A mechanical problem with complicated geometry

[wrobel]

Here I would like to draw colleagues attention to a class of problems that are simple from physics viewpoint but have pure geometric obstacles. I believe such problems can be discussed in basic courses of analytic geometry as well.

A coin of radius $r$ and of mass $m$ is put in a chute such that the plane of the coin can be horizontal. There is no slippery between the coin and the chute. The statement of the problem can be for example as follows: find a frequency of small oscillations. Or we can ask to write down equations of motion in terms of the angle between the coin's plane and the horizontal plane.

 

[mcastillo356]

At first glance, $m\vec{g}=\vec{F_n}$
But I don't know how to solve it if I must consider geometry.
$V=\cfrac{1}{3}\pi\,r^2\,h$
Neither how to deal with $\alpha$ at the bottom: $A=\cfrac{1}{2}\text{base}\times{\text{height}}$, suposing an isosceles triangle.
Interesting. What kind of oscillations? Isn't it static?

 

[Baluncore]

A coin of radius r, and of mass m, is put in a chute such that the plane of the coin can be horizontal.

Is the chute conical, or is it equivalent to a parallel V groove?
What mode of oscillation do you expect?

The only mode I can see is the rocking of the coin, about the two contact points, in a parallel V-groove?

We would need to know the thickness of the coin.
There may be some restriction on the V-angle of the chute.

I would look at, and analyse, the system along the axis that passes through the two contact points. Since the coin's mass is above the two contact points, it may be a bi-stable situation, with small amplitude oscillations either side, or a larger amplitude oscillation, passing through, or crossing the two stable points.

The coin would finally settle in one of the two stable positions, but without friction, it would continue to oscillate. There are three different possible fulcrums for oscillation, left, right, or rocking on both.

 

[wrobel]

 

[Baluncore]

The friction is present since it is said that there in no slippery

There is friction, so there is no slipping. The surfaces are not slippery.
The edge of the coin will roll on the plane sides of the V-groove.

 

[wrobel]

yes


My solution is as follows. Introduce a frame $OXYZ$ such that the axis $OZ$ is directed upwards. The sides of the groove a the planes defined by equations
$$X=Z\tan\alpha ,\quad X=-Z\tan\alpha .$$
The center of the coin has the coordinates $(0,y,z).$ Let $\varphi$ stand for the angle between the coin's plane and the axis $OY$.
The generalized coordinates are $\varphi,y,z$.
The equations of constraints are
$$\dot y=r\dot\varphi \sin\psi\sin\varphi,\quad \dot z=r\dot\varphi\sin\psi\cos\varphi.$$
Here $r$ is a radius of the coin;
the function $\psi=\psi(\varphi)\in(-\pi/2,\pi/2)$ is found from the equation
$$\tan\psi=\sin\varphi\tan\alpha.$$
The Lagrangian is
$$L=\frac{m}{2}(\dot y^2+\dot z^2)+\frac{1}{2}J\dot\varphi^2-mgz,\quad J=mr^2/4.$$
Here $m$ is a mass of the coin.

 

[wrobel]

In particular one can find a distance that the center of the coin passes as the coin makes one complete revolution:
$$\Delta y=r\tan\alpha \int_0^{2\pi}\frac{\sin^2\varphi\,d\varphi}{\sqrt{1+\sin^2\varphi \tan^2\alpha}}.$$
Pure kinematic fact.