0
$\begingroup$

I am currently working on a problem where a particle of mass $m$ can move along the x-axis with no friction. The particle is moving in a gravitational field and the potential energy is given by:

$$E_{pot(x)}=-2Cm\frac{1}{\sqrt{x^2+(ab)^2}}$$

I want to find the equation of motion for this particle by using the law of conservation of energy.

$$E_{pot(x)}+E_{kin(\dot{x})}=constant$$

$$\iff -2Cm\frac{1}{\sqrt{x^2+(ab)^2}}+\frac{1}{2}m(\dot{x})^2=constant$$

I didn't know how to deal with the constant term so I just used the fact that the derivative of any constant is $0$ and differentiated both sides.

$$\implies \frac{\partial}{\partial x}(-2Cm\frac{1}{\sqrt{x^2+(ab)^2}})+\frac{\partial}{\partial x}(\frac{1}{2}m(\dot{x})^2)=0$$

$$\iff Cm\frac{2x\color{red}{\dot{x}}}{(\sqrt{x^2+(ab)^2})^3}+m\dot{x}\color{red}{\ddot{x}}=0$$

Question 1: Am I differentiating correctly here? I am not sure if I have to implicitly differentiate $x$ and $\dot{x}$

Question 2: Does my approach even make sense or am I just wasting my time here?

Edit:

$$m\dot{x}\space (\ddot{x}+\frac{C}{(x^2+(ab)^2)^\frac{3}{2}}x)=0$$

$$\iff \ddot{x}+\frac{C}{(x^2+(ab)^2)^\frac{3}{2}}x=0$$

$\endgroup$
  • $\begingroup$ How do you get $\dot{x}$ in the first term if you are taking the derivatives of $x$? $\endgroup$ – Kyle Kanos Apr 21 '15 at 17:11
  • $\begingroup$ @KyleKanos I differentiated $x^2$ under the square root. This is why I was asking if I need to implicitly differentiate because then I would get $2x\cdot \dot{x}$ $\endgroup$ – qmd Apr 21 '15 at 17:14
  • $\begingroup$ That doesn't make sense. What is $\dot{x}$? $\endgroup$ – Kyle Kanos Apr 21 '15 at 17:17
  • $\begingroup$ I would say that is the velocity. $\endgroup$ – qmd Apr 21 '15 at 17:18
  • $\begingroup$ So how do you get a velocity ($dx/dt$) when taking the derivative with respect to space? $\endgroup$ – Kyle Kanos Apr 21 '15 at 17:18
1
$\begingroup$

You actually don't need to take any time derivatives here. Since the energy, $E$, is a constant, you only need to know it at one moment in time (say at t=0 as an initial condition), and you know it at all other times. Thus, just solve for $\dot{x}(t)$ in terms of the other quantities ($x,E,...)$ in the second equation that you have to get the equation of motion. Like all differential equations, you have to supply initial conditions to get the particular solution or trajectory.

$\endgroup$
  • $\begingroup$ Thanks. So I don't need the second derivative in my equation at all? I always thought they have to be in the form $\ddot{x}+Cx=0$ $\endgroup$ – qmd Apr 21 '15 at 18:43
  • $\begingroup$ An equation of motion is really just any equation involving the trajectory x(t) and its derivatives. $\endgroup$ – mr blick Apr 21 '15 at 18:46
  • $\begingroup$ But when you say "its derivatives", does it have to be all of them ($\ddot{x}(t), \dot{x}(t)$) or can it be just the first derivative $\dot{x}(t)$ $\endgroup$ – qmd Apr 21 '15 at 19:00
  • $\begingroup$ It can just be the first one. Really any combination is an equation of motion. $\endgroup$ – mr blick Apr 21 '15 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.