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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.7">$\leftarrow$Back</a></h3> |
| | | |
| <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> |
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| <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 $2.7.20^*$ |
| </figcaption> | | </figcaption> |
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| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| | | |
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| <center> | | <center> |
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| <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> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| 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 $(\varepsilon = 0)$ |
