Classical mechanics - rigid body rotating

[skrat]

Homework Statement

Around vertical axis ##O## a body on picture below (see attachment) is being rotated with constant angular velocity ##\Omega ##. On the circle we have a body with mass ##m##, that can feels no friction. Find position of that body as function of ##\phi ## and time. Calculate the energy of the system. Is it constant? Also find equilibrium position and frequency for small movements around equilibrium position.

Homework Equations

The Attempt at a Solution

Here is how I started, hopefully not completely wrong.

Firstly, one inertial coordinate system with origin in axis ##0## and axis ##\hat{i} ## pointing to the right and axis ##\hat{j} ## pointed up. Second, non-inertial system positioned in center of circle with radius ##R## where one axis is parallel to the line ##l## and the other of course perpendicular the ##l##.

I used notation ##^{'}## for all coordinates in non-inertial system.

Now ##m{\ddot{r}}'=a_{rel}+2\vec{\Omega }\times v_{rel}+\vec{\Omega }\times(\vec{\Omega }\times {r}')=\vec{F_r}+\vec{F_g}##

where:

##{r}'=R(cos\phi ,sin\phi )##
##\dot{{r}'}=v_{rel}=R\dot{\phi }(-sin\phi ,cos\phi )##
##\ddot{{r}'}=R\ddot{\phi }(-sin\phi ,cos\phi )+R\dot{\phi }(-cos\phi ,-sin\phi )##

##\vec{\Omega }=(0,0,\Omega )##

Now ##2\vec{\Omega }\times v_{rel}=-2\Omega R\dot{\phi}(cos\phi ,sin\phi )## and ##\vec{\Omega }\times(\vec{\Omega }\times {r}')=-\Omega ^2R(cos\phi ,sin\phi )##.

Also ##\vec{F_r}##, which is force on body in radial direction is than ##\vec{F_r}=F_r(cos\phi ,sin\phi )##. Now ##\vec{F_g}=F_g\hat{j}=F_g(sin\Omega t \hat{{i}'}+cos\Omega t \hat{{j}'})=F_g(sin\Omega t ,cos\Omega t )##.

Now writing everything in each directions should give me:

##\hat{{i}'}##: ##-R\ddot{\phi }sin\phi -R\dot{\phi }cos\phi-2\Omega R\dot{\phi }cos\phi -\Omega ^2Rcos\phi =F_rcos\phi +F_Gsin\Omega t## and

##\hat{{j}'}##: ##R\ddot{\phi }cos\phi -R\dot{\phi }sin\phi-2\Omega R\dot{\phi }sin\phi -\Omega ^2Rsin\phi =F_rsin\phi +F_Gcos\Omega t##

That is IF I am not completely mistaken. Before I continue: So, my question here is: Is everything ok so far? Is this the right way to solve the problem OR are there easier ways? Maybe using Lagrangian mechanics or...? Thanks in advance.

 

[kontejnjer]

Usually, when you have a lot of forces to consider, it's wiser to use the Lagrangian approach. This way you can just apply an appropriate coordinate transformation and obtain the equations of motion directly, without having to deal with vectors at all.

 

[skrat]

Usually, when you have a lot of forces to consider, it's wiser to use the Lagrangian approach. This way you can just apply an appropriate coordinate transformation and obtain the equations of motion directly, without having to deal with vectors at all.

I agree, but how would I write the distance of mass from the axis in this case?

Because ##L=T-V##, and ##T=\frac{1}{2}m\dot{\vec{r}}^2## where ##\vec{r}## is a vector from ##O## directly to the body with mass ##m##. The vector's size is not constant also it's direction changes.

Since the Lagrangian approach can only be used in inertial systems, I would assume that this should work:

Let's use ##r## for distance between the mass and ##O##. Than ##r=l+R+Rcos\phi =l+R(1+cos\phi )##.

Now in inertial system this should like something like ##\vec{r}=(l+R(1+cos\phi ))(cos\Omega t, sin\Omega t)## if ##\Omega t## is measured counter clockwise direction with ##0## when ##\vec{r}## is parallel to ##\hat{i}##.

Is this what you imagined?

 

[skrat]

However, the first method, using vectors, if I continue where I stopped:

##-R\ddot{\phi }sin\phi -R\dot{\phi }cos\phi-2\Omega R\dot{\phi }cos\phi -\Omega ^2Rcos\phi =F_rcos\phi +F_Gsin\Omega t## and

##R\ddot{\phi }cos\phi -R\dot{\phi }sin\phi-2\Omega R\dot{\phi }sin\phi -\Omega ^2Rsin\phi =F_rsin\phi +F_Gcos\Omega t##

If I multiply the first one with ##-sin\phi ## and the second one with ##cos\phi ## than sum them together, I get a rather nice equation:

##\ddot{\phi }=\frac{F_g}{R}cos(\phi +\Omega t)## where I can see that units don't match because I completely forgot about ##m##, therefore:

##\ddot{\phi }=\frac{F_g}{mR}cos(\phi +\Omega t)##

For equilibrium:
##\ddot{\phi }=0=\frac{F_g}{mR}cos(\phi +\Omega t)## and so ##cos(\phi +\Omega t)=0##.

