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Point mass rotating about a pole with a spring

  1. Oct 30, 2016 #1

    VMP

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    1. The problem statement, all variables and given/known data
    In the beginning a point mass is rotating in a circle of radius [itex]L[/itex]. The spring is providing the centripetal force ([itex]\vec{F}=-k\vec{r}[/itex]) and the mass rotates with constant speed. At some point in time, a stick of radius [itex]a[/itex] ([itex]a<<L[/itex])lands near the center of the circle in such a way that center of the circle is tangent to the rim of the stick. Point mass is rotating about the stick and while doing so, it gets shorter.

    Question is to find the time it takes for the mass to hit the stick

    Solution in the book:
    [itex]t=\sqrt{\frac{m}{k}}\cdot\frac{4a}{5L}[/itex]
    2. Relevant equations

    Equations of motion in cylindrical coords.

    [tex]\vec{a}=\frac{a^2}{r}\hat{r}-\frac{a \sqrt{r^2-a^2}}{r}\hat{\theta}\\ \;
    \\ \vec{r}\;'=\vec{r}-\vec{a} \\ \; \\ \vec{F}=-k\cdot\vec{r}\;' [/tex]

    Vector [itex]\vec{a}[/itex] represents the radius of the stick, it's useful to have because I can write down the relation of the force to the new origin.

    3. The attempt at a solution

    Okay so first I tried approximating various things like angular momentum since [itex]a<<L[/itex] I assumed the angular momentum is conserved... also later on started using it to simplify some expressions like square roots and alike. As far as I can see, the energy is not conserved since part of the energy is "stuck" on the stick since it never has enough time to fully pull the mass. Other thing that occurred to me is to to look at part of the velocity vector perpendicular to the force. It yielded an equation, but none of these equations are simple equations of motion, I'm stuck with second order nonlinear ODE's.

    Thanks for reading, Cheers.
     
  2. jcsd
  3. Oct 30, 2016 #2

    haruspex

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    Interesting problem.
    Yes, that is probably why that is given, but it is going to be rather off as the mass nears the stick, no?
    Good point.

    Please at least post the working you have. Have you tried dropping the assumption of angular momentum conservation? Unlikely, I know, but it is just possible that the equations are more tractable.
     
  4. Oct 30, 2016 #3

    VMP

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    Yes I did and no success. I have a hunch that the spring is culprit of the problem, so I'm trying to model the spring in following way:

    The speed of the spring near the contact is approximately 0 therefore [itex]ds=L-dr'\;,\;dr'=\frac{rdr}{\sqrt{r^2-a^2}}[/itex] where ds is the amount of the spring wrapped up. The total energy of the system is then [itex]E(t)=E_{o}-\frac{1}{2}ks^2[/itex] I plugged it in, but the result is an integral that only has imaginary value, I've probably made an algebraic mistake.

    Cheers.
     
  5. Oct 30, 2016 #4

    TSny

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    Using ##a \ll L##, can you make an estimate of the time it takes for the mass to make the first orbit around the stick? How does this time compare to the answer in the book for the total time it takes the mass to reach the stick?
     
  6. Oct 31, 2016 #5

    haruspex

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    I think you have to be careful with regard to the spring 'constant'. It cannot be true that the tension is given by kx, where x is the length of the straight part of the spring (the part not yet wrapped around the spindle) and k is a constant. If it were, a solution would be that it keeps going in a circle. So the spring 'constant' effectively increases.
     
    Last edited: Nov 1, 2016
  7. Nov 1, 2016 #6

    VMP

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    Okay, I think I have a solution. If we watch time interval [itex](t,t+dt)[/itex] then length of the spring at time [itex]t+dt[/itex] is:
    [tex]r'(t+dt)=r'(t)-dr'-a\cdot d(\theta-\alpha)\;,\;r'(t+dt)-r'(t)=dr'\;,\;\alpha=arctan(\frac{\sqrt{r^2-a^2}}{a}) \\ \;\\
    r'^2=r^2-a^2\Rightarrow -2\cdot\dot{r}'=\frac{-2r\dot{r}}{\sqrt{r^2-a^2}}=a(\dot{\theta}-\dot{\alpha})\;(1)[/tex]
    Where [itex]\alpha[/itex] is the angle between position vector [itex]\vec{r}[/itex] and vector [itex]\vec{a}[/itex] which I previously mentioned.
    From the force, only radial component is needed since it contains first derivative of [itex]\theta[/itex] which we have in relation (1).
    [tex]\frac{1}{m}\vec{F}\cdot\hat{r}=\omega^2_{0}(\frac{a^2}{r}-r)=\ddot{r}-r\dot{\theta}^2\;,\omega_{0}^2=\frac{k}{m}\;(2)[/tex]
    If we combine (1) and (2) and get rid of [itex]\dot{\theta}[/itex] we get the following differential equation:
    [tex]\ddot{r}-\frac{(2r^2-a^2)^2}{ra^2(r^2-a^2)}\dot{r}^2=-\frac{\omega_{0}^2}{r}(r^2-a^2)[/tex]
    which fortunately is solvable if we substitute:[itex]\ddot{r}=y'y,y=\dot{r},r=x[/itex]
    Unfortunately, I don't know how to go further from the following expression without approximating:
    [tex]\dot{r}^2=a^2\omega_{0}^2(-\frac{a^2}{4r^2}+\frac{3}{4}(\frac{r^2}{r^2-a^2}-3\cdot exp\left \{ \frac{-4r^2}{a^2}\right \})+\frac{1}{4} )\;(3)[/tex]
    I played around with the equation (3) and with approximation the result is [itex]t\approx\frac{L}{a\omega_{0}}[/itex] and is usually slightly less than
    [itex]\frac{L}{a\omega_{0}}[/itex] another thing to note is that for and decent value of [itex]r[/itex] and small value of [itex]a<<1[/itex] equation (3)
    gives [itex]\dot{r}=-a\omega_{0}[/itex], I also played around with numerical integration and the results follow the same pattern.

