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.

Solutions of Savchenko Physics Textbook

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>

    <p>
      $2.2.24^*$
      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.
    </p>
    <center>
      <figure>
        <img src="statement.png"
          loading="lazy"  width="230" />
        <figcaption>
          For problem $2.2.24^*$
        </figcaption>
      </figure>
    </center>

    <h3>Solution</h3>
    <p>
        </p>

Since point $1$ is at rest, it follows that the forces acting on it are compensated $$\vec{F}_{c1}=-\vec{T}_{1}$$ $$T_1=m_1\omega ^2 x$$ Where distance $x$ between point $1$ and center of mass $$x=l\frac{m_2}{m_1+m_2}$$

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 $2$ $$\omega = \frac{v}{l}$$ Now, substitute all of this into the expression for $T_1$ $$T_1=m_1\frac{v^2}{l^2} \cdot l\frac{m_2}{m_1+m_2}$$ $$T_1=\frac{m_1m_2v^2}{(m_1+m_2)l}$$ 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. $$T_1=T_2=T$$ $$\boxed{T=\frac{m_1m_2}{m_1+m_2}\frac{v^2}{l}}$$

    <h4>Answer</h4>
    <p>
        $$F=\frac{m_1m_2v^2}{(m_1+m_2)l}$$
    </p>
  <p style="text-align: right; font-style: italic; font-size: 16;">Almaskhan Arsen</p>
    <h3>Alternative solution</h3>
    <center>
      <figure>
        <img src="draw1.png"
          loading="lazy" width="250" />
        <figcaption>
            Forces acting on the system
        </figcaption>
      </figure>
    </center>
    <p>
      Let's consider this system as different bodies. Using Newton's Second law of motion, we can get:
      $$
      \begin{cases}
      \vec{F}_1 = m_1 \vec{a}_1 \\
      \vec{F}_2 = m_2 \vec{a}_2
      \end{cases}
      \Rightarrow
      \begin{cases}
      \vec{a}_1 = \frac{\vec{F}_1}{m_1} \\
      \vec{a}_2 = \frac{\vec{F}_2}{m_2}
      \end{cases}
      \quad \text{(1)}
      $$
      Let's subtract $\vec{a}_2$ from $\vec{a}_1$:
      $$
      \vec{a}_1 - \vec{a}_2 = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(2)}
      $$
    </p>
    <center>
      <figure>
        <img src="draw2.png"
          loading="lazy" width="180" />
        <figcaption>
          Subtraction of vectors
        </figcaption>
      </figure>
    </center>
    <p>
      As we can see, it looks like the derivative of relative velocity:
      $$
      \vec{v}_{B/A} = \vec{v}_B - \vec{v}_A \quad \text{(3)}
      $$
      Now, let's solve the derivative:
      $$
      \frac{d(\vec{v}_{B/A})}{dt} = \frac{d(\vec{v}_B)}{dt} - \frac{d(\vec{v}_A)}{dt} \Rightarrow \vec{a}_{B/A} = \vec{a}_B - \vec{a}_A \quad \text{(4)}
      $$
      This is why we can say that:
      <div class="scroll-wrapper">
        $$
        \vec{a}_1 - \vec{a}_2 = \vec{a}_{12} = \frac{\vec{F}_1}{m_1} - \frac{\vec{F}_2}{m_2} \quad \text{(5)}
        $$
      </div>
      Due to the weightlessness of the thread and Newton's Third Law of Motion $\vec{F}_1 = -\vec{F}_2 = \vec{F}$:

      Eventually:
      $$
      \vec{a}_{12} = \vec{F} \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = \vec{F} \frac{m_1 + m_2}{m_1 m_2} \quad \text{(6)}
      $$
      $a_{12} = \frac{v^2}{l}$, where $l$ is the length of the thread, and $v$ is the relative velocity$\quad (7)$
      $$
      F = \frac{m_1 m_2}{m_1 + m_2} \frac{v^2}{l} \quad \text{(8)}
      $$
    </p>
  <p style="text-align: right; font-style: italic; font-size: 16;">physicshub</p>
    


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