Savchenko Solutions

Solutions of Savchenko Problems in Physics
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    <h3 id="back-link"><a href="/#2.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.1.45^*.$ A coin is placed on a plane whose slope tangent is equal to the coefficient of friction. In the horizontal direction along the plane, the coin was given a velocity $v$. Find the steady speed of the coin.
    </p>

    <h3>Solution</h3>
    <p>
        <center>

Projecting $m\vec{g}$ on $OY$:

From the figure, we find the support reaction force: $$N=mg \cos \alpha$$ According to the Amontons-Coulomb Law: $$F_{тр}=\mu N=\mu mg \cos \alpha$$ Since $\mu = tg \alpha$ (by condition): $$F_{тр}=\mu mg \cos \alpha$$

$$F_{тр}=mg \cdot tg \alpha \cdot \cos \alpha=mg \sin \alpha$$

$$F=mg \sin \alpha$$
Redraw in the $XY$ plane
At the initial moment $(\varphi=90^{\circ})$:

At the final moment $(\varphi=0^{\circ})$:

Let's consider an arbitrary moment:

Note that $\varphi$ changes from $0^{\circ}$ to $90^{\circ}$:

Let's write Newton's second law: $$\left\{\begin{matrix} \frac{du_x}{dt} = \frac{F \sin \varphi }{m}& \\ \frac{du_y}{dt} = \frac{F (1- \cos \varphi) }{m} & \end{matrix}\right.$$

$$\left{\begin{matrix} \frac{d}{dt} (u \sin \varphi)= \frac{F \sin \varphi }{m}& \ \frac{d}{dt} (u \cos \varphi) = \frac{F (1- \cos \varphi) }{m} & \end{matrix}\right.$$
We solve the system of differential equations:
$$\left{\begin{matrix} \frac{d\varphi}{dt} = \frac{F \sin \varphi }{mu}& (a)\ \frac{du}{dt} = \frac{F (1- \cos \varphi) }{m} & (b) \end{matrix}\right.$$
We divide $(b)$ by $(a)$:
$$\frac{du}{d\varphi} = u\frac{1- \cos \varphi}{\sin \varphi}$$

$$\frac{du}{u} = \frac{1- \cos \varphi}{\sin \varphi} d\varphi$$

$$\frac{du}{u} = tg (\frac{\varphi}{2}) d\varphi$$
We integrate both parts of the equation:

$$\int \frac{du}{u} = \int tg (\frac{\varphi}{2}) d\varphi \; (c)$$

$$\int\frac{du}{u}=ln|u| \; (d)$$

$$\int tg (\frac{\varphi}{2}) d\varphi=-2 ln(\cos(\frac{x}{2})) \; (e)$$
Substitute $(e)$ and $(d)$ into $(c)$:

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