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| + | <meta name="description" content="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?"> | ||
| + | <meta name="author" content="Aliaksandr Melnichenka"> | ||
| + | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
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| + | <header style="text-align:center;"> | ||
| + | <h2>Solutions of Savchenko Problems in Physics</h2> | ||
| + | <p class="author"> | ||
| + | Aliaksandr Melnichenka <br/> | ||
| + | October 2023 | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <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? | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="2.7.20.png" | ||
| + | loading="lazy" width="200" /> | ||
| + | <figcaption> | ||
| + | For problem $2.7.20^*$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | </p> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | |||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="2.7.20_1.png" | ||
| + | loading="lazy" width="270" /> | ||
| + | <figcaption> | ||
| + | Forces acting on the spool | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | |||
| + | 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 $\varepsilon$ | ||
| + | $$F_{fr}\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon\quad(2)$$ | ||
| + | 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: | ||
| + | $$a=\varepsilon\cdot R\quad(3)$$ | ||
| + | Let's substitute $(3)$ into $(1)$ | ||
| + | $$F_{fr}=F\cos\alpha-m\varepsilon R\quad(4)$$ | ||
| + | and put into $(2)$ | ||
| + | $$\left(F\cos\alpha-m\varepsilon R\right)\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon$$ | ||
| + | After algebraic transformations | ||
| + | $$F\left(R\cos\alpha-r\right)=\frac{3mR^2\varepsilon}{2} \Leftrightarrow \boxed{\varepsilon=\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}} \quad(5)$$ | ||
| + | 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 | ||
| + | $$\boxed{\alpha<\arccos\left(\frac{r}{R}\right)}$$ | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer</h4> | ||
| + | <p> | ||
| + | $$\cos\alpha>\frac{r}{R}$$ | ||
| + | </p> | ||
| + | |||
| + | |||
| + | <footer class="row container"> | ||
| + | <br> | ||
| + | <p> | ||
| + | <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | ||
| + | </p> | ||
| + | <p> | ||
| + | <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | ||
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| <html lang="en"> | |||
| <head> | |||
| <meta charset="utf-8"> | |||
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |||
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| <meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda"> | |||
| <meta name="description" content="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?"> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="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?"> | |||
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| </head> | |||
| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <h2>Solutions of Savchenko Problems in Physics</h2> | |||
| <p class="author"> | |||
| Aliaksandr Melnichenka <br/> | |||
| October 2023 | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <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? | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="2.7.20.png" | |||
| loading="lazy" width="200" /> | |||
| <figcaption> | |||
| For problem $2.7.20^*$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| </p> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="2.7.20_1.png" | |||
| loading="lazy" width="270" /> | |||
| <figcaption> | |||
| Forces acting on the spool | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| 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 $\varepsilon$ | |||
| $$F_{fr}\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon\quad(2)$$ | |||
| 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: | |||
| $$a=\varepsilon\cdot R\quad(3)$$ | |||
| Let's substitute $(3)$ into $(1)$ | |||
| $$F_{fr}=F\cos\alpha-m\varepsilon R\quad(4)$$ | |||
| and put into $(2)$ | |||
| $$\left(F\cos\alpha-m\varepsilon R\right)\cdot R - F\cdot r = \frac{mR^2}{2}\cdot \varepsilon$$ | |||
| After algebraic transformations | |||
| $$F\left(R\cos\alpha-r\right)=\frac{3mR^2\varepsilon}{2} \Leftrightarrow \boxed{\varepsilon=\frac{2F\left(R\cos\alpha-r\right)}{3mR^2}} \quad(5)$$ | |||
| 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 | |||
| $$\boxed{\alpha<\arccos\left(\frac{r}{R}\right)}$$ | |||
| </p> | |||
| <h4>Answer</h4> | |||
| <p> | |||
| $$\cos\alpha>\frac{r}{R}$$ | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||