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Kinematics Belt and Pulley Problem

  1. Apr 17, 2017 #1
    1. The problem statement, all variables and given/known data
    I think I made a mistake somewhere..
    Capture.PNG

    2. Relevant equations
    T = Jα
    T = F*R

    3. The attempt at a solution
    A)
    I started with T = Jα
    Since there is no slip, αm = αL

    Thus:
    Tm / Jm = TL / JL

    Plugging in, we find TL = Tm * JL / Jm = 2560

    Now use T = F*R.

    Tm = Fm * Rm
    Plugging in shows Fm = 2666.67.

    Since F has to be equal on both pulleys, we use T = F*R again to solve for RL = 0.96m.
    -----------------------------------------------
    B)
    Now that we have those solved we can solve for t.

    Θ = Θi + 0.5α*t2

    The initial conditions are zero.

    Since α must be the same for either, we can solve for just one scenario.

    m = RLαL
    αm = Tm/Jm
    αL = r*Tm / RL*Jm

    plugging back into this we get
    Θ = 0.5(r*Tm / RL*Jm)*t2

    Which when applying the condition Θ=1rad, we get
    t as the square root of the reciprocal which shows
    t = 0.08sec.

    This is where I'm in trouble though. My velocity vs time for this equation will not have a maximum or a finish. I can get a minimum but my equation of motion Θ is a parabola without a finish line. What did I do wrong here because I can't calculate a maximum from that velocity graph.
    ----------------------------------------------------
    C)

    This is where I'm in trouble
     
  2. jcsd
  3. Apr 18, 2017 #2

    haruspex

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    The same angular acceleration? Why?
     
  4. Apr 18, 2017 #3
    Because each pulley has to rotate the same theta or else there would be slip. They also have to accelerate at the same rate or one old be going faster than the other.
     
  5. Apr 18, 2017 #4

    haruspex

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    This is two pulleys connected by a belt, not sharing an axle, right?
    If each pulley rotates through angle theta, what length of belt has gone around each pulley?
     
  6. Apr 18, 2017 #5
    Okay, I understand now why each pulley cannot have the same alpha. I got stuck in a way of thinking and then I asked someone who was supposedly good at this to look over it and they said that it looked good. Larger wheel has to have a lower angular velocity, rpm's, than the smaller wheel though the length of belt (RΘ) is the same. How then can I adapt my equations to be good for any theta, especially when ω is not constant, but rather increases and decreases with time? My whole foundation was on equating each of them through alpha.
     
  7. Apr 18, 2017 #6

    haruspex

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    By equating the belt movements. That relates displacement, velocity and acceleration in exactly the way.
     
  8. Apr 18, 2017 #7
    I guess I don't see how to relate the acceleration, which based on the maximum torque is T/J and the velocity in a generalized term.
    Relating the diameter of each pulley with the velocity,
    ND = ND, where N is the number of rotations. I guess I'm having trouble now seeing how T/J can have the derivative taken and give me a velocity because it's not very generalized.
     
    Last edited: Apr 18, 2017
  9. Apr 18, 2017 #8

    haruspex

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    If the two pulleys have radii r1 and r2, and rotate through angles θ1, θ2, the belt movement is r1θ1=r2θ2. To relate the angular velocities differentiate once, to relate the angular accelerations differentiate a second time.
     
  10. Apr 18, 2017 #9
    I guess I understand that but here let me show you how i see it,
    I understand that
    r1Θ1 = x
    r1Θ'1 = x'
    r1Θ''1 = x''

    but we have T=Jα, where T and J are given and T is the max torque. That leaves α as T/J = 26,666.67 and if I differentiate it, as it's a constant, will equal zero.
    Conversely, if i integrate r1Θ1 = x then I'm left with Θ = 1/2αt2 + 0. This can't be the answer either as the velocity vs time relationship should ascend and the descend because it starts and stops at 0, unless it immediately stops the instant it reaches the desired position (which would theoretically 'minimize' the amount of time required). So I understand the general relationship between (angular) placement, velocity, acceleration, but I'm having trouble applying it.
     
  11. Apr 18, 2017 #10

    haruspex

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    That equation is only valid for constant acceleration. You have to decide how the acceleration needs to vary with time to satisfy the question.
     
  12. Apr 18, 2017 #11
    You're right, I'm left with an even more generalized solution. r1Θ1 = x
    The question ask me to minimize the time required to move through an angle theta, but it doesn't give me any relationship with time. It just gives me the size, weight and torque of one pulley and then the weight of another.
    If I say r1α1=r1α1 then I could say that α1 = T/J at max torque (and it is assumed that this is the maximum amount of force the pulley can exert, thus minimum time. However, this does not give me a t to manipulate. It just says that r1α1 = (0.1m)(0.0015kg-m^2) / (40N-m) and I have 2 unknowns and 1 expression. I could potentially use the knowledge that r1θ1 = r2θ2 but I don't really see how since the theta is arbitrary.

    I'm sorry to sound like a broken record ( and make you sound like one too) but this deriving things is a serious challenge for me.
     
  13. Apr 18, 2017 #12

    haruspex

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    Right, but you cannot keep it at max torque in the same direction since it has to end stationary. So what can you do instead?
     
  14. Apr 18, 2017 #13
    accelerate it as fast as possible and then decelerate it just as fast at the 50% marker.
     
  15. Apr 18, 2017 #14

    haruspex

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    Exactly
     
  16. Apr 18, 2017 #15
    To me, this would result in a piecewise function, where
    For 0< θ < θ/2
    α = T/J

    For θ/2 < θ < θ
    α = -T/J
     
  17. Apr 18, 2017 #16

    haruspex

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    Right.
     
  18. Apr 18, 2017 #17
    Okay, then this does puts be back at the same problem though.
    At in the first case, 0 < θ < θ/2
    I'm in the original instance of α = t/j but this only shows that r2 = (r * J) / (T *α2)
     
  19. Apr 18, 2017 #18

    haruspex

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    I find it hard to followyour algebra because you don't clearly distinguish the different radii, moments of inertia and torques.

    If the angular acceleration of the load is αL, what torque is being applied? What corresponding torque is the load exerting on the motor pulley? What is the angular acceleration of themotor pulley? What total torque isthe motor therefore generating?
     
  20. Apr 19, 2017 #19
    In order:

    Given:
    r = the radius of the motor
    Jm = the moment of inertia for the motor
    Tm = the torque of the motor
    αm = the angular acceleration of the motor, which = Tm / Jm

    R = the radius of the pulley
    JL = the moment of inertia for the motor
    TL = the torque of the motor
    αL = the angular acceleration of the motor, which = TL / JL

    m = RαL as there is no slip.
    Thus
    αL = rαm / R = rTm / RJm


    Solving for the torque... I have this:

    Tm = Fm * r
    TL = FL *R
    Where F is the tension in the belt, and theoretically has to be the same or else the pulley/motor are moving.

    So Fm = FL
    Thus
    TL = Tm *R /r = JLαL

    If
    αL = rαm / R = rTm / RJm
    then
    Tm *R /r = rTm / RJm
    and
    R2 = r2 / Jm, which is finally solvable.

    The angular acceleration of the motor pulley is Tm / r as stated above
    The total torque would be... the maximum torque given in the problem? I am not quite sure what you mean by this.
     
  21. Apr 19, 2017 #20

    haruspex

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    No, that is what it would be if there were no belt attached.
     
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