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Acceleration of car

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data
    I have a car with a mass (m) and tires having moment of inertia (I). I need to find an equation for the acceleration of the car from what is given. thanks in advance for the help.


    2. Relevant equations
    Not sure what all is relevant here but, M=I[itex]\alpha[/itex] F=ma


    3. The attempt at a solution
    I really have no idea how to even begin this. My attempt got me a=I[itex]\alpha[/itex]/m and im pretty sure this isn't correct
     
  2. jcsd
  3. Jun 16, 2011 #2

    tiny-tim

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    welcome to pf!

    hi flasheddread! welcome to pf! :smile:

    (have an alpha: α :wink:)

    the only external force on the car is the friction, F

    (so the external torque is Fr)

    you'll also need the rolling constraint, a = rα :wink:

    (you'd better assume it's four-wheel drive … two-wheel drive gets rather complicated)​
     
  4. Jun 20, 2011 #3
    :) thanks

    I get that friction is the only external force here. I guess I'm just having trouble understanding how to get the acceleration for the car using the moment of inertia.

    I start off by saying the Fr=Iα, which is the torque created by the tires that will move the car. Then i need to relate that to the acceleration of the car so i say Fr=ma?? Now this is where I think I'm messing up. I say that Iα=ma and rearranging that, I can say that a=Iα/m.

    I know this is completely wrong, since having a bigger moment of inertia for the tires would mean the car would accelerate slower, and this equation proves the opposite. Or would this be correct because the acceleration is on determined by how much force is placed on the tires to rotate? Maybe I'm just thinking about this too hard, because it doesn't seem that difficult.

    Any help please??
     
  5. Jun 20, 2011 #4

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  6. Jun 20, 2011 #5
    Alrighty I think I might have it now!!

    The engine supplies a torque to the tires, which in turn gives them an angular acceleration.
    τ=Iα
    α=a/r

    so, a= τr/I

    This makes much more sense. I have no idea what I was thinking earlier. I'm pretty sure this is correct this time, :smile: and if not I might just drop out of college. :cry:
     
  7. Jun 20, 2011 #6

    tiny-tim

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    wait a mo! :rolleyes:

    you haven't used either the mass of the car, or the friction force F

    don't forget that there are two torques on the wheel (why do you keep saying "tires"? :redface:), the known applied torque τ from the engine and the unknown torque from the friction from the road (which you'll have to eliminate) :wink:
     
  8. Jun 21, 2011 #7
    Oh wow, yea that's just what kind of day I was having yesterday. :frown: Sorry for being so difficult and thanks for sticking around for me.

    Alright so about this friction force. Would we eliminate it by saying the friction force is equal to the mass times acceleration of the car? If so and I am doing this correctly we get.

    τ-Fr=Iα, which then goes to τ-mar=Ia/r

    And from here its just manipulating it to solve for a.
     
  9. Jun 21, 2011 #8

    tiny-tim

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    hi flasheddread! :wink:
    yup, the https://www.physicsforums.com/library.php?do=view_item&itemid=39" on both axles), so F = ma :smile:
    exactly :smile:

    btw, note that if you fivide by r, you get τ = (m + I/r2)a,

    so effectively we can ignore the friction if we pretend that the wheels have an extra "rolling mass" of I/r2 :wink:
     
    Last edited by a moderator: Apr 26, 2017
  10. Jun 21, 2011 #9
    Wouldn't torque be divide by r as well?
     
  11. Jun 21, 2011 #10

    tiny-tim

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    oops!

    oops! :redface:

    yes … τ/r = (m + I/r2)a :smile:
     
  12. Jun 21, 2011 #11
    Sweet, thanks so much for your help! I will keep rolling mass in memory; it sounds pretty useful. :smile:
     
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