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| <h2>Solutions of Savchenko Physics Textbook</h2> | | <h2>Solutions of Savchenko Physics Textbook</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../#2.2">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| | | |
| <p> | | <p> |
| $2.2.24^*$ | | $2.2.24^*$ |
| Two bodies of mass $m_1$ and $m_2$ are connected by a stretched thread of length $l$ and move along a smooth horizontal surface. At some point in time, it turned out that the first body is stationary, and the velocity of the second body, equal to $v$, is perpendicular to the thread. Determine the tension force of the thread. | | Two bodies of mass $m_1$ and $m_2$ are connected by a stretched thread of length $l$ and move along a smooth horizontal surface. At some point in time, it turned out that the first body is stationary, and the velocity of the second body, equal to $v$, is perpendicular to the thread. Determine the tension force of the thread. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $2.2.24^*$ | | For problem $2.2.24^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw4.png" | | <img src="draw4.png" |
| loading="lazy" width="170" /> | | loading="lazy" width="170" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Since point $1$ is at rest, it follows that the forces acting on it are compensated | | Since point $1$ is at rest, it follows that the forces acting on it are compensated |
| $$\vec{F}_{c1}=-\vec{T}_{1}$$ | | $$\vec{F}_{c1}=-\vec{T}_{1}$$ |
| $$T_1=m_1\omega ^2 x$$ | | $$T_1=m_1\omega ^2 x$$ |
| Where distance $x$ between point $1$ and center of mass | | Where distance $x$ between point $1$ and center of mass |
| $$x=l\frac{m_2}{m_1+m_2}$$ | | $$x=l\frac{m_2}{m_1+m_2}$$ |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw3.png" | | <img src="draw3.png" |
| loading="lazy" width="170" /> | | loading="lazy" width="170" /> |
| <figcaption> | | <figcaption> |
| Direction of forces and velocity of the centre of mass | | Direction of forces and velocity of the centre of mass |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Since point $1$ is at rest, the motion is around it. Then the angular velocity of rotation is found through the velocity $v$ of the point $2$ | | Since point $1$ is at rest, the motion is around it. Then the angular velocity of rotation is found through the velocity $v$ of the point $2$ |
| $$\omega = \frac{v}{l}$$ | | $$\omega = \frac{v}{l}$$ |
| Now, substitute all of this into the expression for $T_1$ | | Now, substitute all of this into the expression for $T_1$ |
| $$T_1=m_1\frac{v^2}{l^2} \cdot l\frac{m_2}{m_1+m_2}$$ | | $$T_1=m_1\frac{v^2}{l^2} \cdot l\frac{m_2}{m_1+m_2}$$ |
| $$T_1=\frac{m_1m_2v^2}{(m_1+m_2)l}$$ | | $$T_1=\frac{m_1m_2v^2}{(m_1+m_2)l}$$ |
| According to Newton's third law, since the thread is weightless, the absolute value of tension force of the thread at point $1$ and point $2$ are equal. | | According to Newton's third law, since the thread is weightless, the absolute value of tension force of the thread at point $1$ and point $2$ are equal. |
| $$T_1=T_2=T$$ | | $$T_1=T_2=T$$ |
| $$\boxed{T=\frac{m_1m_2}{m_1+m_2}\frac{v^2}{l}}$$ | | $$\boxed{T=\frac{m_1m_2}{m_1+m_2}\frac{v^2}{l}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <img src="draw1.png" | | <img src="draw1.png" |
