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:

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

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

Improve this question
$\endgroup$
9
  • $\begingroup$ How do you get $\dot{x}$ in the first term if you are taking the derivatives of $x$? $\endgroup$
    – Kyle Kanos
    Apr 21, 2015 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, 2015 at 17:14
  • $\begingroup$ That doesn't make sense. What is $\dot{x}$? $\endgroup$
    – Kyle Kanos
    Apr 21, 2015 at 17:17
  • $\begingroup$ I would say that is the velocity. $\endgroup$
    – qmd
    Apr 21, 2015 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, 2015 at 17:18

1 Answer 1

Reset to default
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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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, 2015 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, 2015 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, 2015 at 19:00
  • $\begingroup$ It can just be the first one. Really any combination is an equation of motion. $\endgroup$
    – mr blick
    Apr 21, 2015 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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