Updated spacing between @ latex expressions

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@@ -6,14 +6,14 @@
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<meta http-equiv="content-language" content="en">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
− <meta name="description" content="A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu. How many revolutions will the cylinder make before it comes to a complete stop?">
+ <meta name="description" content="A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu . How many revolutions will the cylinder make before it comes to a complete stop?">
<meta name="author" content="Aliaksandr Melnichenka">
<meta name="date" content="2023-10" scheme="YYYY-MM">
− <meta property="og:title" content="A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu. How many revolutions will the cylinder make before it comes to a complete stop?">
+ <meta property="og:title" content="A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu . How many revolutions will the cylinder make before it comes to a complete stop?">
<meta property="og:image" content="img/logo.png">
− <meta property="og:description" content="A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu. How many revolutions will the cylinder make before it comes to a complete stop?">
+ <meta property="og:description" content="A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu . How many revolutions will the cylinder make before it comes to a complete stop?">
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− <title>A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu. How many revolutions will the cylinder make before it comes to a complete stop?</title>
+ <title>A thin-walled cylinder of radius R was spun up to an angular velocity \omega and placed in a corner, as shown in the figure. The coefficient of friction between the corner walls and the cylinder is \mu . How many revolutions will the cylinder make before it comes to a complete stop?</title>
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@@ -92,29 +92,29 @@
\end{matrix}\right.$$
Let's find friction force
$$F_{fr}=F_{fr1}+F_{fr2}$$
−$$F_{fr}=\mu (N_1+N_2)=\mu N_2(1+\mu)$$
+$$F_{fr}=\mu (N_1+N_2)=\mu N_2(1+\mu )$$
−Considering, $mg = N_2(1+\mu ^2)$:
−$$N=\frac{mg}{1+\mu ^2}$$
+Considering, $mg = N_2(1+\mu^2)$:
+$$N=\frac{mg}{1+\mu^2}$$
According to rotational motion dynamics, the sum of moments of external forces can be written as
$$M=I\varepsilon = mR^2=F_{fr} R$$
From where, absolute value of angular acceleration is found as
$$\varepsilon = \frac{M}{I} = \frac{F_{fr}}{mR}$$
−$$\varepsilon = \frac{\mu(1+\mu)N_2 }{mR}$$
−$$\varepsilon = \frac{\mu(1+\mu)}{1+\mu ^2} \frac{g}{R}$$
+$$\varepsilon = \frac{\mu (1+\mu )N_2 }{mR}$$
+$$\varepsilon = \frac{\mu (1+\mu )}{1+\mu^2} \frac{g}{R}$$
The angle by which the cylinder will rotate to the moment of total stop is
−$$\varphi=\frac{\omega ^2t}{2\varepsilon }$$
+$$\varphi=\frac{\omega^2t}{2\varepsilon }$$
Until the total stop, the cylinder will make $N$ total turnovers
$$\varphi=2\pi N$$
From where
−$$N=\frac{\omega ^2t}{4\pi \varepsilon }$$
−$$\boxed{N=\frac{1+\mu^{2}}{4\pi \mu(1+\mu)}\frac{\omega^{2}R}{g}}$$
+$$N=\frac{\omega^2t}{4\pi\varepsilon }$$
+$$\boxed{N=\frac{1+\mu^{2}}{4\pi\mu (1+\mu )}\frac{\omega^{2}R}{g}}$$
</p>
<h4>Answer</h4>
<p>
− $$N=\omega^{2}R(1+\mu^{2})/[4\pi g\mu(1+\mu)]$$
+ $$N=\omega^{2}R(1+\mu^{2})/[4\pi g\mu (1+\mu )]$$
</p>
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