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| <meta name="author" content="Aliaksandr Melnichenka"> | | <meta name="author" content="Aliaksandr Melnichenka"> |
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | | <meta name="date" content="2023-10" scheme="YYYY-MM"> |
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| <title>A spool of thread lies on a horizontal plane. The reel is pulled by a thread. At what angles \alpha between the force and the horizontal will the coil accelerate towards the taut thread?</title> | | <title>A spool of thread lies on a horizontal plane. The reel is pulled by a thread. At what angles \alpha between the force and the horizontal will the coil accelerate towards the taut thread?</title> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.7.20^*.$ A spool of thread lies on a horizontal plane. The reel is pulled by a thread. At what angles $\alpha$ between the force and the horizontal will the coil accelerate towards the taut thread? | | $2.7.20^*.$ A spool of thread lies on a horizontal plane. The reel is pulled by a thread. At what angles $\alpha$ between the force and the horizontal will the coil accelerate towards the taut thread? |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.7.20.png" | | <img src="2.7.20.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| For problem $2.7.20^*$ | | For problem $2.7.20^*$ |
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| <p> | | <p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.7.20_1.png" | | <img src="2.7.20_1.png" |
| loading="lazy" width="270" /> | | loading="lazy" width="270" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the spool | | Forces acting on the spool |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
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| The second Newton's law for the horizontal axis | | The second Newton's law for the horizontal axis |
| $$ma=F\cos\alpha-F_{fr}\quad(1)$$ | | $$ma=F\cos\alpha-F_{fr}\quad(1)$$ |
| Let's write a moment of inertia with respect to the axis of rotation of the spool with angular acceleration $\varepsilon$ | | Let's write a moment of inertia with respect to the axis of rotation of the spool with angular acceleration $\varepsilon$ |
| $$F_{fr}\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon\quad(2)$$ | | $$F_{fr}\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon\quad(2)$$ |
| The moment of inertia of the spool in our model (a solid cylinder, symmetry axis) | | The moment of inertia of the spool in our model (a solid cylinder, symmetry axis) |
| $$I=\frac{mR^2}{2}$$ | | $$I=\frac{mR^2}{2}$$ |
| Because there's no slippage: | | Because there's no slippage: |
| $$a=\varepsilon\cdot R\quad(3)$$ | | $$a=\varepsilon\cdot R\quad(3)$$ |
| Let's substitute $(3)$ into $(1)$ | | Let's substitute $(3)$ into $(1)$ |
| $$F_{fr}=F\cos\alpha-m\varepsilon R\quad(4)$$ | | $$F_{fr}=F\cos\alpha-m\varepsilon R\quad(4)$$ |
| and put into $(2)$ | | 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 | | 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 $(\varepsilon = 0)$ | | 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 $(\varepsilon = 0)$ |
| $$\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}>0$$ | | $$\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}>0$$ |
| $$R\cos\alpha-r>0\Leftrightarrow \boxed{\cos\alpha>\frac{r}{R}}$$ | | $$R\cos\alpha-r>0\Leftrightarrow \boxed{\cos\alpha>\frac{r}{R}}$$ |
| The larger the angle $\alpha$, the smaller the value of the trigonometric function $\cos\alpha$, making the expression for the angle | | The larger the angle $\alpha$, the smaller the value of the trigonometric function $\cos\alpha$, making the expression for the angle |
| $$\boxed{\alpha<\arccos\left(\frac{r}{R}\right)}$$ | | $$\boxed{\alpha<\arccos\left(\frac{r}{R}\right)}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\cos\alpha>\frac{r}{R}$$ | | $$\cos\alpha>\frac{r}{R}$$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
| </p> | | </p> |
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