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Rotation and increasing radius

  1. Apr 21, 2003 #1
    Let's say that there is a spoked wheel but the spokes are tubes with freely moving masses inside and that there is no friction. Given a certain fixed angular velocity, what I want to find is the final radius after a certain angle of rotation (assume 270 degrees). An assumption is made that the orbit radius of the weights is held fixed and then suddenly released. What equations can I plug into to find the final radius?

    Thanks for any help.
     
  2. jcsd
  3. Apr 21, 2003 #2
    F = mω2r
     
  4. Apr 21, 2003 #3
    I thought of that but is not "F" a dependent variable on R? Also, I don't think that the value of "m" should matter. It's been a long time since I had physics so excuse my ignorance.

    Thanks--
     
  5. Apr 21, 2003 #4

    dg

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    Since you do not mention it I will not consider any gravity involved or, which is the same, a wheel spinning on an horizontal plane.

    We can simply consider what happen inside one of the frictionless spoke/cylinders.
    Because of the absence of friction, the only force involved along the radial direction is the force that keep in place the mass before it is released. After the mass is released from this constrain it must have 0 radial acceleration!
    Whatever keeps the wheel spinning at constant angular velocity will have to apply force along the only direction allowed (by the friction-free assumption) which is normal to the spoke axis.
    The problem can be simply solved using polar coordinates (see word attachment)

    Really a nice problem!

    Write back for further explanations, Dario
     
  6. Apr 21, 2003 #5
    dg,
    Yes, I had already reached this same conclusion. I am assuming that once the radius is no longer restricted that centripetal acceleration would be zero and the radius would move out by v=wR1. The radius would then continue to change by this velocity and then R2=vT/2Pi + R1 for a (full revolution). Is this assumption correct? Also, how do I retrieve the Word attachment?

    thanks again.
     
    Last edited by a moderator: Apr 21, 2003
  7. Apr 21, 2003 #6

    dg

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    Attachment
    I do not understand what happened to it so I will try to attach it again but I will also give you a link (http://www.geocities.com/dariog71/Documents/Others/PF030421.doc)

    Your assumptions
    Unless I made some major mistake the relation between radial distance and angle (trajectory in polar coordinates) is not immediate and to get there you need to solve a differential equation.
    I have no time now to verify your assumptions, I will write more in a little bit, Dario
     

    Attached Files:

  8. Apr 22, 2003 #7
    dg: i can t read word files, but i would like to read it. when i go to work tomorrow, maybe i will.

    anyway: i fail to see what is wrong with my formula. radial acceleration is given by this formula. you have a constant ω, acceleration goes as r, so you have an exponentially increasing radius for the duration of the rotation.

    this seems believable to me, since the moment of inertia of the system will be increasing quadratically, and to keep the constant angular acceleration, you would have to constant pump in an amount of energy proportional to the moment of inertia.
     
  9. Apr 22, 2003 #8
    dg,
    Thanks for taking the time to work the equations! I have plugged in some trial values and it appears that the rate of change of R is greater than v=rw (time that the constraint is removed). This result suprises me being that "A" along the radius is gone. However, I do not have the mathematical ability that you do so I am obviously overlooking something. I have also attached a free body diagram that I believe somewhat explains what is happening (with some holes to fill in!). Does this resemble the supposition that you worked from to derive your equations?

    Bailey--
     
    Last edited by a moderator: Apr 22, 2003
  10. Apr 22, 2003 #9
    The attachment did not take so here's a link to the diagram.
     
    Last edited by a moderator: Apr 28, 2003
  11. Apr 23, 2003 #10

    dg

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    Hi Bailey,
    I am gonna answer point to point and add some final remarks...
    v=wr gives the tangential velocity not the radial velocity with respect to the center of the wheel. In your case before the mass is liberated there is only a tangential component of velocity, radial velocity is zero because this is equivalent to say that the radial position is constant.
    Well under your assumption no friction implies no radial force and hence no radial acceleration... What makes everything a little hard to imagine is that in general the radial acceleration (with respect to a prefixed origin) is not simply given by the variation of radial position (as it happens when using cartesian coordinates), but it also contain a component depending on pure angular velocity in absence of variations of the distance from the origin (the so-called centripetal acceleration present even in circular motion when the radial position -the radius of the trajectory- does not change)
    Well the free body diagram is mostly correct, what does not really work is the presence of an initial component of velocity in the radial direction.
    Immediately before the constrain is removed, the mass moves with uniform circular motion: constant radial position, constant speed, velocity tangent to the circle and hence perpendicular to the radius, acceleration constant (in modulus) directed radially towards the center of the trajectory. The radial acceleration demands the presence of a force proportional to it along the same direction; this force is given by the constrain only, since the spoke, being frictionless, cannot exert any force tangent to the spoke itself or in other words it cannot exert any radial force.
    Immediately after the constrain is removed, no force is able to keep any longer the radial acceleration so it drops to zero, while all both velocity and radial position have not yet changed. So velocity initially is still purely tangential, which means initial radial velocity is zero.

