Bead sliding inside a paraboloid

[JyJ]

Homework Statement

A bead slides under the influence of gravity on the frictionless interior surface of the paraboloid of revolution $$z = (x^2+y^2)/2a = r^2/2a$$ Find the speed $$v_0$$ at which the bead will move in a horizontal circle of radius $$r_0$$ Find the frequency of small radial oscillations around this circular motion.

Homework Equations

$$F=ma \\ \dot{r} = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k} \\ \ddot{r} = (\ddot{r} - r \ddot{\theta}^2)\mathbf{e}_r+ (r \ddot{\theta}+2 \dot{r} \dot{\theta})\mathbf{e}_{\theta}+\ddot{z}\mathbf{k}$$

The Attempt at a Solution

First, I chose cylindrical coordinates to work with, and deduced equations for each of $$\mathbf{e}_r, \mathbf{e}_{\theta}, \mathbf{k}$$ from $$\mathbf{F} = m\mathbf{a} = \mathbf{N}+ m\mathbf{g}$$. This gave me 3 equation which I then rearranged to eliminate $$N$$ as well as $$z$$ by using the fact that $$\ddot{z} = (\dot{r}^2 - r\ddot{r})/a$$
After all of this, the problem was reduced to just 2 equations:
$$\ddot{r}(a^2+r^2) + \dot{r}^2 r + arg - (a^2h^2/r^3) = 0$$ where $$h = r^2\dot{\theta}$$
Here $$h$$ is angular momentum which is conserved.
Now, from this stage I am not sure if what I am doing is right. For particle moving in a horizontal circle we have that $$z$$ and $$r$$ are unchanged and so I suppose $$\ddot{r} = \dot{r} = 0$$ Plugging into equation gives:
$$arg - a^2h^2/r^3 = 0 \\ r = r_0 = (ah^2/g)^{1/4}$$
Also $$\dot{r} = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k}=r\dot{\theta}\mathbf{e}_{\theta} \\ \dot{r} = v_0 = r\ddot{\theta} = rh/r^2 = g^{1/4}/(\sqrt{h}a^{1/4})$$
As for small radial oscillations I considered small deviations from the circular path by introducing $$r = z + \epsilon$$ with $$z=r_0$$ which I calculated to be $$r_0 = (ah^2/g)^{1/4}$$ After plugging this into the equation of motion and getting rid of powers higher than $$\epsilon$$ I get something not very pretty:
$$\ddot{\epsilon}(a^2+z^2)z^3 + \epsilon (agz^3 + 3az^3) + az^4 - a^2h^2 = 0$$
Without considering the particular integral this has cos and sin in it, so I presume the frequency would be:
$$f = \sqrt{(ag+3a)/(a^2+z^2)}$$ and then of course I can substitute z.

Please advise if my argument is valid. Thank you!

 

[BvU]

Hello JyJ,

I have difficulty understanding your paraboloid equation: to me it looks more like a circle (z=0) or a cylinder ($r=\rho$)...

 

[JyJ]

Hello JyJ,

I have difficulty understanding your paraboloid equation: to me it looks more like a circle (z=0) or a cylinder ($r=\rho$)...

Hi,
I just corrected the equation: it should be z = (x^2+y^2)/(2a) or z = r^2/(2a) in cylindrical coordinates. Sorry for the confusion

 

[BvU]

This gave me 3 equation

Indulge me and show them

problem was reduced to just 2 equations:

and you show one. What is the other ?
$\dot{r} = v_0 = r\ddot{\theta} = rh/r^2 = g^{1/4}/(\sqrt{h}a^{1/4})$ did you check the dimensions ?

Furthermore: an equation like $$
\dot{r} = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k}$$ is very confusing. How about$$
\dot{\vec r} = \dot{\rho}\mathbf{e}_\rho + \rho\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k}\ \ ?$$

 

[JyJ]

Indulge me and show them
and you show one. What is the other ?
$\dot{r} = v_0 = r\ddot{\theta} = rh/r^2 = g^{1/4}/(\sqrt{h}a^{1/4})$ did you check the dimensions ?

