Thinking outside the line

Initially, the kinetic energy of the particle is ( $5^2)/2 = 12.5$ J while the potential energy of the particle is V(1) = 1 J. Hence the total energy of the particle initially is 13.5J. By the conservation of energy, this total energy remains constant throughout since there is no loss of energy to the surroundings.

Now we consider the particle at some arbitrary time t. Its kinetic energy is equal to its total energy less its potential energy. We can also express its kinetic energy using the formula $mv^2/2$ . Equating these two expressions, we have $(v^2)/2 = 13.5 - V(x)$ , so $(v^2)/2 = 13.5 - 1/(x^2)$ . Rearranging this equation gives us $v = x/\sqrt{27x^2 - 2}$ . Since its initial velocity is positive, we take v to be positive.

The relationship between v and x is given by v = dx/dt. Substituting this into our previous equation, we have $dx/dt = x/\sqrt{27x^2 - 2}$ . We proceed to solve this differential equation, keeping in mind that x = 1 when t = 0. This gives us $t = (\sqrt{27x^2 - 2} - 5)/27$ .

Therefore, when t = 5, we will have $x^2 = 726$ . Taking x to be positive (since the initial position is positive), we have x = 26.94.

The total energy E of this system is just the sum of the potential and kinetic energies, so $E = \frac{1}{x^{2}} + \frac {mv^{2}}{2} = \frac{1}{x^{2}} + \frac {m}{2} \left( \frac{dx}{dt} \right)^{2}$ . E is constant (conservation of energy) and can be calculated from the initial conditions. So we have a 1st order differential equation, and our goal is to solve for $x \left( t \right)$ . Isolating the above equation for $\frac {dx}{dt}$ gives

$\frac {dx}{dt}$ = $\sqrt {\frac {{2}}{{{m}}}}\sqrt {{\frac {E{x}^{2}-1}{{x}^{2}}}}$

The next step is to isolate the x's and t's and integrate. Doing so gives

$\int \sqrt {{\frac {{x}^{2}}{E{x}^{2}-1}}} {dx} = \int \sqrt{ \frac{2}{m}} {dt}$

The right hand side easily evaluates to $t\sqrt{\frac{2}{m}} + C$ , where C is our constant of integration. We can evaluate the integral on the left by making the change of variable $x = \frac{sin(u)}{\sqrt{E}}$ . Then, $dx = \frac{cos(u)}{\sqrt{E}} {du}$ and our integral becomes $\frac{1}{E} \int \sqrt {{\frac { \sin^{2}(u)}{-(1-\sin^{2}(u))}}} \cos \left( u \right) {du}$ . Using the Pythagorean identity, this simplifies to $\frac{1}{E} \int \ \sqrt{-\sin^{2}(u)} {du}$ or $\frac{1}{E} \int \ i\sin(u) {du}$ where $i = \sqrt{-1}$

This integral is easy, and evaluates to $\frac{-i\cos(u)}{E}$ . Now, if $sin(u)= x\sqrt{E}$ (our original change of variables) then $\cos(u) = \sqrt{1-Ex^{2}}$ , and our expression becomes $\frac {-i \sqrt{1-Ex^{2}}}{E}$ . Pulling out an i from the radical gives $\frac {-i^{2} \sqrt{Ex^{2} -1 }}{E} = \frac {\sqrt{Ex^{2} -1 }}{E}$ .

Going all the way back to our original equation with t, we finally have $\frac {\sqrt{Ex^{2} -1 }}{E} = t\sqrt{\frac{2}{m}} + C$ . We decide to plug in our initial conditions before solving for x, in order to simplify the algebra. x = 1 m when v = 5 m/s, so $E = 1 + \frac{25}{2} = \frac{27}{2}$ J. Also, when t = 0, x = 1 m and v = 5 m/s also. Substituting this information into our solution gives $C = \frac {\sqrt{\frac{27}{2}1^{2} -1 }}{\frac{27}{2}} = \frac{\sqrt{50}}{27}$ . Now that we have E and C, we need to solve our solution for x. After some algebra, we obtain $x(t) = \sqrt {1+27\,{t}^{2}+10\,t}$ , which is the explicit time dependence of x we are looking for. At t = 5 we have $x(5) = \sqrt{726} \approx26.94$ meters, which is our answer.

