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Drag race...

  1. Oct 29, 2015 #1
    Skjermbilde 2015-10-29 kl. 21.45.15.png

    1. The problem statement, all variables and given/known data

    A 4-wheel drive car of mass M = 2100 kg accelerates from rest to 100 km/h

    in 3.40 seconds, and we will assume that the acceleration is constant. In the

    following, ignore resistance from the air and the rolling friction of the wheels

    against the surface. The wheels are rolling without sliding, and they have

    a diameter of D = 50.0 cm. Each wheel has a mass of m = 18.0 kg, and

    has a moment of inertia around their centre of mass corresponding to a disc,

    I = 1/2mR2. The mass of the wheels is included in the total 2100 kg. The

    wheels are taken to be identical and carry the same amount of weight. The

    gravitational constant is g = 9.80 m/s2.

    a) What is the required torque that the engine has to provide for each wheel,

    to have this acceleration? Provide both the algebraic expression, and the

    numerical result.
    b) What is the required coefficient of static friction between road and wheel,

    to avoid spinning? Provide both the algebraic expression, and the numerical

    result.

    c) What is the total kinetic energy of the car as it reaches 100 km/h?


    3. The attempt at a solution
    Can somebody check on my solutions, All answers appreciated!!

    a)
    a=(vf-vi)/(tf-ti)=8.17 m/s^2
    F=ma=17157 N
    F=mrα
    α=F/MR=32.68 rad/s^2
    τ= Iα=1/2mR^2*α= 2144.63 N*m
    τ/4=536.16 N*m

    b)
    μ=a/g=0.834

    c) this is how far i got
    K=1/2Mv^2=29166.7 J
     
  2. jcsd
  3. Oct 29, 2015 #2

    gneill

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    Staff: Mentor

    The linear and angular accelerations look fine, as does the net force for the linear acceleration. But I'm not seeing how you arrived at that value of torque with the given values. It looks as though you've used the mass of the whole vehicle in the expression for the rotational inertia of a wheel, and I don't see how you've accommodated the torque that's required to provide the linear acceleration of the vehicle.
    What about the rotational KE of the wheels?
     
  4. Nov 2, 2015 #3
     
  5. Nov 2, 2015 #4
    I have recalculated and i got 0.5625 Nm as the torque that the engine has to provide for each wheel, is this correct?
     
  6. Nov 2, 2015 #5

    gneill

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    Staff: Mentor

    It doesn't look likely to me. That's a pretty small torque to be responsible for moving the vehicle and spinning up the wheels. I'd expect something closer to a thousand Nm per wheel. Can you show your calculations in detail?
     
  7. Nov 2, 2015 #6
    for c i now got 1264074.1J?
     
    Last edited: Nov 2, 2015
  8. Nov 2, 2015 #7
    i use this equation
    T=Ia
    mR^2*a/2
    (18kg*0.25^2*32.68)/2
    i actually get 18.3825 Nm
     
  9. Nov 2, 2015 #8

    gneill

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    Staff: Mentor

    That looks like a reasonable value, although it needs to be trimmed to the appropriate number of significant figures.

    That would account for the torque required to accelerate one wheel if it were free to rotate and that was the only torque involved. But the wheel is touching the ground and "pushing" the car forward. So there's another torque. Perhaps you should draw a free body diagram of a wheel showing the engine's torque and the torque due to the road/wheel interface.
     
  10. Nov 4, 2015 #9
    How did you end up with c) 824,1 kJ?

    I think this formula is correct? K = 0.5*I*w^2+0,5*Mv^2 ?
    (0,5*0,56*8,18^2+0,5*2100*27,8^2 = 811,50kJ)
    I don't get the same solution as 824,1kJ
     
  11. Nov 4, 2015 #10

    gneill

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    Staff: Mentor

    How did you determine your value for ω? Also, keep in mind that there are four wheels.
     
  12. Nov 4, 2015 #11
  13. Nov 4, 2015 #12
    I used w=a = 8,18

    The correct formula should be? 0,5*m*v^2 for car and (0,5*I*w^2+0,5*M*v^2) *4 for wheels
    Car = 810,31kJ
    Wheels = (0,5*18*27,78^2+0,5*0,56*8,18^2) *4 = 27,86kJ
     
    Last edited: Nov 4, 2015
  14. Nov 4, 2015 #13

    gneill

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    Staff: Mentor

    ω is not the same as a. a is the linear acceleration, while ω is the angular velocity. Totally different beasts. You should be able to determine ω from the final velocity (100 kph) of the vehicle and the dimensions of the wheels.

    Also, the problem states that the masses of the wheels are included in the 2100 kg of the car, so no need to write a separate linear KE expression for their individual masses.
     
  15. Nov 4, 2015 #14
    w = v/r = 27,78/0,25 = 111,12

    0,5*2028*27,78^2=782,53kJ
    (0,5*0,56*111,12^2 + 0,5*18*27,78^2)*4 = 41,6kJ

    = 824,13 kJ
     
  16. Nov 5, 2015 #15
    I added T(to make wheel spin) and T(to make car move). And got: btw a= alpha here
    T(total) = T(wheel) + T(car) = (1/2*4m*r^2*a) + (M*r^2*a) = 73.53+4289.25 = 4362.78
    Divide this by 4 and the torque for each wheel is 1090.70Nm.
    Can anyone tell me if this is correct?
     
  17. Nov 5, 2015 #16
    what angular a are you using?
     
  18. Nov 5, 2015 #17
    Looks good.

    Should be your 32.68 rad/s2 from part a).
     
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