##\phi +\Omega t=-\frac{\pi }{2}+n\pi ##

Since we expect equilibrium when ##\Omega t=-\frac{\pi }{2}##

##\phi _0=0##

Now for small movements around equilibrium position:

##\ddot{\phi }=\frac{F_g}{mR}cos(-\frac{\pi}{2}-\phi )## where ##\phi ## is now very small, therefore

##\ddot{\phi }=\frac{F_g}{mR}sin(\phi )=\frac{F_g}{mR}\phi ##

This now gives me ##\omega ^2=\frac{F_g}{mR}=\frac{g}{R}##. And I doubt that the result can be that simple. :(

 

[kontejnjer]

I agree, but how would I write the distance of mass from the axis in this case?

Because ##L=T-V##, and ##T=\frac{1}{2}m\dot{\vec{r}}^2## where ##\vec{r}## is a vector from ##O## directly to the body with mass ##m##. The vector's size is not constant also it's direction changes.

Since the Lagrangian approach can only be used in inertial systems, I would assume that this should work:

Let's use ##r## for distance between the mass and ##O##. Than ##r=l+R+Rcos\phi =l+R(1+cos\phi )##.

Now in inertial system this should like something like ##\vec{r}=(l+R(1+cos\phi ))(cos\Omega t, sin\Omega t)## if ##\Omega t## is measured counter clockwise direction with ##0## when ##\vec{r}## is parallel to ##\hat{i}##.

Is this what you imagined?

Actually, if you pick the center of the circle as the origin of the Cartesian coordinates with the z-axis pointing in the direction of the angular velocity, after a coordinate transformation to spherical coordinates the Lagrangian becomes simple to evaluate, with only one non-cyclic coordinate. From there, it's easy to find the effective potential and evaluate the (stable) equilibrium points.

As for the vector approach, I haven't done the calculations myself, but I get rather different equilibrium positions and frequency of small oscillations from the E-L equations.

 

[skrat]

Actually, if you pick the center of the circle as the origin of the Cartesian coordinates with the z-axis pointing in the direction of the angular velocity, after a coordinate transformation to spherical coordinates the Lagrangian becomes simple to evaluate, with only one non-cyclic coordinate. From there, it's easy to find the effective potential and evaluate the (stable) equilibrium points..

Can I really do that? I am asking because I know we said that Lagrangian method can only be written for inertial systems, and this circle, which rotates around axis ##O##, is therefore non-inertial.

 

[kontejnjer]

The E-L equations work for inertial as well as non-inertial systems, since technically the forces that appear in non-inertial systems aren't "real" forces to begin with, so the most general form of the E-L equations isn't necessary here, the coordinate transformations account for them.

 

[skrat]

Than

##T=\frac{1}{2}m(\dot{r}^2+\dot{\varphi }^2\dot{r}^2sin^2\theta +\dot{\theta }^2r^2)##

##T=\frac{1}{2}mr^2\dot{\theta }^2##

and

##V=mgcos\theta ##

?

 

[kontejnjer]

Than

##T=\frac{1}{2}m(\dot{r}^2+\dot{\varphi }^2\dot{r}^2sin^2\theta +\dot{\theta }^2r^2)##

##T=\frac{1}{2}mr^2\dot{\theta }^2##

and

##V=mgcos\theta ##

?

Well, almost. In your case, [itex]\dot{\varphi}[/itex] isn't zero since the circle is rotating in that plane with constant angular velocity [itex]\Omega[/itex]. Also, the second term in the kinetic energy isn't [itex]\dot{\varphi }^2\dot{r}^2sin^2\theta[/itex], rather [itex]\dot{\varphi }^2r^{2}sin^2\theta[/itex] (no time derivative of r).

EDIT: didn't register it at first, but your potential energy expression doesn't have units of energy, since the transformation for z is [itex]z=rcos\theta[/itex] you probably forgot the r term.

 

[skrat]

HOW in earth!?

Njah, I don't get it. I simply don't. I believe you, but I don't get it. Spherical coordinates in general are http://en.wikipedia.org/wiki/File:3D_Spherical.svg written for the coordinate system with origin in the center of the circle. Now how in Earth is ##\varphi =\Omega t##? How can you possibly see that?

How do you see anything spherical here where everything happens in 2D, that I do not know.