    By the way the correct solution is [itex]\frac{4L}{5a\omega_{0}}[/itex] and I have no idea where does [itex]\frac{4}{5}[/itex] come from.

    Maybe better modeling of the spring could lead to this solution...

    Cheers.
     
    Last edited: Nov 1, 2016
  8. Nov 1, 2016 #7

    VMP

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    P.S. I'll try to model the spring "constant".
     
  9. Nov 1, 2016 #8

    TSny

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    OK. My post #4 was intended to show that the answer as given in the OP couldn't be correct.
     
  10. Nov 2, 2016 #9

    haruspex

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    Sorry, I am not even able to understand your first equations. How are you defining r'? The very first two equations seem to contradict each other.

    Here's what i did:
    x is the length of the straight portion of string (the unwound part). ω Is the rotation rate.
    In time Δt, a length aωΔt wraps onto the spindle, stretching the straight part to that extent.
    At the other end, the distance from mass to spindle changes. The new tension is a result of these two movements.
    ##\Delta T=ka\omega \Delta t+k\dot x\Delta t##
    ##\Delta x=\dot x\Delta t##
    ##\Delta \dot x=\ddot x\Delta t##
    The acceleration of x is the difference between the tension and the centripetal force.
    ##\ddot x=-kx/m+x\omega^2##
    Even though a<<initial length, that becomes wrong as the mass approaches the spindle, so angular momentum is not conserved.
    The new rotation rate is given by
    ##m(x+\Delta x)^2(\omega+\Delta \omega)=mx^2\omega-aT\Delta t##
    The effective new string constant is implied by the new length and the new tension:
    ##k+\Delta k=(T+\Delta T)/(x+\Delta x)##

    I have not attempted to turn those into differential equations, but I did put them in a spreadsheet to simulate it. I put all constants to 1 except a, which I set to 0.1. I think that is a valid normalisation, i.e. varying a is the only parameter of interest.

    Numerically, I got 2 time units, not the 0.08 seconds from the supposed formula.
    Note that a/L makes no sense. The greater the ratio the longer it will take.
    ##\sqrt{\frac mk}\frac{L}{5a}## fits what I found.
     
    Last edited: Nov 2, 2016
  11. Nov 2, 2016 #10

    VMP

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    Okay, I've been a bit vague. Imagine you've got an origin ##\mathcal{O}## at the center of the stick and there is a vector ##\vec{a}##
    that is given by the contact point P of the spring and the origin. This vector ##\vec{a}## is always perpendicular to the vector ##\vec{r}'##.
    Vector ##\vec{r}'## is given by the contact point P and location of the mass. In other words norm of ##\vec{r}'## is the length of the spring than is unwound (straight as you called it). What I'm saying in my first few equations is that ##r'(t+dt)## is the sum of following terms:

    ##r'(t)## length of straight part at time t.

    ##-dr'## is the amount that spring force shortens it.

    ##-ad(\theta-\alpha)## is the amount that is wrapped up. The angle ##(\theta-\alpha)## can be found by a little bit of geometry.

    I hope this makes it clear.

    Cheers

    Edit: I just realized that that I set up the equation involving differentials wrong. It should be like this:

    ##r'(t+dt)-|dr'|-ad(\theta-\alpha)=r(t)## The differential ##dr'## is negative that is why it has absolute value.
     
    Last edited: Nov 2, 2016
  12. Nov 3, 2016 #11

    haruspex

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    Update to post #9.
    I found a bug in the equations as typed into my spreadsheet. Having correctly coded the equations I laid out in post 9, I now observe numbers matching ##\sqrt{\frac mk}\frac{4L}{5a}## fairly accurately up to a/L=0.3:
    Code (Text):

    a/L         t         expected
    0.01      80      80.00
    0.02      40      40.00
    0.04      20.1    20.00
    0.06      13.4    13.33
    0.07      11.5    11.43
    0.10      8.13     8.00
    0.15      5.49     5.33
    0.20      4.19     4.00
    0.25      3.43     3.20
    0.30      2.9      2.67
    0.35      2.55     2.29
     
    I tried dropping the angular momentum change, i.e. making it constant, but the match was much poorer.
    So it would seem the only flaw in the original question was a typo which put L/a up the wrong way.
     
  13. Nov 3, 2016 #12

    haruspex

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    This is not a good idea. If r' is a variable defined in some way, dr' should be the net change in time dt, from all causes.
    Found from what?
    I suggest you read through my equations in post #9, understand them, and try to turn them into differential equations.
     
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