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Homework Help: Accelerating a car including the moment of inertia of the wheels

  1. Nov 5, 2017 #21

    TSny

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    The torque equations can be used to get expressions for Ffront and Frear in terms of m and α. Use your diagrams for the wheels to help set up these equations.
     
  2. Nov 5, 2017 #22
    Could it be FrearR = Iα?
    Is it the same at the front?
     
  3. Nov 5, 2017 #23

    TSny

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    Yes. Can you identify the type of force represented by Frear?
    No, the front wheel also has the torque of the engine that you need to deal with.
     
  4. Nov 5, 2017 #24
    It's a contact force. Frictional force, i guess.

    Is it:
    τ - FfrontR = Iα?
     
  5. Nov 5, 2017 #25

    TSny

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    Yes

    Yes.
     
  6. Nov 5, 2017 #26
    Cool! :partytime:
    Thank you for all the help. :thumbup:
     
  7. Nov 5, 2017 #27

    TSny

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    Sure. Good work!
     
  8. Dec 3, 2017 #28
    Umm, guys, i just did this using:
    2Ffront - 2Frear = a(M + 4m),
    τ - FfrontR = Iα,
    FrearR = Iα,
    but it gave me ##a = \frac {2τ} {R(M+2m)}## instead of ##a = \frac {2τ} {R(M+6m)}##

    Is something missing?
     
  9. Dec 3, 2017 #29

    TSny

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    These equations look correct.

    Check your work. The equations should yield ##a = \frac {2τ} {R(M+6m)}##.
     
  10. Dec 4, 2017 #30
    I made a mistake, all good now. =]
     
  11. Jul 7, 2018 #31
    Can the OP or someone else explain how that equation was simplified to get a factor of 3 in the denominator for "m"?
    The original equation involves a term with rotational kinetic energy (1/2Iω2).

    Was that "I" for the object's inertia converted to some kind of equivalent mass?
     
    Last edited: Jul 7, 2018
  12. Jul 7, 2018 #32

    TSny

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    Each wheel is treated as a uniform, solid cylinder for which I = (1/2)mR2. Also, for rolling without slipping, ω = v/R.
     
  13. Jul 7, 2018 #33
    Thank you.

    I was using I = mR2, hence why it didn't come out the same.
     
  14. Jul 7, 2018 #34

    TSny

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    OK. Glad it makes sense now.
     
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