However, putting that aside;

##T=\frac{1}{2}m(\dot{r}^2+\dot{\varphi }^2r^2sin^2\theta +\dot{\theta }^2r^2)## and

##L=T-V=\frac{1}{2}m(\dot{r}^2+\dot{\varphi }^2r^2sin^2\theta +\dot{\theta }^2r^2)-mgcos\theta ##

Which gives me ##mgsin\theta +mR^2\Omega ^2sin\theta cos\theta -mR^2\ddot{\theta }=0##

and frequency ##\omega ^2=\frac{g}{R}+\Omega ^2##

 

[kontejnjer]

HOW in earth!?

Njah, I don't get it. I simply don't. I believe you, but I don't get it. Spherical coordinates in general are http://en.wikipedia.org/wiki/File:3D_Spherical.svg written for the coordinate system with origin in the center of the circle. Now how in Earth is ##\varphi =\Omega t##? How can you possibly see that?

How do you see anything spherical here where everything happens in 2D, that I do not know.

However, putting that aside;

##T=\frac{1}{2}m(\dot{r}^2+\dot{\varphi }^2r^2sin^2\theta +\dot{\theta }^2r^2)## and

##L=T-V=\frac{1}{2}m(\dot{r}^2+\dot{\varphi }^2r^2sin^2\theta +\dot{\theta }^2r^2)-mgcos\theta ##

Which gives me ##mgsin\theta +mR^2\Omega ^2sin\theta cos\theta -mR^2\ddot{\theta }=0##

and frequency ##\omega ^2=\frac{g}{R}+\Omega ^2##

Well, like I said, if you place the coordinate system so the z axis points in the direction of the angular velocity, then from the Wiki picture, it should be clear that [itex]\Omega=\frac{d\varphi}{dt}=\dot{\varphi}[/itex] (z rotates with a constant angular velocity). Generally speaking, we can choose any coordinate system origin and orientation we want, so it's usually best to choose the one which greatly simplifies the calculations. As for the 2D/3D part: the circle itself is rotating around an axis passing through its diameter, so its surface of revolution will obviously be a sphere, hence it is easiest to use spherical coordinates for the problem (as the mass cannot leave the circle). In general, it is necessary to use 3 coordinates since it is the minimum amount required to fully specify its position of a point mass in space (which is 3 dimensional). It just so happens that for this particular problem, in spherical coordinates two of them will be cyclic, making the problem easiest to solve.

This picture might clear things up:
http://www.seos-project.eu/modules/laser-rs/images/coordinates-spherical.png

Also, I'm somewhat puzzled as to how you concluded that [itex]\omega^{2}=\frac{g}{R}+\Omega^{2}[/itex], I don't see how it could come from the differential equation above, which is again missing an [itex]r[/itex] that comes from the potential energy term.

 

[skrat]

##L=T-V=\frac{1}{2}m(\dot{\varphi }^2r^2sin^2\theta +\dot{\theta }^2r^2)-mgrcos\theta##

##\frac{\partial L}{\partial \theta }-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{\theta }}=0##

##\frac{1}{2}m\Omega ^2R^22sin\theta cos\theta +mgRsin\theta -\frac{\mathrm{d} }{\mathrm{d} t}(mR^2\dot{\theta })=0##

##\ddot{\theta }-\frac{1}{2}\Omega ^2sin(2\theta )-\frac{g}{R}sin\theta =0##

For small ##\theta ##:

##\ddot{\theta }-\Omega ^2\theta -\frac{g}{R}\theta =\ddot{\theta }-(\Omega ^2+\frac{g}{R})\theta =0##

Right?

 

[kontejnjer]

That approximation is valid only for small angles, but you still haven't found the equilibrium points, which might not be small at all, so it's unjustified. Furthermore, the differential equation you've obtained doesn't have a periodic solution at all, so how could it then have a frequency in the first place?

One way to find the equilibrium positions is to write the Lagrangian in the form:
[itex]L=T(\dot{\theta})-V_{eff}(\theta)[/itex]
that is, separate the terms that explicitly contain [itex]\dot{\theta}[/itex] from those that contain [itex]\theta[/itex] (this is possible if there are no mixed terms, i.e. [itex]\theta\dot{\theta}[/itex]), and then treat the particle as if it is moving under the potential [itex]V_{eff}[/itex]. You can then easily find the equilibrium points of this effective potential, and determine their stability.

 

[skrat]

Ok If I understand you correctly (which is probably not the case because it does not seem to be easy at all) than ##\frac{1}{2}m\Omega ^2R^2sin^2\theta -mgrcos\theta=0## to find the equilibrium points?

