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Torque Plus Power In Relation to Velocity

  1. Jan 22, 2015 #1
    1. The problem statement, all variables and given/known data

    The maximum torque output from the engine of a new experimental car of mass m is τ . The
    maximum rotational speed of the engine is ω. The engine is designed to provide a constant power
    output P. The engine is connected to the wheels via a perfect transmission that can smoothly
    trade torque for speed with no power loss. The wheels have a radius R, and the coefficient of
    static friction between the wheels and the road is µ.
    What is the maximum sustained speed v the car can drive up a 30 degree incline? Assume no
    frictional losses and assume µ is large enough so that the tires do not slip.

    (A) v = 2P/(mg)
    (B) v = 2P/(√3mg)
    (C) v = 2P/(µmg)
    (D) v = τω/(mg)
    (E) v = τω/(µmg)

    2. Relevant equations


    3. The attempt at a solution
    I don't even know how to start on this. I'm supposing that the first step is finding a relationship between the torque, power, and velocity , but I don't know how to do that. Thoughts?
     
  2. jcsd
  3. Jan 22, 2015 #2

    Bystander

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    Can you write a relationship for two of the three variables you've listed?
     
  4. Jan 22, 2015 #3
    P=F*v*cos(theta).
     
  5. Jan 22, 2015 #4

    Bystander

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    And "F" is what?
     
  6. Jan 22, 2015 #5
    Force
     
  7. Jan 22, 2015 #6

    haruspex

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    Well, yes, I think we all guessed that, but which force?
     
  8. Jan 22, 2015 #7
    The force is the one that the car provides.
     
  9. Jan 22, 2015 #8

    haruspex

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    The car's engine provides a torque, not a force.
    I could guess you mean the propulsive force up the hill provided by friction, but then I don't know where the cos(theta) comes from.
     
  10. Jan 22, 2015 #9
    I looked up the torque-velocity relation ,was I supposed to derive it somehow?
     
  11. Jan 22, 2015 #10

    haruspex

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    If you mean P=Fv, no. Just explain exactly what F is in the present context and how cos theta comes into it.
     
  12. Jan 22, 2015 #11
    F is the force produced from the power of the engine and the cos theta comes from the angle of the incline, the bigger the angle the less velocity.
     
  13. Jan 22, 2015 #12

    haruspex

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    You are still not explaining what you mean by "the force from the engine". Could you point to it on a diagram? As I wrote, the engine produces a torque, not a force.
    When you apply a standard equation like P=Fv, you need to understand what relationship those entities must have for the equation to be applicable. In this case, F is a force applied to and driving the motion of an object, and v is the velocity of the object in the direction of that force. (And both need to be constant.)
    Assuming you mean the v as given in the question, that's the velocity up the plane. If your F, when you have defined it, is acting up the plane also then it's going to be P=Fv, no role for theta.
     
  14. Jan 22, 2015 #13
    Ok so F is the component of the weight that is parallel to the incline, is that right?
     
  15. Jan 22, 2015 #14

    haruspex

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    Yes!
    Small correction -it's equal and opposite to the component of the weight parallel to the plane.
     
  16. Jan 22, 2015 #15
    Ok, so now what?
     
  17. Jan 23, 2015 #16

    haruspex

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    So what is the component of the weight parallel to the incline, in terms of the given data?
     
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