| loading="lazy" width="250" /> | | loading="lazy" width="250" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the system | | Forces acting on the system |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Let's consider this system as different bodies. Using Newton's Second law of motion, we can get: | | Let's consider this system as different bodies. Using Newton's Second law of motion, we can get: |
| $$ | | $$ |
| \begin{cases} | | \begin{cases} |
| \vec{F}_1 = m_1 \vec{a}_1 \\ | | \vec{F}_1 = m_1 \vec{a}_1 \\ |
| \vec{F}_2 = m_2 \vec{a}_2 | | \vec{F}_2 = m_2 \vec{a}_2 |
| \end{cases} | | \end{cases} |
| \Rightarrow | | \Rightarrow |
| \begin{cases} | | \begin{cases} |
| \vec{a}_1 = \frac{\vec{F}_1}{m_1} \\ | | \vec{a}_1 = \frac{\vec{F}_1}{m_1} \\ |
| \vec{a}_2 = \frac{\vec{F}_2}{m_2} | | \vec{a}_2 = \frac{\vec{F}_2}{m_2} |
| \end{cases} | | \end{cases} |
| \quad \text{(1)} | | \quad \text{(1)} |
| $$ | | $$ |
| Let's subtract $\vec{a}_2$ from $\vec{a}_1$: | | Let's subtract $\vec{a}_2$ from $\vec{a}_1$: |
| $$ | | $$ |
| \vec{a}_1 - \vec{a}_2 = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(2)} | | \vec{a}_1 - \vec{a}_2 = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(2)} |
| $$ | | $$ |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw2.png" | | <img src="draw2.png" |
| loading="lazy" width="180" /> | | loading="lazy" width="180" /> |
| <figcaption> | | <figcaption> |
| Subtraction of vectors | | Subtraction of vectors |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| As we can see, it looks like the derivative of relative velocity: | | As we can see, it looks like the derivative of relative velocity: |
| $$ | | $$ |
| \vec{v}_{B/A} = \vec{v}_B - \vec{v}_A \quad \text{(3)} | | \vec{v}_{B/A} = \vec{v}_B - \vec{v}_A \quad \text{(3)} |
| $$ | | $$ |
| Now, let's solve the derivative: | | Now, let's solve the derivative: |
| $$ | | $$ |
| \frac{d(\vec{v}_{B/A})}{dt} = \frac{d(\vec{v}_B)}{dt} - \frac{d(\vec{v}_A)}{dt} \Rightarrow \vec{a}_{B/A} = \vec{a}_B - \vec{a}_A \quad \text{(4)} | | \frac{d(\vec{v}_{B/A})}{dt} = \frac{d(\vec{v}_B)}{dt} - \frac{d(\vec{v}_A)}{dt} \Rightarrow \vec{a}_{B/A} = \vec{a}_B - \vec{a}_A \quad \text{(4)} |
| $$ | | $$ |
| This is why we can say that: | | This is why we can say that: |
| <div class="scroll-wrapper"> | | <div class="scroll-wrapper"> |
| $$ | | $$ |
| \vec{a}_1 - \vec{a}_2 = \vec{a}_{12} = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(5)} | | \vec{a}_1 - \vec{a}_2 = \vec{a}_{12} = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(5)} |
| $$ | | $$ |
| </div> | | </div> |
| Due to the weightlessness of the thread and Newton's Third Law of Motion $\vec{F}_1 = -\vec{F}_2 = \vec{F}$: | | Due to the weightlessness of the thread and Newton's Third Law of Motion $\vec{F}_1 = -\vec{F}_2 = \vec{F}$: |
| | | |
| Eventually: | | Eventually: |
| $$ | | $$ |
| \vec{a}_{12} = \vec{F} \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = \vec{F} \frac{m_1 + m_2}{m_1 m_2} \quad \text{(6)} | | \vec{a}_{12} = \vec{F} \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = \vec{F} \frac{m_1 + m_2}{m_1 m_2} \quad \text{(6)} |
| $$ | | $$ |
| $a_{12} = \frac{v^2}{l}$, where $l$ is the length of the thread, and $v$ is the relative velocity$\quad (7)$ | | $a_{12} = \frac{v^2}{l}$, where $l$ is the length of the thread, and $v$ is the relative velocity$\quad (7)$ |
| $$ | | $$ |
| F = \frac{m_1 m_2}{m_1 + m_2} \frac{v^2}{l} \quad \text{(8)} | | F = \frac{m_1 m_2}{m_1 + m_2} \frac{v^2}{l} \quad \text{(8)} |
| $$ | | $$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p> | | <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p> |
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