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Finding the Time of Max and 0 Angular Speed

  1. Jul 30, 2017 #1
    20170730_164233.jpg 1. The problem statement, all variables and given/known data
    Mod note: Fixed thread title. OOPS NOT Acceleration, Speed! (Thread title is incorrect)

    2. Relevant equations
    w=d∅/dt
    v=rw
    x={+,-√(b2-4ac)}/2a


    3. The attempt at a solution
    I solved this with help from Chegg study, however, I'm still not entirely sure what I am doing. Obviously, for part a, you differentiate the position to get the angular velocity which is w=5t-1.8t2. At that point I struggled. Chegg answers simply differentiated twice to find the time of maximum angular speed as 1.39 s. Is that because ∝=5-3.6t gives the time of maximum acceleration? and therefore max speed?

    My first instinct was to perform the quadratic of the speed equation to give me t= 2.78 s and 1.39 s. One is the time of max speed and one is when it is zero (aka part c for when it will reverse direction)... How do I know which is which other than having ruled 1.39 out in part a? For part c, Chegg used the equation t(5t-1.8t2)=0 to find the time.But I'm not sure where that equation came from. Where are they taking the t out front from? Is there a better way to do this? Once I can get the time, I am fine to solve the rest.
     
    Last edited: Jul 30, 2017
  2. jcsd
  3. Jul 30, 2017 #2

    Mark44

    Staff: Mentor

    Although you could differentiate again, in this case it isn't required. Your angular velocity, ##\omega(t)##, is a quadratic polynomial whose graph is a parabola, so to find the times where ##\omega## is zero, simply solve the equation ##5t - 1.8t^2 = 0##. Note that there will be two times.
    To find the max. angular velocity, locate the vertex of the parabola.
    See what I wrote above. If you recognize that the graph of the angular velocity is a parabola, it should be straightforward to figure out which one is the max. point, and which is a point where ##\omega## is zero.
     
  4. Jul 30, 2017 #3
    I suppose that means I should start taking my graphing calculator to exams! Once I graphed it, I saw it immediately. Thank you! How do I solve this via equations, though? So differentiating twice solves the time of max speed. Is there any other method than graphing to find when the speed is 0? I mean I see how they solved t(5t-1.8t2), however I have no clue where they got the t out front from.
     
    Last edited: Jul 30, 2017
  5. Jul 30, 2017 #4

    Mark44

    Staff: Mentor

    A graphing calculator isn't necessary, but recognizing a quadratic function is necessary. To find the ##\omega## intercepts, solve the equation ##5t - 1.8t^2 = 0##, which you can do by factoring to ##t(5 - 1.8t) = 0##. To find the maximum point complete the square in ##\omega = 5t - 1.8t^2##. This will result in ##\omega = -1.8(t - a)^2##, where a is the coordinate of the vertex.
     
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