^2\ R\sqrt{1+\mu^2}}\right).

To calculate the acceleration we can first decompose the forces into x and y components and write down Newton’s second law

F_{\mathrm{friction}} = \mu\ \mathrm{N} = \mu\,(M\mathrm{g} + T\cos\varphi)

M\,\mathrm{a}(\varphi) = T\sin\varphi - F_{\mathrm{friction}} = T\sin\varphi - \mu\,(M\mathrm{g} + T\cos\varphi)

Then

V(\varphi) = \int a(\varphi)dt = \int\frac{a(\varphi)d\varphi}{\omega}

Integration gives us:

V(\varphi) = -\left(\frac{T}{\omega M}(\cos\varphi + \mu\sin\varphi) + \frac{\mu g\varphi}{\omega}\right) + \left(\frac{T}{\omega M}(\cos\varphi_0 + \mu\sin\varphi_0) + \frac{\mu g\varphi_0}{\omega}\right).

a) [1point] Then

V_{\mathrm{max}} = \mathrm{v}(\psi) = 2\frac{T}{\omega M}\sqrt{1+\mu^2-\mu^2\left(\frac{Mg}{T}\right)^2}\ -\ \frac{\mu g}{\omega}\left(\pi - 2\arcsin\ \frac{\mu Mg}{T\sqrt{1+\mu^2}}\right)

b) [1 point] The condition for $\theta$ : $V(\theta) = 0$

\frac{T}{\omega M}(\cos\varphi_0 + \mu\sin\varphi_0) + \frac{\mu g\varphi_0}{\omega} = \frac{T}{\omega M}(\cos\theta + \mu\sin\theta) + \frac{\mu g\theta}{\omega}

The values are such that $\dfrac{Mg}{T} = \dfrac{1}{\mu} = 1/(\dfrac{\pi}{2} - 1)$ .

a) [0.5 point] Using this equality we get :

\varphi_0 = 90^{0},\ \psi = 90^{0} + 2\operatorname{arctg}(0.57) \approx 150^{0},\ \theta = 180^{0}

b) [0.5 point]

V_{\mathrm{max}} = 2\frac{\mu g}{\omega}(\mu - \operatorname{arctg}\mu) \approx 6*10^{-3}\mathrm{m/s} \ll V_0

c) [1 point] Displacement is

\mathrm{S} = \int\frac{V(\varphi)d\varphi}{\omega}.

Plug in

\mu = \frac{\pi}{2} - 1,\ \varphi_0 = \frac{\pi}{2},\ \theta = \pi,

We get

\mathrm{S} = \left(\frac{\pi}{2} - 1\right)\left(2 + \frac{\pi^2}{8} - \pi\right)\mathrm{g}/\omega^2 \approx 5.3*10^{-5}\mathrm{m}.

d) [1 point] The change in object’s velocity occurs because the kinetic energy is lowered due to work by friction force. In first nonvanishing order, the change in velocity is determined from the Law of change in energy, where the work of friction force is calculated assuming constant absolute velocity of the object:

mV_0\Delta V = \int F_{friction}ds = -\int_{\varphi_0}^{\theta}\mu(\mathrm{Mg} + \mathrm{T}\cos\varphi)\,\mathrm{v}(\varphi)\frac{d\varphi}{\omega}.

By order of magnitude

\frac{\Delta V}{V_0} \propto -\frac{\mu MgS}{mV_0^{\,2}} \propto -5*10^{-4}.

The exact value: $-3*10^{-4}$

Оригинал в Архива: IZhO-2007-Theory_eng_sf.doc · официални решения: IZhO-2007-Theory_eng_sol_sf.doc