0
$\begingroup$

I was trying to obtain the brachistochrone as a function of time, and I failed several times because I wrongly assumed that the $v=\sqrt{2gy}$ vector points downwards (vertical vector). However, in order to get the solution, the correct assumption was that this vector was the tangential velocity vector, the one pointing in the direction of movement through the curve (as I assumed it was instead vertical, I wrongly stated that $v_T = v \sin{(\arctan{\frac{dy}{dx}})}$). So I don't understand why it is actually $v_T=v$. I guess I had my reasons to believe $v$ was the vertical vector because $$\frac{dv}{dt}= \frac{dv}{dy} \frac{dy}{dt} = \frac{dv}{dy} v_y = g \Rightarrow \\ v_y dv = g \cdot dy, v_{y_0} = 0, y_0=0 \Rightarrow \frac{1}{2} v_y^2 = gy \Rightarrow v_y=\sqrt{2gy}$$ and you also get to $v=\sqrt{2gy}$ by conservation of energy. So that's why I assumed $v=v_y$. As I said, this should be wrong. Can someone then explain why is $v$ the tangential velocity so that $v^2 = v_x^2 + v_y^2 , v_x \neq 0$?

Edit: Thanks for the answers. Both helped me. If anyone is interested I'll write my conclusions here. So first I was told that the velocity is always pointing towards the direction of motion, so it makes sense for it to be tangential to the curve. But this made me wonder what I did wrong in my calculations. It was indeed wrong as someone stated that the force applied was not vertical. The force I have to consider is $mg \sin{\alpha}$, where $\alpha= \arctan{\frac{dy}{dx}}$, so this sine becomes $\frac{dy}{\sqrt{dx^2 + dy^2}}= \frac{dy}{ds}$, where s is the arc length. Then, by Newton's second law, $$m \frac{dv}{dt} = m \frac{dv}{ds} \frac{ds}{dt} = m \frac{dv}{ds} v = mg \frac{dy}{ds} \Rightarrow v \cdot dv = g \cdot dy \Rightarrow \\ \Rightarrow \dots \Rightarrow v= \sqrt{2gy}$$ And we got the same result as from energy-conservation. This one made it clear for me that this was indeed the tangential velocity, as I used the $\frac{ds}{dt}$ definition.

$\endgroup$
2
$\begingroup$

I am not entirely sure, but I think you made two wrong assumptions in your calculations. First of all, dv/dt is not equal to g. In general, for an object in free fall, the derivative of the velocity vector is equal to the gravitational acceleration vector, which points downwards and has the magnitude of g. But an object in the brachistochrone is not in free fall, as it travels along a path, resulting in a normal force. So in addition to the g vector, you need a vector pointing along this force, and with a magnitude F/m, in the equation for the derivative of the v vector.

Also, it seems like you made a mistake when solving the differential equation (dv/dy)v_y=g. When you integrate (v_y)dv, as you did, you don't get (1/2)(v_y)^2, as you're integrating with respect to v, not v_y.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Hey I really understood that. It seems that I just considered the velocity vector to be vertical only, so I interchanged $v$ and $v_y$ freely, but as you said that doesn't make sense because here we are also moving in the $x$ direction. Now I'll try to correct those equations and get a better understanding. Anyway, you found what I did wrong, so you solved my problem. Thank you. $\endgroup$ – user267998 Jun 20 at 14:13
0
$\begingroup$

The brachistochrone is the trajectory of the falling mass as it slides along the curve. Velocity is always tangent to the trajectory.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I am sorry, I study math, so my physics intuition might not be very good. I could understand that velocity is always tangent to the trajectory, but then, what's wrong with the $v_y = \sqrt{2gy}$ calculation? Why won't it work here? What wrong assumption did I make? $\endgroup$ – user267998 Jun 20 at 13:40
  • $\begingroup$ if your object is sliding along a friction-less curve, that formula gives the magnitude of the velocity, not the y component. $\endgroup$ – R.W. Bird Jun 21 at 16:31

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.