3
$\begingroup$

This is my solution to finding angular displacement/velocity/acceleration on cone so far.

Consider a cone, with an apex of half-angle $\psi$ pointing down, and a height of $h$. If I roll a ball into the tangent of the top of the cone (the top of the cone is parrallel with the ground, as is the initial path of the ball) with velocity $u$. We will call this 'initial velocity'.

I am using the relationship $$\alpha = {a_t \over r} = {F\over mr}$$ to start off. Where alpha is angular acceleration.

Radius is given by trigonometry as: $$r = (h-Z)\tan (\psi)$$ where $Z$ is the displacement from the top of the cone.


the expressions:

Given that the only tangential force should be caused by friction (and negative, because it is acting against the motion), and integrating with respect to time we get the three equations: $$\alpha = - {\mu N \over m(h-Z)\tan (\psi)}$$ $$\omega = - {\mu Nt \over m(h-Z)\tan(\psi)}+{u \over h\tan(\psi)}$$ $$\theta = - {\mu Nt^2 \over 2m(h-Z)\tan(\psi)} + {ut \over h\tan(\psi)}$$


Z:

$Z$ is displacement from the top of the cone at time $t$. I first imagine the overall acceleration down the cone, $a$, as the hypotenuse of a triangle, and the vertical acceleration $a_v$ as the adjacent when using $\psi$. Therefore $$a_v = a\cos(\psi)$$ and through integration, the vertical displacement $Z$ is ... $$Z = {1\over 2} at^2 \cos(\psi)$$ we can then place these in for $Z$ in the above expressions!


summary of what I don't understand:

The parts I don't understand are the parts I have left! an expression for $N$, the normal force, and for $a$ the acceleration towards the apex of the cone. At first I thought these would be simple, like when you get a question on a mass on an inclined plane, or a banked turn. As it turns out it is a lot harder because there should be a force going up the slope (on top of friction), induced by $u$! (right?) When not considering this part, the equations were almost right, except the path of the ball (when viewed on the top) did a sort of half turn around the apex before hitting it. Whereas in a simulation I did in unity (and as you would expect in real life) went round the apex more and more as it got closer, until terminating. (I'll get some pictures to show you). So if I add this part to the overall force/acceleration then it should keep rolling for longer. I am also not sure, does the tangential velocity $v_t$ ever change from $u$? this may be useful to know if the force going up is proportional to $v_t$.

Thank you in advance!

The answer that @ja72 gave me (which was fantastic, but not sure where exactly to take it, may end up using some of it to solve this :D) was for an older version of this question. (I don't know if you can view my edits for this, if you can you should be able to see the old version).

$\endgroup$
  • $\begingroup$ Notice the problem will be two-dimensional since the initial velocity is in the radial direction. It should essentially be a ball rolling down a ramp. $\endgroup$ – Brian Moths Oct 14 '14 at 18:08
  • $\begingroup$ It will not be the same, the tangential speed will increase by conservation of angular momentum in a way not described by a mass just going down a ramp $\endgroup$ – Wolphram jonny Oct 14 '14 at 18:21
  • $\begingroup$ @julianfernandez I know this, but consider a mass going down a slope. There is no component of force, other than friction that acts up the slope. In the case of my problem, there is, and I don't know how to find out what this is. If I maybe drew a diagram, showing what I wanted would that make it simpler? $\endgroup$ – Sam Walls Oct 14 '14 at 18:28
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Not neccesarily? Shouldn't the force going tangent to the circle and the force acting up the slope be proportional in some way? This is really annoying, for every direction i am considering there is an overall force, a normal force and a force of friction, it boggles my mind anyway. :) $\endgroup$ – Sam Walls Oct 14 '14 at 18:31
  • $\begingroup$ Sam, I was answering to the comment made by Now... To answer your question I need to draw it first, I'll do it after I pick up my kids from school if nobody else did. $\endgroup$ – Wolphram jonny Oct 14 '14 at 18:31
1
$\begingroup$

I set a coordinate system on the cone apex, with $+\hat{y}$ pointing up and $+\hat{x}$ pointing radially out at an azimouth (location) angle $\theta=0$. The position of the ball is defined by the distance from the apex up the slope to the ball $r$ and the longitude around the cone $\theta$

$$\vec{p} = {\rm Rot}(\hat{y}, \theta) \begin{pmatrix} r \sin \psi \\ 0 \\ r \cos \psi \end{pmatrix} = \begin{pmatrix} r \sin \psi \cos \theta \\ r \cos \psi \\ -r \sin \psi \sin \theta \end{pmatrix} $$