F = 2 / (x^3) m (V dV/dx)=2/(x^3) m=1 kg; so, VdV=2 (dx/(x^3)) integrating on both sides, we get, (V^2)/2=(-1)/(x^2)

applying limits,
V=5 when x=1, we get, (V^2)-25=2-(2/(x^2))

or,

V = \surd((27(x^2)-2)/ (x^2)) writing V as dx/dt, and integrating on both sides, we get, (\surd((27*(x^2))+2))/27=t

applying limits, x=1 when t=0 and x=x when t=5 we get x=26.94438

The question is really simple if you work on the basis of conservation of energy. At x=1 we have the sum of Potential and Kinetic Energy to be equal to 27\2 At any other point x,consider the velocity to be V and the potential energy will be 1\x^2 So,from conservation of energy we have (V^2)\2 +1\x^2=27\2 So,V^2=27-(2\x^2) Write V as dx\dt and integrate keeping limits of x as x=1 to x=x and t=0 till t=5 Solve the equation that you get,and you will get the answer to be around 27metres. Note: The velocity was in the forward direction,so we have assumed x to be increasing.

Conversation of energy: V(x)+K(x)=Const => \­(\frac{1}{x^2}+\frac{v^2}{2}=\frac{1}{x 0^2}+\frac{v 0^2}{2}=13.5) => $v=sqrt{2(13.5-\frac{1}{x^2}}$ We obtain the curve of v(x). By observing this curve, we see that the speed after the initial point (1,5) is approximately fixed at value 5.2 m/s. Therefore, the distance that the point moves after 5s is L= 5x5.2=26 (m). And we obtain the postion of the particle at t=5s: x=x0+L=27 (m)

Before we try to solve the above problem in a creative way with an extra dimension, let's review the method of reducing the dimension in a central potential (which is the inverse of the technique for this problem and more familiar). We start with 2-dimensional polar coordinates. Since a central potential is only depends on the radial distance between the object and the center of the field (homogeneity in the angular part), any rotational transformation that rotates the whole space around the field's center will not change the dynamical properties of the problem , so we have a symmetry (rotational symmetry) in the angular dimension. We can therefore eliminate the angular dimension, since one can store the information about the motion of the object in that dimension by the conserved quantity that pops out from the symmetry. In the above case the angular momentum $L=mv_{\theta}r$ ( $v_{\theta}=r \dot{\theta}$ is the angular component of the velocity) around the field's center is conserved.

In order to keep the dynamical properties in the remaining radial dimension the same as in the original problem, we have to change the equation of the potential $U(r)$ a bit - adding the energy that's stored inside the eliminated angular dimension $\Delta E_{\theta}$ and then the whole thing will be given a name: the effective potential $U_{eff}(r)=U(r)+\Delta E_{\theta}$ . Since the potential does not depend on the angle, the angular dimension only stores kinetic energy. The kinetic energy of an object in a given dimension is equal to the square of the canonical momentum in that dimension divide by 2 times the inertia property of that dimension. For angular dimension, the canonical momentum is the angular momentum $L$ and the inertial property of that dimension is nothing else but the moment of inertial around the field's center $I=mr^2$ , and we get $\Delta E_{\theta} = L^2/2I = L^2/2mr^2$ . The equation for the effective potential is exactly the same as in some standard Mechanics textbooks $U_{eff}(r)=U(r)+\frac{L^2}{2mr^2}$ .