Furthermore: an equation like $$
\dot{r} = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k}$$ is very confusing. How about$$
\dot{\vec r} = \dot{\rho}\mathbf{e}_\rho + \rho\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k}\ \ ?$$

For the 3 component equations I obtained:
$$ m(\ddot{r} - r(\dot{\theta})^2) = -\frac{Nr}{\sqrt(r^2+a^2)} \\
\frac{m} {r} d/dt(r^2\dot{\theta}) = 0 \\
m\ddot{z} = \frac{Na}{\sqrt(r^2+a^2)} - mg $$
where N is the magnitude of the normal force.
After eliminating N and z, I indeed showed only one (oops) but in fact I got 2:
$$ \ddot{r} - r(\dot{\theta})^2 = -\frac{r}{a} (\frac{(\dot{r})^2 - r\ddot{r}} {a} + g) \\
\frac{1}{r} \frac{d}{dt}(r^2\dot{\theta}) = 0 $$
and it is the second equation that I used to deduce that angular momentum h is conserved. I then substituted $$ \dot{\theta}^2 = h^2/r^4 $$ into the first equation to get rid of theta.
As for dimensions, something is wrong as I get the following:
$$ \frac {(ms^{-2})^{1/4}} {m\sqrt(\frac{radians} {s}) m^{1/4}} = m^{-1}radians^{-1/2} $$
I can't really interpret it as velocity :(

 

[JyJ]

Furthermore: an equation like $$
\dot{r} = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k}$$ is very confusing. How about$$
\dot{\vec r} = \dot{\rho}\mathbf{e}_\rho + \rho\dot{\theta}\mathbf{e}_{\theta} + \dot{z}\mathbf{k}\ \ ?$$[/QUOTE]

Yes I agree, it should be a vector - I cannot edit that in for some reason

 

[BvU]

Are we still busy finding $v_0$ when $r_0$ is given ? Isn't it easier there to set $\dot \rho = \dot z = 0$ ?

By the way, I thought angular momentum also has a factor mass

 

[JyJ]

For the 3 component equations I obtained:
$$ m(\ddot{r} - r(\dot{\theta})^2) = -\frac{Nr}{\sqrt(r^2+a^2)} \\
\frac{m} {r} d/dt(r^2\dot{\theta}) = 0 \\
m\ddot{z} = \frac{Na}{\sqrt(r^2+a^2)} - mg $$
where N is the magnitude of the normal force.
After eliminating N and z, I indeed showed only one (oops) but in fact I got 2:
$$ \ddot{r} - r(\dot{\theta})^2 = -\frac{r}{a} (\frac{(\dot{r})^2 - r\ddot{r}} {a} + g) \\
\frac{1}{r} \frac{d}{dt}(r^2\dot{\theta}) = 0 $$
and it is the second equation that I used to deduce that angular momentum h is conserved. I then substituted $$ \dot{\theta}^2 = h^2/r^4 $$ into the first equation to get rid of theta.
As for dimensions, something is wrong as I get the following:
$$ \frac {(ms^{-2})^{1/4}} {m\sqrt(\frac{radians} {s}) m^{1/4}} = m^{-1}radians^{-1/2} $$
I can't really interpret it as velocity :(

Are we still busy finding $v_0$ when $r_0$ is given ? Isn't it easier there to set $\dot \rho = \dot z = 0$ ?

By the way, I thought angular momentum also has a factor mass

Yes, I guess so since in what I have so far the units give something rather bizarre. When I was calculating v_0 I had: $$\vec{v_0} = r\dot{\theta}^2\mathbf{e}_\theta$$ in which I assumed $$\dot{r} = \dot{z} = 0$$
As for the angular momentum I don't quite see how factor m can appear in my derivation from:
$$\frac{m}{r} \frac{d}{dt}(r^2\dot{\theta}) = 0 $$
since $$\frac{m}{r} $$ is never equal to zero and so the terms inside d/dt must be constant.

 

[BvU]

Can you post a simple free body diagram of the bead (in the simple $v_0, \rho_0\$ case) ?

 

[JyJ]

Can you post a simple free body diagram of the bead (in the simple $v_0, \rho_0\$ case) ?

Yes of course! Hope this will be sufficient:

 

[BvU]

Very good. This way the relationship between the forces ( $\vec N + m\vec g = \vec F_c$ ) becomes clear. An angle and a magnitude in equation form is the next step.