Using D' Alembert's Principle on an inverted cone.

[carllacan]

Homework Statement

A masspoint finds itself under the influence of gravity and constrained to move on a (inverted) circular cone. Using D'Alembert's Principle find the equations of motion on cylindric coordinates.

Homework Equations

D'Alembert's Principle: ($\vec{F_a}$ -m·$\vec{a}$)·$\delta$$\vec{r}$=0

The Attempt at a Solution

Chose as generalized coordinates l and m, which measure, respectively, "how high" is the particle on the cone and the angle coordinate.

Write F = -mgz, where z is the vector for the vertical cylindric coordinate. Write $\delta$$\vec{r}$ as the total differential dr minus the dt term, i.e. $\delta$$\vec{r}$ = $\stackrel{d\textbf{r}}{dl}$ $\delta$l + $\stackrel{d\textbf{r}}{dm}$ $\delta$m

Then, as the generalized coordinates are independent we can equate the coeficients of $\delta$l and $\delta$m to zero. Which should give us the equations of motion. Th problem is that I obtain one equation according to which the angular acceleration is 0 (as expected) and another one that reads: a_r ·tg($\alpha$)+a_z = -g, where a_r and a_z are the radial and vertical coordinates of the acceleration.

 

[carllacan]

Anyone?

 

[TSny]

Hello carllacan. I think you'll need to show more of your work so that we can see how you are getting your results. You should find that the angular acceleration is not zero in general for this problem. [EDIT: By angular acceleration I am thinking $\ddot{\theta}$. That will not be zero in general. But, the azimuthal component of acceleration $a_\theta$ will be zero. That's probably what you are saying.]

Can you clarify "inverted cone"? Does that mean like an ice-cream cone "opening upward" or does it mean opening downward?

Also, it might be less confusing if you use some other letter to denote the angle coordinate since m is used for the mass.

 

[carllacan]

Yeah, an ice-crem would be a good description.

I wanted to write my arithmetic, in case the problem was there, but the editor is so slow and wonky that I desisted.

And yes, I should have used another letter, my bad. I will edit.

 

[carllacan]

As I can't edit my post anymore I will rewrite the excercise here.

Homework Statement

A masspoint finds itself under the influence of gravity and constrained to move on a (inverted) circular cone with angle α. Using D'Alembert's Principle find the equations of motion on cylindric coordinates.

Homework Equations

D'Alembert's Principle: ($\vec{F_a}$ -m·$\vec{a}$)·$\delta$$\vec{r}$=0

The Attempt at a Solution

Chose as generalized coordinates u and v, which measure, respectively, "how high" is the particle on the cone and the angle coordinate.

Write F = -mgz, where z is the vector for the vertical cylindric coordinate. Write $\delta$$\vec{r}$ as the total differential dr minus the dt term, i.e. $\delta$$\vec{r}$ = $\stackrel{d\textbf{r}}{du}$ $\delta$u + $\stackrel{d\textbf{r}}{dv}$ $\delta$v

Then, as the generalized coordinates are independent we can equate the coeficients of $\delta$u and $\delta$v to zero. Which should give us the equations of motion. The problem is that I obtain one equation according to which the angular acceleration is 0 (as expected) and another one that reads: a_r ·tg($\alpha$)+a_z = -g, where a_r and a_z are the radial and vertical coordinates of the acceleration.

 

[carllacan]

There, I've uploaded a foto of my handwritten solution. The quality is poor, but I think it can be read.
http://www.subirimagenes.com/fotos-img20140123002826-8784660.html

 

[TSny]

OK. You are using the notation $(z, r, \theta)$ for the cylindrical coordinates. And it appears that you are using two coordinates $(u, v)$ to located points on the cone where I believe $u$ is the z-coordinate of a point on the cone and $v$ is the azimuthal angle. So, as you indicated, the polar coordinates of a point on the cone may be written $(z, r, \theta) = (u, u\tan\alpha, v)$ where $\alpha$ is the half-angle of the cone.

However, your expression for a small displacement $\delta\vec{ r}$ is not correct. All components of $\delta\vec{ r}$ should have dimensions of length. Your last term $(0,0,1)\delta v$ does not have dimensions of length since $v$ is an angle.

You might want to review how to write position, velocity, and acceleration in cylindrical coordinates. For example, see http://www.maths.ox.ac.uk/system/files/coursematerial/2013/1115/77/CylCoords.pdf where they use $\rho$ instead of $r$ for the radial cylindrical coordinate. [Using this notation avoids confusing the $r$ in $(z, r, \theta)$ with the magnitude of the position vector $\vec{r}$ which locates a point relative to the origin.] (Also, they use $\phi$ instead of $\theta$ for the azimuthal angle.)

The expression for an infinitesimal displacement is at the bottom of the first page of the link, and an expression for acceleration is in the middle of the next page.

 

[carllacan]

Oh god, you're right, thank you! I totally forgot about the acceleration vector being different.

I don't have any way to check my new result, but apart from that I think my procedure was the right one, so if I carried well the arithmetic it should be correct, shouldn't it?

 

[TSny]

Yes. Your overall method looks correct.