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Rotational Kinematics - The differential gear of a car axle

  1. Nov 21, 2009 #1
    Note - pi is not meant to be an exponent, not sure why it looks like one...

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

    The differential gear of a car axle allows the wheel on the left side of a car to rotate at a different angular speed than the wheel on the right side. A car is driving at a constant speed around a circular track on level ground, completing each lap in 23.7 s. The distance between the tires on the left and right sides of the car is 1.50 m, and the radius of each wheel is 0.350 m. What is the difference between the angular speeds of the wheels on the left and right sides of the car?

    t = 23.7 s

    r = 0.350 m

    d = 1.50 m

    s = 2[tex]\pi[/tex]

    [tex]\vartheta[/tex] = ?

    [tex]\omega[/tex] = ?

    2. Relevant equations

    [tex]\vartheta[/tex] = s / r

    [tex]\omega[/tex] = [tex]\Delta[/tex][tex]\vartheta[/tex] / [tex]\Delta[/tex]t

    3. The attempt at a solution

    I'm really drawing a blank on this...I think that I want to find out how many times the wheels spin during the 23.7 s it takes for one lap. Then that amount can be entered into the angular velocity equation to get their speed. As the question states that one wheel moves faster than the other, I'm thinking that the distance between the two wheels would help me figure out the difference...but I'm not sure how to work that distance in.

    So, first I started out by trying to find the angular displacement [tex]\vartheta[/tex].

    [tex]\vartheta[/tex] = s / r

    [tex]\vartheta[/tex] = 2[tex]\pi[/tex][STRIKE]r [/STRIKE]/ [STRIKE]r[/STRIKE]

    [tex]\vartheta[/tex] = 2[tex]\pi[/tex]

    [tex]\vartheta[/tex] = 6.2831 rad


    Then I attempted to find the angular velocity [tex]\omega[/tex].

    [tex]\omega[/tex] = [tex]\Delta[/tex][tex]\vartheta[/tex] / [tex]\Delta[/tex]t

    [tex]\omega[/tex] = 6.2831 rad / 23.7 s

    [tex]\omega[/tex] = 0.26336 rad/s

    After this though, I'm not sure where to go...assuming what I've done so far is correct. Any thoughts, suggestions, etc, would be appreciated.

    Thank you!
     
  2. jcsd
  3. Nov 21, 2009 #2

    Delphi51

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    I think you will have better luck if you stick with linear quantities until you get to the tire. Say the radius of the circle to the near tire is R and to the far one R+1.5. Then the velocities are 2πR/T and 2π(R + 1.5)/T for the two wheels, where T is the time for one circle. The next step would be to convert this into the number of turns or radians per second for each of the tires using their radius.

    Interesting that the radius of the circular path R is not needed to get the answer.
     
  4. Nov 22, 2009 #3
    Thank you.

    Using that advice, and hopefully interpreting it accurately, this is what I've done...

    R = Radius of circle. Value is unknown but probably in meters...

    Left Tire

    2[tex]\pi[/tex]R / t

    = 0.26511 m/s

    Right Tire

    2[tex]\pi[/tex](R + 1.50m) / t

    2[tex]\pi[/tex]*R + 2[tex]\pi[/tex]*1.50m / t

    = 0.66277 m/s

    So now that I have the linear velocities of the tires, I have to convert that to rad/s.

    [tex]\vartheta[/tex] = s / r

    s = r[tex]\vartheta[/tex]

    s / t = r ([tex]\vartheta[/tex] / t)

    Vt = r([tex]\omega[/tex]) ---> Vt is the Tangential Velocity.

    [tex]\omega[/tex] = r / Vt

    Left Tire

    [tex]\omega[/tex] = r / Vt

    = 0.350m / 0.26511 m/s

    = 1.3202 _/s <----------- Shouldn't this end up being rad/s? As the meters cancel, do we just put rad in there?

    Right Tire

    [tex]\omega[/tex] = r / Vt

    = 0.350m / 0.66277 m/s

    = 0.52814 _/s <------------- Same thing, should be rad/s....

    Assuming this is correct, we would subtract the two angular velocities to find the difference.

    1.3202 _/s - 0.52814 _/s = 0.79206 _/s <--- should be rad/s

    Umm...how does this look?
     
  5. Nov 22, 2009 #4

    Delphi51

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    I think you have ω = r / Vt wrong. Should be ω = V/r.
    I left the numbers till the end:
    v = 2πR/T, ω = v/r = 2πR/(rT) for one tire and 2π(R+1.5)/(rT) for the other.
    and I can't find a number for either because R is unknown.
    However, their difference does not have an R in it.
     
  6. Nov 22, 2009 #5
    Right, I solved Vt = rw incorrectly. Should have been: w = V / r

    Left tire

    w = 2[tex]\pi[/tex]R / t / r

    w = 2[tex]\pi[/tex]R / t * 1 / r

    w = 2[tex]\pi[/tex]R / rt


    Right Tire

    w = 2[tex]\pi[/tex](R + 1.50m) / t / r

    w = 2[tex]\pi[/tex](R + 1.50m) / t * 1/r

    w = 2[tex]\pi[/tex](R + 1.50m) / rt


    Would this be considered the end so that I can put numbers in?

    Left Tire

    w = 2[tex]\pi[/tex]R / rt

    w = 6.2831 R / 8.295 ms

    Right Tire

    w = 2[tex]\pi[/tex](R + 1.50m) / rt

    w = 2[tex]\pi[/tex]R + 2[tex]\pi[/tex] * 1.50m / 8.295 ms

    w = 15.707m R / 8.295 ms

    Ugh........this still doesn't feel right...
     
  7. Nov 22, 2009 #6

    Delphi51

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    Difficult when you don't know R!
    Better to subtract first.
    difference = 2π(R + 1.50m) / (rt) - 2πR/(rt)
    If you factor out the 2π/(rt), you'll find the subtraction a snap. And no R in the answer.
     
  8. Nov 22, 2009 #7
    *slaps forehead* Touche...

    2[tex]\pi[/tex]R / rt - 2[tex]\pi[/tex](R + 1.50m) / rt

    2[tex]\pi[/tex] / rt (R - R + 1.50)

    2[tex]\pi[/tex] / rt (1.50m)

    6.2831 rad / 8.295 [STRIKE]m[/STRIKE]s (1.50[STRIKE]m[/STRIKE])

    = 1.1361 rad/s or 1.14 rad/s with significant figures

    Umm...yes?
     
  9. Nov 22, 2009 #8

    Delphi51

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    Right on, Crono!
     
  10. Nov 23, 2009 #9
    Ugh...this problem kicked my butt! Thank you so much for your help and patience!
     
  11. Nov 23, 2009 #10

    Delphi51

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    Most welcome! Do try to practise doing more with the formulas before putting the numbers in. You are less likely to make a calculation mistake, it is easier to write and see, and I think you will soon run into a lot of problems that must be done that way.
     
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