In the above the angle $\psi$ is fixed and $r$, $\theta$ are variable. Hence

$$ \vec{v} = \dot{\vec{p}} = \frac{\partial \vec{p}}{\partial r} \dot{r} + \frac{\partial \vec{p}}{\partial \theta} \dot{\theta} \\ \vec{v} ={\rm Rot}(\hat{y}, \theta) \begin{pmatrix} \dot{r} \sin \psi \\ \dot{r}\cos\psi \\ -r \dot{\theta} \sin \psi \end{pmatrix} = \begin{pmatrix} \sin\psi (\dot{r} \cos\theta - r \dot{\theta} \sin \theta) \\ \dot{r} \cos \psi \\ -\sin\psi (\dot{r} \sin \theta + r \dot{\theta} \cos\theta ) \end{pmatrix}$$

Similarly

$$ \vec{a} = {\rm Rot}(\hat{y}, \theta) \begin{pmatrix} \sin\psi (\ddot{r}-r \dot{\theta}^2) \\ \ddot{r} \cos\psi \\ -\sin\psi(r \ddot{\theta}+2 \dot{r} \dot{\theta}) \end{pmatrix} $$

If you ignore the rotational inertia and spin motion and focus on the path of the ball then $\sum \vec{F} = m \vec{a}$ with the net forces acting from gravity $\vec{W} = \begin{pmatrix} 0 \\ -m g \\ 0 \end{pmatrix}$ and contact normal force $\vec{N} = {\rm Rot}(\hat{y},\theta) \begin{pmatrix} -N \cos\psi \\ N \sin\psi \\ 0 \end{pmatrix} =\begin{pmatrix} -N \cos\psi \cos\theta \\ N \sin\psi \\ N \cos\psi \sin\theta \end{pmatrix} $

$$ \vec{W} + \vec{N} = m \vec{a} $$

$$ \boxed{ \begin{aligned} N & = m g \sin\psi + m r \dot{\theta}^2 \sin\psi\cos\psi \\ \ddot{r} & = r \dot{\theta}^2 \sin^2\psi - g \cos\psi \\ \ddot{\theta} & = - \frac{2 \dot{r} \dot{\theta}}{r} \end{aligned} } $$

You initial conditions are at $t=0$, $\theta=0$, $r = \frac{h}{\cos\psi}$, $\dot{\theta} = \frac{u}{h \tan \psi}$ and $\dot{r}=0$. I do not know of an analytical solution, so I would use a numerical simulation to get results.


Notes

The $\dot{\boxed{\,}}$ notation is the time derivative, and ${\rm Rot}(\hat{y},\theta)$ is the standard 3×3 rotation matrix about the y axis.

$\endgroup$
  • $\begingroup$ WOW, this is excellent! However some of it is a little beyond me. Just explain for me: What does $\dot{x}$ or $\dot \theta$ mean? (or what are you using it for?) I assume it is the derivative, but in terms of what?; $Rot(y, \theta )$, does this mean a rotation of $\theta$ around $y$ ? If so, how do you multiply a rotation by the vector like you did?; what is $z$ in this case?; finally, this might be a big question, but what do I need to do for a numerical solution? just run simulations, or do experiments, then use the data to find expressions? $\endgroup$ – Sam Walls Oct 15 '14 at 19:30
  • $\begingroup$ @ja72 I do not get why the friction cancels out and how do you estimated that the effects of inertia and rotation are negligible and can be ignored. Is your background physics? $\endgroup$ – Wolphram jonny Oct 15 '14 at 22:04
  • $\begingroup$ To consider the rotational inertia complicates the equations because the only thing providing a torque is friction. My background is in robotics and dynamics. $\endgroup$ – ja72 Oct 16 '14 at 0:27
  • $\begingroup$ Ok, after considering this for a while, I think it can be done easier. My equations are almost right, and look almost like the simulations I have done in unity. Imagine you are on the flat of a velladrome, as you go into the bend, if you do not make an effort to turn with the bend, reaction forces cause you to go into the bend and up the slope. What I need (I think) to complete my equation, is the force going up the cone. I have a feeling that there should be a solution, something similar to the way you resolve forces on a banked turn, to find this force. I like where u are going though. $\endgroup$ – Sam Walls Oct 16 '14 at 21:51

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.