1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Start-up acceleration, max speed and torque of a transfer car

  1. Apr 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A transfer car moves concrete slabs.
    Given values are wheel diameter D, shaft diameter d, rolling resistance FL , speed v, car weight mv , load weight mk , output power P at rotation speed n, average torque Mm , motor moment of inertia Jm , fan moment of inertia Jt and efficiency of the mechanical transmission system η.

    a) Calculate start-up acceleration and distance (up to maximum speed).
    b) Calculate the torque of the driving shaft of the gear at start.

    2. Relevant equations
    None given... (this is the main problem). I am studying a course abroad which has support material in a foreign language.
    I have found these so far:
    P=Mm⋅ω
    a=R⋅α
    α=dω/dt

    3. The attempt at a solution
    I have tried to find the angular acceleration using power, torque, radius and speed, however, these don't work in the start-up case since v=0. As for torque, I think I first need to find the acceleration and the just input into torque=moment of inertia⋅angular acceleration
    I also don't know how and what forces to sum to calculate the torque.
     
  2. jcsd
  3. Apr 20, 2016 #2

    billy_joule

    User Avatar
    Science Advisor

    Show your attempt so we can see where you're stuck.
    You can find acceleration at v=0 by assuming average torque is constant. Power is irrelevant when v=0 as no work is being done.
     
  4. Apr 20, 2016 #3
    @billy_joule The problem is I don't really know where to start, therefore, I don't really have an attempt as such. I have only come up with the conclusion that torque=angular acceleration x moment of inertia, but even then, which of the two moments of inertia am I supposed to use? I don't even know why I have two different moments of inertia :frown:
    Once I solve that I will have the acceleration and from there I just use uniform acceleration formulas to find the distance reached at top speed.
    As for start-up torque, I believe I need to calculate the sum of all weight plus rolling resistance forces and calculate the torque needed to move that? But then, with what formula do I relate angular with linear forces?
     
  5. Apr 20, 2016 #4

    billy_joule

    User Avatar
    Science Advisor

    Is that the entire problem statement word for word?


    Well, do the fan and motor both need to be accelerated? Or to put it another way, would a fan or motor that didn't spin be any use at all?

    Yes, this approach will ignore P @ n and assume average torque is constant from v = 0 to vmax. There is really no way the given power variable can be used as it's only for a single RPM.

    T=Fr
    The motor and fan are accelerated rotationally, the car and load are accelerated linearly. These two accelerations are constrained together as the linear velocity of the car depends on the rotational velocity of the motor ( ωengine ∝ v car ). To solve for linear acceleration, you need to form an expression that includes this fact.
    It'll require Newtons second law:
    F = ma
    and it's rotational analogue:
    T = Iα
    You'll also need to account for torque loss in the transmission.
     
  6. Apr 20, 2016 #5
    I have attached a screenshot of the original problem.
    My thoughts are both moments of inertia will only be used for the calculation of the torque on the shaft. They add each other?
    I agree power won't be used for a) but isn't it used to find one more torque to add to b) with the formula P=Mm⋅ω?
    How do I find torque loss? Does that involve rolling resistance and efficiency of the mechanical transmission system?
     

    Attached Files:

  7. Apr 21, 2016 #6

    billy_joule

    User Avatar
    Science Advisor

    Yes they add, in just the same way you'd add masses constrained together to find linear acceleration.
    eg
    F = ma = (mvehicle + mload)a
    T = Iα = (Imotor + Ifan


    Well you're only given the power at a single RPM so it's not much use.
    The question is rather vague, I think there are multiple way to interpret the situation:

    1) The transmission is a 1:1 right angle drive (Fair assumption as transmission ratio isn't mentioned) - in which case the motor RPM is the same as the wheel RPM and is much lower than the given power value RPM making the provided power value completely irrelevant (as the motor never reaches that speed).

    2) We are expected to assume the motor is running at the given power value RPM when travelling at top speed, this will give use a gear ratio that is not 1:1.
    And it introduces a significant problem: The given power output does not match that found if we use the average torque value for that RPM eg

    We are told P @ 2780RPM = 0.6 kW

    and we know P = Tω (and 2780RPM = 291 rad/s) and are told average motor torque is 3.9 Nm
    so P = (3.9 Nm) * (291 rad/s) = 1.13 kW

    Which begs the question: When is the given motor torque valid? And when is the given motor power valid?
    In conclusion, the question is very poor, I'd suggest skipping it or asking your tutor to clarify.

    You multiply the torque input into the transmission by efficiency (and the gear ratio) to find the torque output.

    The rolling resistance is a resistive, external force on the car as a whole. So, we have two forces, the forward force due to the driving torque on the wheel:
    T = Fr
    Twheel shaft=Fwheel-road contactrwheel radius
    so
    Fwheel-road contact = Twheel shaft / rwheel radius

    and the rearward rolling resistance force Frolling resistance
    If we apply Newtons second law we can solve for the net force

    F = ma = Fwheel-road contact - Frolling resistance = ma
    Solve for a

    a = (Fwheel-road contact - Frolling resistance ) / (mvehicle + mload)
     
  8. Apr 21, 2016 #7
    I had also delved around this thought, it does not make any sense...

    According to this you say and some calculations I managed to make last night, I came up with an equation, I hope it works for calculating the torque on the shaft once I find the angular acceleration:
    Ms=(Jm+Jt-((D-d)/2)⋅FL)ηα (considering transmission efficiency not only affects power)
    The problem now would be finding either the angular or linear acceleration since the equation you gave needs for me to find the shaft torque. Is it possible to find the angular acceleration from the fact that I know the moment of inertia and torque of the motor? I would simply solve Mm=(Jm+Jt)α, or α=Mm/(Jm+Jt) but then again, that is not affecting the shaft or wheels...

    Edit: what if I use I=mr2/2 using the total mass of the cart plus slabs, and then find α=Torque of motor/I ?
     
  9. Apr 21, 2016 #8

    billy_joule

    User Avatar
    Science Advisor

    Well, which shaft? If we assume the power flow goes motor > Fan > transmission > Wheels and there is a shaft between each step then torque will decrease at each step (assuming no gearing).

    That equation is not dimensionally consistent. You can't add a moment of inertia to a torque. And 'd' should not appear in your working, the wheel diameter is the only relevant distance. Also, FL does not produce a torque about the wheels.

    That would be the correct solution of the motor were only accelerating itself and it's fan, but we know that's not the case.

    The cart and slab aren't rotating so their α is zero and moment of inertia (about any axis) isn't relevant.

    I'll post a more complete explanation of the correct approach tomorrow to get you on the right track.
     
  10. Apr 22, 2016 #9
    The question says on the driving shaft of the gear. I assume this means the point of the shaft between the wheels and motor-fan set-up.

    You are right, I made a mistake there, this should be the equation I meant Ms=(Jm+Jt)ηα-((D-d)/2)⋅FL and I think the shaft diameter does affect since the moment created by the rolling resistance times the radial distance (torque=force x distance) starts from the end of the shaft's radius until the end of the wheels right?

    Ok then, if F=ma, Torque=F x distance and the mass is given, a=Torque / (Distance x m). Is this any help?
     
  11. Apr 27, 2016 #10
    Ok after analyzing your past messages I came up with a new formula.
    a=(T/r-FL)⋅mη
    This considers Fwheel-road contact = Twheel shaft / rwheel radius, however T=moment of inertia x angular acceleration; how do i find the latter?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Start-up acceleration, max speed and torque of a transfer car
Loading...