Правка раздела «Solution»

Аноним правка от 7 сент. 2025 г., 16:27

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			### Statement

			$2.7.6.$ 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?

			![ For problem $2.7.6$ \|372x281, 31%](../../img/2.7.6/statement.png)

			### Solution

			![ Forces acting on the cylinder \|924x709, 39%](../../img/2.7.6/draw1.png)

			Conditions of equilibrium:

			$$
			\left\\{\begin{matrix} O'X\text{: }F_{fr2} = N_1 \\\ O'Y\text{: }F_{fr1}+N_2 = mg \\\ \end{matrix}\right.
			$$

			From where

			$$
			\left\\{\begin{matrix} \mu N_2=N_1 \\\ \mu N_1+N_2=mg \\\ \end{matrix}\right.
			$$

			Let's find friction force

			$$
			F_{fr}=F_{fr1}+F_{fr2}
			$$

			$$
			F_{fr}=\mu (N_1+N_2)=\mu N_2(1+\mu )
			$$

			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

			$$
			M=I\varepsilon = mR^2\varepsilon=F_{fr} R
			$$

			From where, absolute value of angular acceleration is found as

			$$
			\varepsilon = \frac{M}{I} = \frac{F_{fr}}{mR}
			$$

			$$
			\varepsilon = \frac{\mu (1+\mu )N_2 }{mR}
			$$

			$$
			\varepsilon = \frac{\mu (1+\mu )}{1+\mu^2} \frac{g}{R}
			$$

		@@ -59,7 +59,7 @@Solution
			The angle by which the cylinder will rotate to the moment of total stop is

			$$
		−	\varphi=\frac{\omega^2t}{2\varepsilon }
		+	\varphi=\frac{\omega^2}{2\varepsilon }
			$$

			Until the total stop, the cylinder will make $N$ total turnovers

			$$
			\varphi=2\pi N
			$$

		@@ -71,7 +71,7 @@Solution
			From where

			$$
		−	N=\frac{\omega^2t}{4\pi\varepsilon }
		+	N=\frac{\omega^2}{4\pi\varepsilon }
			$$

			$$
			\boxed{N=\frac{1+\mu^{2}}{4\pi\mu (1+\mu )}\frac{\omega^{2}R}{g}}
			$$

			#### Answer

			$$
			N=\omega^{2}R(1+\mu^{2})/[4\pi g\mu (1+\mu )]
			$$
		ещё строк без изменений 8

	### Statement		### Statement

	$2.7.6.$ 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?		$2.7.6.$ 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?

	![ For problem $2.7.6$ \|372x281, 31%](../../img/2.7.6/statement.png)		![ For problem $2.7.6$ \|372x281, 31%](../../img/2.7.6/statement.png)

	### Solution		### Solution

	![ Forces acting on the cylinder \|924x709, 39%](../../img/2.7.6/draw1.png)		![ Forces acting on the cylinder \|924x709, 39%](../../img/2.7.6/draw1.png)

	Conditions of equilibrium:		Conditions of equilibrium:

	$$		$$
	\left\\{\begin{matrix} O'X\text{: }F_{fr2} = N_1 \\\ O'Y\text{: }F_{fr1}+N_2 = mg \\\ \end{matrix}\right.		\left\\{\begin{matrix} O'X\text{: }F_{fr2} = N_1 \\\ O'Y\text{: }F_{fr1}+N_2 = mg \\\ \end{matrix}\right.
	$$		$$

	From where		From where

	$$		$$
	\left\\{\begin{matrix} \mu N_2=N_1 \\\ \mu N_1+N_2=mg \\\ \end{matrix}\right.		\left\\{\begin{matrix} \mu N_2=N_1 \\\ \mu N_1+N_2=mg \\\ \end{matrix}\right.
	$$		$$

	Let's find friction force		Let's find friction force

	$$		$$
	F_{fr}=F_{fr1}+F_{fr2}		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, $mg = N_2(1+\mu^2)$:		Considering, $mg = N_2(1+\mu^2)$:

	$$		$$
	N=\frac{mg}{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		According to rotational motion dynamics, the sum of moments of external forces can be written as

	$$		$$
	M=I\varepsilon = mR^2\varepsilon=F_{fr} R		M=I\varepsilon = mR^2\varepsilon=F_{fr} R
	$$		$$

	From where, absolute value of angular acceleration is found as		From where, absolute value of angular acceleration is found as

	$$		$$
	\varepsilon = \frac{M}{I} = \frac{F_{fr}}{mR}		\varepsilon = \frac{M}{I} = \frac{F_{fr}}{mR}
	$$		$$

	$$		$$
	\varepsilon = \frac{\mu (1+\mu )N_2 }{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 )}{1+\mu^2} \frac{g}{R}
	$$		$$

	@@ -59,7 +59,7 @@Solution
	The angle by which the cylinder will rotate to the moment of total stop is		The angle by which the cylinder will rotate to the moment of total stop is

	$$		$$
	\varphi=\frac{\omega^2t}{2\varepsilon }		\varphi=\frac{\omega^2}{2\varepsilon }
	$$		$$

	Until the total stop, the cylinder will make $N$ total turnovers		Until the total stop, the cylinder will make $N$ total turnovers

	$$		$$
	\varphi=2\pi N		\varphi=2\pi N
	$$		$$

	@@ -71,7 +71,7 @@Solution
	From where		From where

	$$		$$
	N=\frac{\omega^2t}{4\pi\varepsilon }		N=\frac{\omega^2}{4\pi\varepsilon }
	$$		$$

	$$		$$
	\boxed{N=\frac{1+\mu^{2}}{4\pi\mu (1+\mu )}\frac{\omega^{2}R}{g}}		\boxed{N=\frac{1+\mu^{2}}{4\pi\mu (1+\mu )}\frac{\omega^{2}R}{g}}
	$$		$$

	#### Answer		#### Answer

	$$		$$
	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 )]
	$$		$$
	ещё строк без изменений 8