Solved Problem on Dynamics

Solution 2 of the problem:

The equation of velocity as a function of position, v(x), is given by

\[ \begin{gather} \bbox[5px,border:5px solid #99CCFF] {v^2(x)=-4x^2+100} \tag{I} \end{gather} \]

Solution 3 of the problem:

The equations of motion and velocity are given as functions of time

\[ \begin{gather} x(t)=5\sin 2t \tag{II} \end{gather} \]

\[ \begin{gather} v(t)=10\cos 2t \tag{III} \end{gather} \]

solving the equation (II) for time t

\[ \begin{gather} t=\frac{1}{2}\arcsin \frac{x}{5} \end{gather} \]

substituting this value into equation (III), we obtain the equation of velocity as a function of position v(x)

\[ \begin{gather} v=10\cos 2t\\[5pt] v=10\cos\left(\cancel 2\times\frac{1}{\cancel 2}\arcsin \frac{x}{5}\right) \end{gather} \]

\[ \begin{gather} \bbox[5px,border:5px solid #99CCFF] {v(x)=10\cos\left(\arcsin \frac{x}{5}\right)} \tag{IV} \end{gather} \]

Let’s show that equation (IV) is equal to equation (I)..
In equation (IV), defining θ as

\[ \begin{gather} \theta=\arcsin \frac{x}{5}\\[5pt] \sin \theta=\frac{x}{5}\\[5pt] \sin \theta=\frac{\frac{x}{5}}{1} \end{gather} \]

θ is the angle of a right triangle with a hypotenuse equal to 1 and legs equal to \( \left(\frac{x}{5}\right) \) and b (Figure 1).
Applying the Pythagorean Theorem, we find the value of leg b

\[ \begin{gather} h^2=a^2+b^2\\[5pt] 1^2=\left(\frac{x}{5}\right)^2+b^2\\[5pt] b=\sqrt{1-\left(\frac{x}{5}\right)^2\;} \end{gather} \]

Figure 1

The cosine of angle θ is given by (Figure 2)

\[ \begin{gather} \cos\theta=\frac{b}{1}=\frac{\sqrt{1-\left(\frac{x}{5}\right)^2\;}}{1}\\[5pt] \cos\left(\arcsin \frac{x}{5}\right)=\sqrt{1-\frac{x^2}{25}\;} \tag{V} \end{gather} \]

Figure 2

Substituting the value of the cosine from (V) into equation (IV)

\[ \begin{gather} v=10\sqrt{1-\frac{x^2}{25}\;}\\[5pt] v^2=\left(10\sqrt{1-\frac{x^2}{25}\;}\right)^2\\[5pt] v^2=100\left(1-\frac{x^2}{25}\right)\\[5pt] v^2=100\left(\frac{25-x^2}{25}\right)\\[5pt] v^2=4(25-x^2)\\[5pt] \qquad\qquad v^2=100-4x^2\qquad (:2)\\[5pt] \frac{v^2}{2}=\frac{100}{2}-\frac{4x^2}{2}\\[5pt] \frac{v^2}{2}=\frac{-{4x^2}}{2}+50 \end{gather} \]

The trigonometric equation (IV) is equal to the algebraic equation (I), and they represent the same physical system.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .