# Homework Help: Rigid body (mass sliding over a cone)

1. Mar 4, 2012

### fluidistic

1. The problem statement, all variables and given/known data
There's a problem that I don't even know really where to start. It seems extremely complex to me.
Basically there's a reverted cone of mass M with a mass of mass m that can slides over it without friction. But there's a constraint of motion for the particule, it must stay on a trail over the cone, of step L. The cone is totally free to rotate (it's on a frictionless surface). The system is put under the gravitational field.
1)Express the velocity of the particule with respect to an inertial system.
2)Write down the Lagrangian and the conserved quantities.

2. Relevant equations

Not sure.

3. The attempt at a solution
First of all the center of mass of the cone will move and its axis of rotation too (I think). So that it looks like extremely difficult to decide where to put my coordinate system. Let's say I put the original of my coordinate system on a point on the ground. Now I don't know what coordinate system is more appropriated since I'm dealing with a cone. Let's take cylindrical coordinates $(r, \theta, z)$. I don't know how to proceed further. :(
What should I do here?

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2. Mar 4, 2012

### tiny-tim

hi fluidistic!

i may be wrong , but my guess is that the question intends that it is only free to rotate (not free to translate) …
i'd try it that way first!

(and if i am wrong, i really don't see how cylindrical or spherical coordinates will help)

3. Mar 4, 2012

### fluidistic

Let's assume you are right. Do you mean that the cone will rotate around a vertical axis passing by its center of mass?

4. Mar 4, 2012

### tiny-tim

if we're lucky!

5. Mar 4, 2012

6. Mar 5, 2012

### fluidistic

Ok tiny-tim, the cone is rotating around a fixed axis, as we assumed (I asked the professor). Rigid body problems aren't my forte. :(
I realize that there are 2 degrees of freedom, so that I should end up with a Lagrangian involvign 2 generalized coordinates.
Using cylindrical coordinates $(r, \theta , z)$, I get that $z(r)=H \left ( 1- \frac{r}{R} \right )$ where H is the height of the cone. I also reached $z(\theta )=H-\frac{\theta L}{2\pi }$.
To build the Lagrangian, the total kinetic energy will be the one of the cone $T_2=\frac{I \omega ^2}{2}$ plus the one of the particle, $T_1=\frac{mv^2}{2}$. The potential energy of the Lagrangian will be the one of the particule only (I can discard the one of the cone right? Since it's a constant), namely $V_1(\theta )=mg \left ( H- \frac{\theta L}{2\pi} \right )$.
Now my next task is to find out the kinetic energy in terms of the 2 generalized coordinates. Theta seems to be one of them but I'm totally unsure about the other generalized coordinate we could choose to make the algebra simpler.
Is this correct so far? How can I continue?

7. Mar 6, 2012

### tiny-tim

oooh, i never got the hang of lagrangians …

i was rather hoping that once i got you spinning on the right axis, you'd show me how to do it!

let me see … if one coordinate is basically height ,then i suppose the other must be basically angle …

so how about angular momentum of the mass m?

(i'm really only stringing words together in the hope that they look wise )

8. Mar 6, 2012

### fluidistic

hehehe, I have absolutely no experience on this kind of problem.
Now that I think, I should take theta as first generalized coordinate. For my other generalized coordinate, I think I should consider $\Theta$, the angle the cone moves with respect to my inertial reference frame system. You can visualize this angle better if you take a fixed point on the cone (say on the trail).
I already have $z(\theta )=H-\frac{\theta L}{2\pi }$.
So I have $\omega = \dot \Theta$. $\Theta$ is totally independent of $\theta$, however using intuition, $\dot \theta$ must depend on the velocity of the particle. Anyway I'm asked (first), the velocity of the particle. In cylindrical coordinates the speed is worth $\sqrt {\dot r ^2+r^2 \dot \theta ^2 +\dot z^2}$.
I need to express these terms in function of $\theta$ only. I already have z. Now $r(\theta )=\frac{RL\theta}{2\pi H}$.
So that $v = \sqrt { \left ( \frac{RL \dot \theta }{2\pi H} \right ) ^2+\left ( \frac{RL\theta}{2\pi H} \right ) ^2\dot \theta ^2 +\left ( H-\frac{L\theta }{2\pi} \right ) ^2}$. I hope this simplify a bit somehow. What do you think so far about my thoughts?

Edit: $v=\sqrt {\frac{R^2L^2 \dot \theta ^2 }{4\pi ^2 H^2}(1+\theta ^2)+\left ( H-\frac{L\theta }{2\pi} \right ) ^2}$. This result does not seem to make sense to me :( Do you spot any problem?

Last edited: Mar 6, 2012