Правка разделов «Statement», «Solution», «Answer»

Jorge Llorente 08 правка от
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en/3.1.13.md +83 −78
@@ -1,103 +1,108 @@
### Statement
−$3.1.13.$ [Insert the problem statement]
+$3.1.13.$ A small weight of mass ∆m is placed on the body of mass m in problem 3.1.12
+(a) and released. Find the maximal velocity of the body with the weight.
+
### Solution
−![For problem $3.1.13$ |1054x697, 31%](../../img/3.1.13/Imagen1.png)
+According to Problem 3.1.13, in relation to Problem 3.1.12, we find the following configuration: a small charged body of mass \(m\) can slide freely on a vertical spoke, at the lower point of which there is a charge of the same sign as the charge of the body. The equilibrium position of the body is at a distance \(R\) from this charge. Then a small mass \(\Delta m\) is added jointly to the charged body of mass \(m\).
−According to Problems~3.1.12 and~3.1.13, we consider the following configuration.
+Evidently, this will alter the value of \(R\) in the interaction of forces, first established to be:
−A small charged body of mass $m$ can slide freely on a vertical spoke. At the lower end of the spoke there is a fixed charge of the same sign as that of the body. The equilibrium position of the body is located at a distance $R$ from the fixed charge. A small additional mass $\Delta m$ is then rigidly attached to the charged body.
−This modification alters the equilibrium distance. Initially, the balance of forces gives
−\begin{equation}
−\frac{k Q q}{r^2} = mg,
−\end{equation}
−from which
−\begin{equation}
−r = \sqrt{\frac{k Q q}{m g}}.
−\end{equation}
+\[
+K\frac{Qq}{r^{2}} = mg
+\qquad \Rightarrow \qquad
+r = \sqrt{\frac{KQq}{mg}} .
+\]
−After adding the mass $\Delta m$, the equilibrium condition becomes
−\begin{equation}
−\frac{k Q q}{r'^2} = (m + \Delta m) g,
−\end{equation}
−leading to
−\begin{equation}
−r' = \sqrt{\frac{k Q q}{(m + \Delta m) g}}.
−\end{equation}
−The simplest way to solve the problem is by using the principle of conservation of mechanical energy, since the system is conservative:
−\begin{equation}
−\Delta \Sigma E = 0.
−\end{equation}
−Thus,
−\begin{equation}
−E_{\text{el}}(r) + E_{\text{grav}}(r)
−=
−E_{\text{el}}(r') + E_{\text{grav}}(r') + E_{\text{kin}}.
−\end{equation}
+After adding the mass:
−Expanding each term, we obtain
−\begin{equation}
−(m+\Delta m) g r + \frac{k Q q}{r}
−=
−(m+\Delta m) g r' + \frac{k Q q}{r'} + \frac{1}{2}(m+\Delta m) v^2.
−\end{equation}
−Substituting the expression
\[
−r' = \sqrt{\frac{k Q q}{(m + \Delta m) g}},
+K\frac{Qq}{r'^2} = (m+\Delta m)g
+\qquad \Rightarrow \qquad
+r' = \sqrt{\frac{KQq}{(m+\Delta m)g}} .
\]
−the equation becomes
−\begin{equation}
−(m+\Delta m) g r + \frac{k Q q}{r}
−=
−(m+\Delta m) g \sqrt{\frac{k Q q}{(m + \Delta m) g}}
−+
−\frac{k Q q}{\sqrt{\frac{k Q q}{(m + \Delta m) g}}}
−+
−\frac{1}{2}(m+\Delta m) v^2.
−\end{equation}
−Solving for the velocity, the maximum speed is found to be
−\begin{equation}
−v_{\max}
−=
−\sqrt{
−2 g R
−\left(
−1 + \frac{m}{m+\Delta m}
−- 2 \sqrt{\frac{m}{m+\Delta m}}
−\right)
−}.
−\end{equation}
+The simplest way to solve this problem is with the Universal Energetic Conservation Law, by which \(\Delta \Sigma m = 0\), allowing:
−This result can also be justified by analyzing the motion itself. Since no dissipative forces (such as air resistance) are present, the system is conservative, and the motion is therefore simple harmonic motion (SHM).
+\[
+E_{el,pot}(R) + E_{gr,pot}(R) = E_{el,pot}(R') + E_{gr,pot}(R') + E_{kin}
+\]
−For SHM, the velocity is given by
−\begin{equation}
−v = \omega \sqrt{A^2 - x^2},
−\end{equation}
−where $A$ is the amplitude and $\omega$ the angular frequency. The velocity reaches its maximum value at the equilibrium position, where $x = 0$, yielding
−\begin{equation}
−v_{\max} = A \omega.
−\end{equation}
−The amplitude corresponds to the difference between the original equilibrium position and the new one:
−\begin{equation}
−A = r - r'
−= \sqrt{\frac{k Q q}{m g}}
−-
−\sqrt{\frac{k Q q}{(m + \Delta m) g}}.
−\end{equation}
−Substituting this expression into the SHM formulas yields an equivalent result to that obtained via energy conservation. Although this approach is equally valid, it requires a Taylor expansion, which unnecessarily complicates the analysis.
+Expanding:
+\[
+(m+\Delta m) g r + \frac{KQq}{r} = (m+\Delta m) g r' + \frac{KQq}{r'} + \frac{(m+\Delta m)v^{2}}{2}
+\]
+
+
+Since
+
+
+\[
+r' = \sqrt{\frac{KQq}{(m+\Delta m)g}},
+\]
+
+
+the equation becomes:
+
+\[
+(m+\Delta m) g r + \frac{KQq}{r} = (m+\Delta m) g \sqrt{\frac{KQq}{(m+\Delta m)g}} + \frac{KQq}{\sqrt{\frac{KQq}{(m+\Delta m)g}}} + \frac{(m+\Delta m)v^{2}}{2}
+\]
+
+
+
+Solving for \(v\), the maximum velocity is:
+
+
+\[
+v_{max} = \sqrt{ 2gR \left( 1 + \frac{m}{m+\Delta m} - 2\sqrt{\frac{m}{m+\Delta m}} \right) }
+\]
+
+
+
+This can be proved by examining the motion it will produce. As no air drag or other forces act, we are working with a conservative force field, implying the body will describe an unrestrained Simple Harmonic Motion (SHM). According to the SHM velocity formula:
+
+
+\[
+v = \omega \sqrt{A^{2} - x^{2}},
+\]
+
+
+the maximum speed occurs at \(x=0\), so:
+
+
+\[
+v_{\max} = A\omega .
+\]
+
+
+
+The amplitude is the difference between the original equilibrium position and the new one:
+
+
+\[
+A = r - r' = \sqrt{\frac{KQq}{mg}} - \sqrt{\frac{KQq}{(m+\Delta m)g}}
+\]
+
+
+Substituting this into the SHM formulas yields an equivalent expression to the one obtained using the Universal Energy Conservation Law, although doing so directly would require a Taylor expansion, complicating the problem.
+
+
+
#### Answer
−[Insert a concise answer or boxed result]
+![For problem $3.1.13$|1054x697, 50%](../../img/3.1.13/Imagen1.png)
+
+\[
+v_{max} = \sqrt{ 2gR \left( 1 + \frac{m}{m+\Delta m} - 2\sqrt{\frac{m}{m+\Delta m}} \right) }
+\]