    After that the radial velocity will increase (exponentially) because of the absence of a radial component of acceleration which -at constant angular speed- translates into a second derivative of radial distance proportional to the the distance itself.

    I have no idea how to solve the problem without differential equations...

    Ask more if you want, Dario
     
    Last edited by a moderator: Apr 23, 2003
  12. Apr 23, 2003 #11
    Dario,
    Thanks for elaborating more here. I jumped the gun on the radial velocity. Instead, what I had intended to point out was that for an initial incremental time (dt) there would be a v tangential (as you just pointed out) and this would have a radial component. This is because there would be a resulting change in distance to the center after dt because of the tangential movement of the object. The new distance from the object to the center would be R2=R1/cos(wt) and therefore, dR=(R1/cos(wt))-R1. Another way of defining it is by using the incremental tangential movement (ds)therefore, dR=(ds/sin(wt))-R1.

    However, as you pointed out, V tangential is changing with the radius (as the radius changes by dR=(R1/cos(wt))-R1 above). If I could find this A tangential, I could determine ds above and from that get dR. Of course, you have already solved it using differential equations. If it could be done, the only purpose would be to check the two results.

    Thanks--
    Bailey--
     
    Last edited by a moderator: Apr 23, 2003
  13. Apr 23, 2003 #12
    lethe's method works fine, the centripetal acceleration would be r*w^2=v^2/r to keep the masses in uniform circular motion; so they will accelerate outward at this rate. you can check it with the Lagrangian formulation of the problem

    V=0
    T=m/2 ( r'^2 + r^2*theta'^2)
    L=T-V=T

    d/dt (m*r') = m*r*theta'^2 --> r''=r*theta'^2 --> r''=r*w^2

    So r(t)=A*exp(w*t)+B*exp(-w*t), r(0)=A+B, r'(0)=Aw-Bw=0, so A=B=r0/2

    r(t)=r0/2*(exp(wt)+exp(-wt))= r0*cosh(wt)
    r(theta) = r0*cosh(theta)
     
  14. Apr 23, 2003 #13
    lethe, damgo

    I am trying to understand here. How can there be a=w^2r when there is no more force (in this case the radial constraint)? If a=f/m what is the force that your speaking of that keeps accelerating the object? There's something about this that I am not getting.

    Thanks--
     
  15. Apr 24, 2003 #14
    it is the pseudoforce that any object in an accelerating frame feels (like in a rotating frame). it acts radially outward, pushing the mass away from the center.
     
  16. Apr 24, 2003 #15

    dg

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    Hi lethe (Bailey and Damgo, please read!),
    today I have plenty of time and I will dedicate some time to you to...
    Without any remark on the direction and origin of this force this is just a 'raw' formula. I could not understand what you meant and this is the main reason why I have not commented on it before.
    nobody said your formula is wrong, it just needed some explanations.
    It seems like you want to solve the problem in the non-inertial frame that rotates with the spoke. And this is fine by me, it is actually a beautiful alternative solution that gives further insight in the physics of rotations but you have to explain it...
    ...just saying that there is a "radial" acceleration is a statement that need some clarification: in the inertial frame the acceleration along the radial direction is zero(!) in this problem because of the absence of friction, nevertheless there is a non-zero second derivative of the radial position! This is because of the way derivatives transform under a change of coordinates like the cartesian to polar I used.
    When you change reference frame to the co-rotating one (same origin in the center of the wheel) the radial acceleration appears as a result of the fictitious centrifugal force. This is equivalent in the equation of motion in polar coordinates to move one of the terms that give the total radial acceleration on the force side of the equation:
    Frad=m*Arad
    0=m*(d^2r/dt^2-r*(dθ/dt)^2)
    becomes
    m*r*(dθ/dt)^2=m*(d^2r/dt^2)
    clear!?
    Moment of inertia will increase quadratically with respect to position but this is by its definition. I guess you mean "to keep constant angular speed"!?
    If you want to go into energy considerations is much better if you consider the variation of the total energy in either the inertial or non-inertial frame.
    In the inertial frame there is only kynetic energy and, yes, its variation depends only on the variation of the moment of inertia. But in the reference frame you have chosen you have to include the potential of the centrifugal force!!