Note that the derivation we use for that equation is purely based on theoretical arguments in Physics, which is different than the normal method of force's integration along the radial direction (mathematical based). Also, we can see how can we store information of a symmetric dimension in its conserved quantity: the possible values of $L$ itself create something similar to a "dimension" ( $L$ can be any real value), and in general, different $L$ means different $U_{eff}(r)$ and different dynamical property - so there's no symmetry in that "dimension". Theoretically, we just choose a new "dimension" to describe our problem: we change from a symmetric dimension (with the value of the position of the object in this dimension $\theta$ is changed) to a non-symmetric "dimension" (with the value of the position of the object in this "dimension" $L$ is unchanged).

Let's go back to our 1-dimensional problem and ask ourselves a question: what if this problem is just a modification of a 2-dimensional central potential problem (with $x$ corresponds to the radial dimension $r$ , as $x=r$ ) and the given potential is nothing else but an effective potential? Let's try to add one more dimension with one more symmetry (angular symmetry), and I will show you how to make a more simple problem out of that method. Because of the form of our 1-dimensional potential is positive ( $A>0$ ) and proportional to $1/x^2$ , one can choose the angular momentum of the 2-dimensional central potential problem to satisfy $L=\sqrt{2mA}$ ( $L$ is unchanged with time, as expected from a conservation that comes out from the symmetric nature of the angular dimension) so that the bare potential in 2 dimensions is just equal to 0, as the object is just keep moving straight with constant velocity:

$U(r)=U_{eff}(r)-\frac{L^2}{2mr^2}=\frac{A}{r^2}-\frac{L^2}{2mr^2}=0$

The initial velocity in the angular direction of the object in 2-dimensional problem can be found from the initial conditions in the radial direction $r_o=x_o$ , $\dot{r_o}=\dot{x_o}$ :

$v_{\theta o}=\frac{L}{mr_o}=\frac{\sqrt{2mA}}{mx_o}=\sqrt{\frac{2A}{mx_o^2}}$

The motion of the object in the new 2-dimensional problem is just a straight line, so one can easily write down the equation of the distance $r$ between it and the field's center (which is exactly the equation for the position $x$ in the original 1-dimensional problem):

$r(t)=\sqrt{(r_o + \dot{r_o} t)^2+(v_{\theta o}t)^2}=\sqrt{(x_o + \dot{x_o} t)^2+\frac{2A}{mx_o^2} t^2}$

$\Rightarrow x(t)=\sqrt{(x_o + \dot{x_o} t)^2+\frac{2A}{mx_o^2} t^2} = 26.94(m)$

As we can see, without any fancy calculations, we solve the problem quickly and elegantly.

(Some addition)

Although adding dimensions and symmetries may simplify the original given problem, but it can only be used in some special cases. For example, if the potential in our 1-dimensional problem is negative ( $A<0$ ), one can not use the same trick to try to solve it, since the imaginary value of the angular momentum ( $L=\sqrt{2mA}=i\sqrt{2m|A|}$ )doesn't have any physical meaning and cannot be created. Despite the fact that we can hardly apply this method in Classical Mechanics, the similar method - the Holographic method - is quite popular in Theoretical Physics. One of the examples for increasing dimensions and imposing additional symmetries to make a problem easier is how to calculate the conductivity of a given material (basically, finding the correlation between currents) with different fermionic species in some extreme cases of String Theory (usually with the number of species goes to infinity, and the coupling constant is very high - about the same as the coupling constant in graphene, which is $\approx 2.5$ ) - basically you put in the problem one more dimension and a new $U(1)$ gauge symmetry, and with the conserved current that pops out from that gauge symmetry, one can rewrite and simplify the dynamics. The basic idea, as we can clearly see, is pretty much the same with our Classical Mechanics problem in this paper. Beside the goal to show the readers how to modify a special Classical Mechanics problem by increase dimension and solve it with ease, this paper can also serve the purpose to show what is the theoretical meaning of adding extra dimensions and symmetries.