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BPO 2023, Теория — Задача 2. Plank

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5th Balkan Physics Olympiad, BPO 5 - Albania, 24 June 2023, Durrës, ALBANIA — Problems · 24 юни 2023 г.

Условие

A plank of mass M and length l lies on a smooth horizontal surface, and it can move across a rough area (i.e. along the x axis), characterized by its length l and the sliding friction coefficient $\mu$ , the same as the static one. The initial position of the plank is that depicted in the figure below.

A. The plank is launched with the unknown initial, horizontal speed $v_0$ , towards the rough area. Derive the minimum initial speed of the plank for which: [6,3 т.]

A horizontal x axis. A grey plank labelled M lies on the axis between −l and 0, on the smooth region marked μ=0. From 0 to l the axis is hatched (rough region, marked μ); beyond l the region is again marked μ=0; the axis arrow is labelled x.

A1. it fully enters the rough area. [5 т.]

A2. it completely surpasses the rough area. [1,3 т.]

B. The plank starts from rest as illustrated in the figure above but is pulled to the right by a constant horizontal force $F_0$ , permanently acting on it. The purpose is to pull the plank on the rough area. [3,7 т.]

B1. Determine the minimum value of the force for the plank to completely enter the rough area. [2,5 т.]

B2. Derive the maximum value of the plank’s speed during its motion analyzed at B1, for the minimum value found for $F_0$ . [1,2 т.]

C. The plank starts from rest as illustrated in the figure above but is pulled to the right by a constant horizontal force $F_0$ , permanently acting on it. The purpose is to make the plank surpass the rough area for a minimum value of $F_0$ . [15 т.]

C1. Make a graph representation of the net force acting on the plank versus the coordinate x of its front end for $x \in [0, 3l]$ . [3,7 т.]

C2. Determine the minimum value of the force for the plank to completely exit the rough area. It is known that the minimum value of the plank kinetic energy is very small, i.e. a very small fraction $\varepsilon$ of the maximum value of the kinetic energy the plank had until it reached the minimal value. From the mathematical point of view, the fact that $\varepsilon \ll 1$ means that its algebraic powers higher than one can be neglected. The value of $\varepsilon$ is known. [5 т.]

C3. Derive the maximum value of the plank’s speed during its motion on the interval $x \in [0, 2l]$ . Plot the graph of the plank’ speed as a function of $x$ , for $x \in [0, 3l]$ . [6,3 т.]

Note: If useful, you can use the Bernoulli’s approximation $(1+x)^n \cong 1 + nx$ , if $|x| \ll 1$ .

Отговори

Решение

A1. Solution: (5 p)

When the plank has a portion of length $x$ on the rough area, the friction force acting on it is:

F_f(x) = \mu\left(\frac{M}{l}x\right)g.

(1 p)

Graphing this force (1 p), we get the adjacent representation, the area below which is the work of the friction force.

Graph of the friction force F_f versus x: a straight line rising from the origin; the triangular area under the line up to some x is hatched in blue.

So, writing the kinetic energy theorem for the plank, it reads:

\frac{Mv^2(x)}{2} - \frac{Mv_0^2}{2} = -\frac{\mu M g}{l}\frac{x^2}{2},

(1 p)

from which it follows that

v_0 = \sqrt{v^2(x) + \frac{\mu g}{l}x^2}.

(0.6 p)

This speed is minimal when

v(l) = 0,

(0.7 p)

v_{0,\min} = \sqrt{\mu g l}.

(0.7 p)

A2. Solution: (1.3 p)

For the plank to completely exit the rough area, the energy lost by friction doubles (1 p), so

v_{0,\min} = \sqrt{2\mu g l}.

(0.3 p)

B1. Solution (2.5 p):

In this case, the kinetic energy theorem is

\frac{Mv^2(x)}{2} = F_0 x - \frac{\mu M g}{2l}x^2.

(1.5 p)

Again, the force has the minimum value when

v(l) = 0,

(0.5 p)

F_{0,\min} = \frac{\mu M g}{2}.

(0.5 p)

B2. Solution (1.2 p):

The plank’s speed is a quadratic function of $x$ , having the maximum value when

x = \frac{F_0 l}{\mu M g} = \frac{l}{2}.

(0.5 p)

For this value, the maximum speed is

v_{max} = \frac{1}{2}\sqrt{\mu g l}.

(0.7 p)

C1. Solution (3.7 p):

For the plank’s entrance on the rough area, i.e. $x \in [0, l]$ , the net force acting on it is

F(x) = F_0 - \frac{\mu M g}{l}x.

(0.8 p)

For the plank’s exit from the rough area, i.e. $x \in [l, 2l]$ , the friction force acting on it is

F_f(x) = \frac{\mu M g}{l}[l - (x - l)] = \frac{\mu M g}{l}(2l - x),

(0.8 p)

so, the net force acting on it is

F(x) = F_0 + \frac{\mu M g}{l}(x - 2l).

(0.4 p)

The graph is depicted below (1.2 p): (for each value on the graph, the quantities represented on the axis and each line segment, there will be given 0.1 p).