    Any remark? :)

    Hope this helped, Dario
     
    Last edited by a moderator: Apr 24, 2003
  17. Apr 24, 2003 #16

    dg

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    Here we have to clarify a little bit what we mean by tangential (velocity). If by it you mean tangent to the trajectory, this is redundant since velocity is tangent to the trajectory by definition of trajectory. What I wanted to stress in my reply is what happens in two orthogonal directions (radial and azimuthal) at the beginning, while mass is still constrained into circular motion. At this time tangential refers to two different concepts: one is being tangent to the trajectory, which is obvious; the other is the direction orthogonal to the radial one, tangent to the circle, which I will call azimuthal direction. After the mass is released, it is better to break the motion into two components along the radial and azimuthal directions (which is tangent to the circle going through the instantaneous position of the mass with center in the center of the wheel) and yes, at this point velocity will have both a radial and an azimuthal component: the first is growing exponentially according to the function hyperbolic sine, the second is growing according to v=ω*r and hence still exponentially but with a hyperbolic cosine law.
    Sometime I might have used tangential instead of azimuthal since at the beginning there is no difference between the two. Hope I did not!

    Dario
     
    Last edited by a moderator: Apr 24, 2003
  18. Apr 24, 2003 #17

    dg

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    Well there is no centripetal acceleration in this problem, lethe's formula works only in a non-inertial rotating frame where there would be at most a centrifugal acceleration created by the the fictitious centrifugal force...
    So here you are solving the problem in the inertial frame... (no potential energy...)
    It seems to me that you got the numbers right but I cannot agree on your interpretation of the results... ;)
    You should read your equation results in the original reference frame in three space, configuration space can be tricky! The result, in an inertial frame simply tells that the component of acceleration vector along the radius is zero! All your equations tell indeed is that there is a second derivative of a lagrangian coordinate that is non-zero...
    Anyway the interpretation can be made closer to lethe's assertions interpreting the results in the non-inertial frame... (see previous replies)

    I will be happy to clarify any doubt, Dario
     
    Last edited by a moderator: Apr 24, 2003
  19. Apr 26, 2003 #18
    ^^^ That's worked in a standard inertial frame, with polar coords -- r is exactly the radius here.

    My first blurb was a bit handwavy, but in fact works: the instantaneous centrifugal pseudoforce always has the same magnitude as the centripetal force at that w, and so in the absence of other radial forces, this will give the radial acceleration. If the relationship of w_instantaneous to r is not simple, we can't easily getting motion from this, but here they are completely decoupled.

    ??? I think I see the confusion: my r''(t) is the acceleration of the radial coordinate. The r'' that equals zero is the acceleration in an fixed direction momentarily parallel to the radial direction.
     
  20. Apr 27, 2003 #19

    dg

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    centripetal acceleration (not force) is present in circular motions meaning by this motion along a circular trajectory (constant radius), but here there is no such thing!
    Since there are no pseudoforces in an inertial frame (by definition of pseudoforce), I would be really curious to read how you would explain your solution to the problem in plain Newton equations in the inertial frame, with forces, reaction and so on...
    Could you detail this?
    Well there is no second derivative of anything that is equal to zero, it is one component of a (vector) second derivative that is zero.

    Beside that many of our consideration about energy so far (including mine) are wrong and energy does not increase simply by the moment of inertia since this does not consider the contribution of the "center of mass" speed:

    E=m/2*(ω2*r2+(dr/dt)2)

    and only the first term goes as we said not the second one...
     
    Last edited by a moderator: Apr 27, 2003
  21. Apr 27, 2003 #20
    From what I saw, the way you worked out the problem with free-body diagrams looked fine -- that's how I'd grunge through it with Newton's eqns if I had to. You came up with the same diffeq, right?
    OK - anytime the instantaneous radial force is zero, the equation r''=r*theta'^2 holds. So, if we are given theta(t) as part of the problem -- it could be any function of t -- we can just formally integrate this equation to get r(t).

    It's also doable in other cases, say if some torque tau(t) is specified to act on the wheel/mass. We have angular momentum L=r x p=m*r^2*theta', and tau=dL/dt, so

    L=int(tau(t))
    m*r^2*theta'=int(tau(t))
    theta' = int(tau(t)) / (m*r^2)

    Substituting this in our first diffeq,
    r''= int(tau(t))^2 / (m^2*r^3)
    which again involves only r and known functions of t. In the simple case where the wheel is "let go," tau=0 so L=const and theta' ~ r^-2, so r''= const/r^3 .

    But, if there is some complex relationship between r and theta -- like a torque that varies with r or theta -- then we will in general need to use the other Lagrangian equation d/dt (m*r^2*theta') = r*F_theta -- and it's no longer obvious how to get decoupled diffeqs.

    ---

    The CM energy won't affect anything, as long as the axis of the rotating wheel is fixed (or if the wheel masses are symmetrically distributed, of course).
     
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