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Classical mechanics - rigid body rotating

  1. Apr 21, 2014 #1
    1. The problem statement, all variables and given/known data
    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.

    et.JPG

    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Apr 21, 2014 #2
    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.
     
  4. Apr 21, 2014 #3
    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?
     
  5. Apr 21, 2014 #4
    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. :(
     
  6. Apr 21, 2014 #5
    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.
     
  7. Apr 22, 2014 #6

    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.
     
  8. Apr 22, 2014 #7
    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.
     
  9. Apr 22, 2014 #8
    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 ##

    ?
     
  10. Apr 22, 2014 #9
    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.
     
  11. Apr 22, 2014 #10
    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##
     
  12. Apr 22, 2014 #11
    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.
     
  13. Apr 22, 2014 #12
    ##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?
     
  14. Apr 22, 2014 #13
    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.
     
  15. Apr 22, 2014 #14
    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?
     
  16. Apr 22, 2014 #15
    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)?
     
  17. Apr 22, 2014 #16
    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##?
     
  18. Apr 22, 2014 #17
    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.
     
  19. Apr 22, 2014 #18
    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?
     
  20. Apr 22, 2014 #19
    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?
     
  21. Apr 23, 2014 #20
    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?
     
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