 

[kontejnjer]

Mathematically, what is the necessary requirement for a function (in this case, [itex]V_{eff}=V_{eff}(\theta)[/itex]) to have a stationary point (which can physically be interpreted as an equilibrium point)?

 

[skrat]

When ##\frac{\partial }{\partial \theta }V_{eff}(\theta )=0##

Which gives me ##sin\theta (\Omega ^2Rcos\theta +g)=0## therefore ##\theta= n\pi ##.

If that's correct, than equilibrium is at ##\theta =0##?

 

[kontejnjer]

Alright, that is one solution. Note that, while [itex]\theta=0[/itex] corresponds to the point mass being in the highest point on the circle, you can also have [itex]\theta=\pi[/itex] which would correspond to the lowest point on the circle. However, there might be an extra solution since you also have a factor [itex]\Omega^2Rcos\theta+g[/itex], which should be taken into consideration.

 

[skrat]

From ##\Omega^2Rcos\theta+g## also##\theta =-arccos(\frac{g}{\Omega ^2R})+n\pi ##.

Amm... Is it not exactly the opposite? When ##\theta =0## the mass is the furthest from the axis ##O## and therefore in the lowest point of the circle?

 

[kontejnjer]

Since we defined the z axis to point in the "up" direction, then [itex]\theta[/itex] is the angle the radius vector makes with the positive z axis (see picture from a previous post). From this, [itex]\theta=0[/itex] corresponds to the highest point on the circle and [itex]\theta=\pi[/itex] corresponds to the lowest point. You can of course make a substitution [itex]\phi=\pi-\theta[/itex] (since this [itex]\phi[/itex] corresponds to the one in your original post) which would then match your reasoning.

Okay, now you found all of the equilibrium points. The question that remains now is figuring out whether they are stable (local/global minimum) or unstable (local/global maximum or saddle point), how would you mathematically determine that?

 

[skrat]

Am...

Do I do that by simply inserting ##\theta _i## into ##V_{eff}(\theta )## ? I can do that for all three ##\theta _i## and than compare the values of ##V_{eff}##, meaning the greater the potential the greater stability.

Would this be it?

 

[kontejnjer]

Well, not quite. When thinking about potential energy, the best analogy (classically anyway) would be in terms of hills and valleys. When a particle is on top of a hill (its potential energy is the highest), then it would roll down under the influence of a small disturbance (force), meaning it wouldn't have a tendency to return to its original state, so that a "hill" is unstable. However, if it is in a valley, then after a slight disturbance it would have a tendency to go back to its original position, and it would oscillate back and forth around it, hence that position would be called stable.

Now, back to the original problem. In order to figure out the nature of the obtained stationary points, you need to apply the second derivative test:
http://en.wikipedia.org/wiki/Second_derivative_test

 

[skrat]

I seriously think that we complicated this way to much... Firstly, I can see no reason why even bother with potential energy when axis is perpendicular to the circle plane.

Here is my solution, for 2D motion and hopefully it is the right one.

##L=T-V##

##T=\frac{1}{2}mv^2## where ##\vec{v}=v_{rel}+\Omega \times \vec{r}^{'}+\Omega \times \vec{R_k}## where ##\vec{R_k}## is a vector from the origin of coordinate system in the center of the circle to the axis, where fixed coordinate system is.

Now ##\vec{r}^{'}=(Rcos\phi ,Rsin\phi )## and ##\Omega =(0,0, \Omega)## and ##\vec{R_k}=(l+R,0,0)## all expressed in system with the origin in the center of the circle.

Everything together gives me ##\vec{v}=(-Rsin\phi(\dot{\phi}+\Omega),Rcos\phi (\dot{\phi}+\Omega)+l+R)## and accordingly

##T=\frac{1}{2}m[R^2(\dot{\phi}+\Omega)^2+\Omega ^2(l+R)^2+2R\Omega cos\phi(\dot{\phi}+\Omega)(l+R)]##

and ##V=0##.

Now Euler-Lagrangian equations leave me with

##\ddot{\phi }+\frac{l+R}{R}\Omega ^2sin\phi =0##

Therefore equilibrium position at ##\phi _0=0## and ##\phi _0=\pi ## for unstable position.

Since now we know that ##\phi ## is small around equilibrium position, let's use Taylor expansion and write

##\ddot{\phi }+\frac{l+R}{R}\Omega ^2\phi =0##

From where ##\omega =\Omega \sqrt{1+l/R}##I mean, don't get mad at me but it says axis is VERTICAL and... yeah.

 

[kontejnjer]

Well, it seems that an apology is in order, at least from my side, as I seem to have misinterpreted your picture...a lot . Yes, the results are indeed correct.

 

[skrat]

No apology is needed kontejnjer. I learned a lot more like this than if we solved this in two posts therefore I would like to thank you for all the help!

Cheers