Graph of the net force F versus x: starting at F_0 at x = 0, falling linearly through zero at x_1 to the minimum F_0 − μMg at x = l, rising linearly through zero at x_2 back to F_0 at x = 2l, then constant F_0 up to x = 3l. Dashed guide lines mark F_0, F_0 − μMg, l, 2l and 3l.

If $F_0$ is to be minimized, the lowest point on the graph should be below zero, which means that

F_0 < \mu M g.

(0.5 p)

C2. Solution (5 p):

The maximum value of the kinetic energy is reached when

x = x_1 = \frac{F_0 l}{\mu M g}

(0.5 p)

and the minimum value when

x = x_2 = 2l - \frac{F_0 l}{\mu M g}.

(0.5 p)

The kinetic energy theorem gives:

W_{k,\max} = \frac{F_0 x_1}{2} = \frac{F_0^2 l}{2\mu M g}

(0.8 p)

and

W_{k,\min} = \frac{F_0 x_1}{2} - (\mu M g - F_0)\frac{x_2 - x_1}{2} =

(0.7 p)

= \frac{F_0^2 l}{2\mu M g}\left[1 - 2\left(\frac{\mu M g}{F_0} - 1\right)^2\right].

(0.6 p)

Since

W_{k,\min} = \varepsilon W_{k,\max},

(0.6 p)

from the above two equations it follows that

F_0 = \frac{\mu M g}{1 + \sqrt{\frac{1-\varepsilon}{2}}} = \frac{\sqrt{2}\mu M g}{\sqrt{2} + (1-\varepsilon)^{\frac{1}{2}}} \cong \frac{\sqrt{2}\mu M g}{\sqrt{2} + 1 - \frac{1}{2}\varepsilon} =

(0.7 p)

= \frac{\sqrt{2}}{\sqrt{2}+1}\mu M g\left[1 - \frac{\varepsilon}{2(\sqrt{2}+1)}\right]^{-1} \cong

\cong (2-\sqrt{2})\mu M g\left(1 + \frac{\sqrt{2}-1}{2}\varepsilon\right).

(0.6 p)

Note: If $\varepsilon = 0$ , then the speed and acceleration are concomitantly zero, so the plank will finish its motion in $x = x_2 < 2l$ , which is not what the problem asks.

C3. Solution (6.3 p):

For $x \in [0, l]$ , the maximum speed of the plank is

v_{\max,1} = v(x_1) = \frac{F_0}{M}\sqrt{\frac{l}{\mu g}},

(0.7 p)

v_{\max,1} = (2-\sqrt{2})\sqrt{\mu g l}\left(1 + \frac{\sqrt{2}-1}{2}\varepsilon\right).

(0.5 p)

For $x \in [l, 2l]$ , the maximum speed of the plank is $v_{\max,2} = v(2l)$ , which is given by the kinetic energy theorem:

\frac{Mv_{\max,2}^2}{2} - \frac{Mv_{\min}^2}{2} = \frac{F_0}{2}(2l - x_2),

(0.8 p)

but

\frac{F_0}{2}(2l - x_2)\frac{F_0^2 l}{2\mu M g} = \frac{Mv_{\max,1}^2}{2},

(0.6 p)

\frac{Mv_{\max,2}^2}{2} = \frac{Mv_{\max,1}^2}{2} + \frac{Mv_{\min}^2}{2} = (1+\varepsilon)\frac{Mv_{\max,1}^2}{2}.

(0.6 p)

From here we get

v_{\max,2} = \sqrt{1+\varepsilon}\,v_{\max,1} \cong \left(1 + \frac{\varepsilon}{2}\right)(2-\sqrt{2})\sqrt{\mu g l}\left(1 + \frac{\sqrt{2}-1}{2}\varepsilon\right) \cong

\cong (2-\sqrt{2})\sqrt{\mu g l}\left(1 + \frac{\sqrt{2}}{2}\varepsilon\right),

(0.7 p)

which is the maximum speed reached by the plank on the considered interval.

For plotting the requested graph, the value of the plank’ speed for $x = 3l$ is needed, which can be obtained again, using the kinetic energy theorem:

v_3 = \sqrt{v_{\max,2}^2 + \frac{2F_0 l}{M}} \cong

(0.7 p)

\cong \sqrt{2\mu g l(5 - 3\sqrt{2})\left[1 + \frac{3}{14}(3\sqrt{2} - 2)\varepsilon\right]} \cong

\cong \sqrt{2\mu g l(5 - 3\sqrt{2})}\left[1 + \frac{3}{28}(3\sqrt{2} - 2)\varepsilon\right].

(0.7 p)

The graph is represented below (1 p):

Graph of the plank speed v versus x: rising from v_min at x = 0 to the maximum v_max,1 at x_1, falling past l to a minimum v_min at x_2, rising steeply to v_max,2 at 2l (slightly above v_max,1, blue dashed line), then increasing more slowly to v_3 at 3l. Axis marks x_1, l, x_2, 2l, 3l and v_min, v_max,1, v_max,2, v_3.

The same figure as in the statement: the plank M on the smooth region between −l and 0 (μ=0), the hatched rough region from 0 to l (μ), and the smooth region beyond l (μ=0) along the x axis.

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf