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I Maximum tangential velocity from maximum linear velocities

  1. Feb 17, 2017 #1
    I have a particle that moves along a circular arc centered at origin in 2D plane.

    I have the following angular displacement function for time T0 through T3 and the following acceleration and velocity constraints in Cartesian coordinates. At t = T0, theta = 0, velocity = 0. At t = T3, velocity = 0.

    I want to calculate the maximum A_T and V_T in terms of the linear A_max and V_max quantities (A_max and V_max put a limit on A_T and V_T). I want to eventually be able to calculate T1, T2, T3.

    [tex]
    \forall t \in [T_0, T_1]
    \begin{cases}
    \frac{d^2\theta}{dt^2}(t) = A_T \\
    \frac{d\theta}{dt}(t) = \dot{\theta}_{0} + A_T(t-T_0) \\
    \theta(t) = \theta_{0} + \dot{\theta}_{0}(t-T_0)+\frac{1}{2}A_T(t-T_0)^2 \\
    \end{cases}

    \\

    \forall t \in [T_1, T_2]
    \begin{cases}
    \frac{d^2\theta}{dt^2}(t) = 0 \\
    \frac{d\theta}{dt}(t) = V_T \\
    \theta(t) = \theta(T_1)+V_T(t-T_1) \\
    \end{cases}

    \\

    \forall t \in [T_2, T_3]
    \begin{cases}
    \frac{d^2\theta}{dt^2}(t) = -A_T \\
    \frac{d\theta}{dt}(t) = \dot{\theta}(T_2) - A_T(t-T_2) \\
    \theta(t) = \theta(T_2) + \dot{\theta}(T_2)(t-T_2) - \frac{1}{2}A_T(t-T_2)^2 \\
    \end{cases}
    \\
    \\
    \vec{A_{max}} = \begin{pmatrix}
    A_{max,x} \\
    A_{max,y} \\
    \end{pmatrix}
    \\
    \vec{V_{max}} = \begin{pmatrix}
    V_{max,x} \\
    V_{max,y} \\
    \end{pmatrix}
    \\
    \vec{Position} =
    \begin{pmatrix}
    x \\
    y \\
    \end{pmatrix} =
    \begin{pmatrix}
    \cos(\theta(t)) \\
    \sin(\theta(t)) \\
    \end{pmatrix}
    [/tex]

    At first I figured I could solve
    [tex]\vec A =
    \begin{pmatrix}
    \frac{d^2}{dt}[\cos(\theta(t))] \\
    \frac{d^2}{dt}[\sin(\theta(t))]\\
    \end{pmatrix}
    [/tex]

    for A_T at t = T0 with all of the initial conditions applied but I'm afraid this is not correct.

    How do I approach this problem?

    Thanks
     
  2. jcsd
  3. Feb 17, 2017 #2

    BvU

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    Your sequence is: constant angular acceleration in ##[T_0, T_1]##, uniform angular motion in ##[T_1, T_2]## and constant angular deceleration in ##[T_2, T_3]## with the same angular acceleration, but opposite sign..
    You don't mention the radius ##r## of the circle.
    Your expressions provide these numbers: maximum angular velocity ##A_T (T_1-T_0) ## at ##T_1##. Linear acceleration at that point (value unknown if ##r## is unknown) and linear velocity (idem) both at maximum.
    etc etc.
     
  4. Feb 18, 2017 #3
    Yes, sorry. I wanted to make the radius R be variable and I forgot to include it both the description and the equations.

    I think I should have stated my question better. I'm looking for the largest magnitudes of A_T and V_T such that
    [tex]|A_x| =|\frac{d^2}{dt^2}(R\cos(\theta(t)))| \leq A_{max,x}\\
    |A_y| =|\frac{d^2}{dt^2}(R\sin(\theta(t)))| \leq A_{max,y}\\
    |V_x| =|\frac{d}{dt}(R\cos(\theta(t)))| \leq V_{max,x}\\
    |V_y| =|\frac{d}{dt}(R\sin(\theta(t)))| \leq V_{max,y}\\[/tex]

    If I manage to come up with something, I'll update the post.
    Thanks
     
    Last edited: Feb 18, 2017
  5. Feb 18, 2017 #4

    BvU

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    Well, you have an expression for ##\theta(t)## so you can differentiate it.
     
  6. Feb 18, 2017 #5
    For
    [tex]t \in [T_0,T_1], \dot{\theta}(T_0) = 0, \theta(T_0) = 0, R > 0, A_{max,x} > 0[/tex]
    Analyzing A_x. After a bit of simplification, I arrive at
    [tex]|(-1)(A_T\sin(\frac{1}{2}A_T(t-T_0)^2)+A_T^2(t-T_0)^2\cos(\frac{1}{2}A_T(t-T_0)^2))| \leq \frac{|A_{max,x}|}{R} = \frac{A_{max,x}}{R}[/tex]
    I'm not quite sure how to get an upper bound for A_T from this.

    I would then like to perform the same analysis for
    [tex]t \in [T1,T2], t \in [T2,T3][/tex]
    and combine the estimates. I'm not sure if this approach will work
     
    Last edited: Feb 18, 2017
  7. Feb 18, 2017 #6

    BvU

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    Is ##A_{max,x}## different from ##A_{max,y}## ? Is seems to me that this treatment in cartesian coordinates makes things unnecessarily complicated; that's why I ask.
    Can you show your differentiation steps ? Doesn't look right dimension-wise !

    If I do ##{d^2\over dt^2 } \left ( R\cos\theta\left (t \right )\right)## I get (with a shorthand dot for time derivative): $$
    {d\over dt } \left ( R\cos\theta\left (t \right )\right) = -R\sin\theta\; \dot \theta \Rightarrow $$ $$
    {d^2\over dt^2 } \left ( R\cos\theta\left (t \right )\right) = - R\cos\theta\; \dot \theta^2 - R\sin\theta\; \ddot \theta$$with ## \ddot \theta = A_T## and ## \dot \theta = A_T(t-T_0)##
     
  8. Feb 18, 2017 #7
    Yes, the maximum acc/vel magnitudes in cartesian coordinates are not ideal to work with. This is how the world (2d plane) in which my particle moves is described unfortunately. [tex]A_{max,x}[/tex] could be different from [tex]A_{max,y}[/tex] which is why I feel I need to break this down into components.

    For [tex]t\in[T_0,T_1][/tex]
    I get
    [tex]A_x = -R((A_T(t-T_0))^2\cos(0.5A_T(t-T_0)^2) + A_T\sin(0.5A_T(t-T_0)^2))[/tex]

    I expect to find A_T in terms of A_x then in terms of A_y and settle for the smaller of the values so that neither A_max,x nor A_max,y are ever exceeded as the particle travels around the circular arc.
     
    Last edited: Feb 18, 2017
  9. Feb 19, 2017 #8

    BvU

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    Our ##A_x## agree. So (in a mixed notation) ##A_x = - \omega^2 x - \alpha y\quad ## with ##(\alpha = A_T)##. and so on.
    In uniform circular motion the amplitudes of ##A_x## and ##A_y## are equal, but in uniformly accelerated motion both grow at the same rate, but individual extrema occur at different moments, so the value of ##T_1## comes into play.
    It depends: if ##T_1## is large and ##A_T## is not, then during one revolution the extrema won't differ much; that's when you can simplify and use ##\min (A_x, A_y)## as constraint.
    But if you accelerate strongly and change to coasting quickly it's different.

    Comparable reasoning for ##T_2, T_3##.

    Numerical simulation (or simply drawing ##x(t)## and ##y(t)## from ##T_0## to ##T_3##) might give easy insight
     
  10. Feb 19, 2017 #9
    Thanks for taking the time to respond BvU.
    I plotted the x''(t) and y''(t) on the domain from T0 to T1 with some arbitrary numbers and that gave me some interesting insight. I see the amplitudes peaking at time T1. In general I see the dependence on time.

    The problem I'm facing is that only T0 is known, theta at T0 and T3 is known but T1, T2, T3 are not known. I'm not sure how to calculate these without knowing A_T and V_T.
     
  11. Feb 20, 2017 #10

    BvU

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    I understand. Is it clear that the number of revolutions during the cycles plays a role in estimating whether the difference between ##A_x{\rm,\; max}## and ##A_y{\rm,\; max}## is relatively unimportant ? (in particular during the stage ##[T_1, T_2]## ?

    Can you formulate the optimization problem (apart from what's already in post #1, I mean):
     
  12. Feb 20, 2017 #11
    I should have mentioned that:
    [tex]|\theta_{T_3} - \theta_{T_0}| \leq 2\pi[/tex]

    ie. the particle will never make more than one full revolution around the origin (both thetas are known).

    The particle needs to reach the coasting stage or transition straight from the acceleration to deceleration stage as quickly as possible.

    I need to think about everything that was posted so far again from the beginning.
     
    Last edited: Feb 20, 2017
  13. Feb 20, 2017 #12

    BvU

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    Yes. Helps understand why x and y can be different :smile:
    that too ! The constraint I've been fishing for ...
    Perhaps you can consider only Ay in first order if theta(0) = 0 ?
     
  14. Feb 21, 2017 #13
    At [tex]\theta(0)=0 \implies A_y = RA_T[/tex]

    Sorry if I misunderstood the question
     
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