Updated greek laters @ latex compiling

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правка #10135 предыдущая #9587 GitHub dbb31ba ← раньше позже →
en/2.7.20.md +6 −6
@@ -81,23 +81,23 @@
The second Newton's law for the horizontal axis
$$ma=F\cos\alpha-F_{fr}\quad(1)$$
−Let's write a moment of inertia with respect to the axis of rotation of the spool with angular acceleration $\varepsilon$
−$$F_{fr}\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon\quad(2)$$
+Let's write a moment of inertia with respect to the axis of rotation of the spool with angular acceleration $\varepsilon $
+$$F_{fr}\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon \quad(2)$$
The moment of inertia of the spool in our model (a solid cylinder, symmetry axis)
$$I=\frac{mR^2}{2}$$
Because there's no slippage:
−$$a=\varepsilon\cdot R\quad(3)$$
+$$a=\varepsilon \cdot R\quad(3)$$
Let's substitute $(3)$ into $(1)$
$$F_{fr}=F\cos\alpha-m\varepsilon R\quad(4)$$
and put into $(2)$
−$$\left(F\cos\alpha-m\varepsilon R\right)\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon$$
+$$\left(F\cos\alpha-m\varepsilon R\right)\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon $$
After algebraic transformations
−$$F\left(R\cos\alpha-r\right)=\frac{3mR^2\varepsilon}{2} \Leftrightarrow \boxed{\varepsilon=\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}} \quad(5)$$
+$$F\left(R\cos\alpha-r\right)=\frac{3mR^2\varepsilon }{2} \Leftrightarrow \boxed{\varepsilon =\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}} \quad(5)$$
For the angular acceleration to be clockwise, and consequently the motion to be co-directional with the direction of $\vec{F}$, the external force moment from equation $(2)$ must be positive, and consequently the expression $(5)$ must be greater than 0. The moment of external forces in equation $(2)$ must be positive $(\varepsilon = 0)$
$$\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}>0$$
$$R\cos\alpha-r>0\Leftrightarrow \boxed{\cos\alpha>\frac{r}{R}}$$
The larger the angle $\alpha$, the smaller the value of the trigonometric function $\cos\alpha$, making the expression for the angle
−$$\boxed{\alpha<\arccos\left(\frac{r}{R}\right)}$$
+$$\boxed{\alpha<\arc\cos\left(\frac{r}{R}\right)}$$
</p>
<h4>Answer</h4>
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