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Finding Tension in Horizontal circular motion with friction

  1. Nov 17, 2015 #1
    1. The problem statement, all variables and given/known data
    A car of mass (m) is attached to a cable of length (L) to a rotating pole. The car will speed up and travel in a circle. If the maximum speed of the car is vmax, what should the minimum strength of the cable connecting the car to the pole be? Note the surface is not frictionless.


    m = 137.7 kg, L = 5.5 m, μK(road-tires) = 1.13, μS(road-tires) = 1.70, vmax = 10.7m/s

    ** I'm confused because the car is attached by a cable to the pole, but I also have values for static and kinetic friction. If the car is attached to the pole, does friction even matter? Since Ff=Fc and Tension=Fc, doesn't Ff=Tension? And where does kinetic friction factor into this?

    2. Relevant equations
    Fc=mv2/r, Fc=Ff, Fc=Tension, ac=v2/r

    3. The attempt at a solution
    My FBD attempt. (Still confused by the friction).
    http://imgur.com/asxof5h
     
  2. jcsd
  3. Nov 17, 2015 #2

    CWatters

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    Hints
    On a car (without a pole) what normally provides the centripetal force needed to go around a corner?
    What limits how fast the car can go around the corner?
    If the car was to go faster than that limit it will skid because there is insufficient centripetal force - what could provide the additional centripetal force?
     
  4. Nov 17, 2015 #3
    -The force of friction allows it to go around a corner.
    -The value of static friction limits the speed of the car. (Higher coefficient of static friction allows for the turn to be navigated at greater speed).
    -If the car exceeds the max speed allowed by the force of friction, it starts to skid and the force is kinetic friction...and kinetic friction points opposite to the direction of motion.

    *I'm still not sure how the cable tension is involved in this?
     
  5. Nov 17, 2015 #4

    CWatters

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    It's hard to give another hint without giving the whole answer.

    See last line of my post above.
     
  6. Nov 17, 2015 #5

    CWatters

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    What could provide the additional centripetal force to stop it skidding when going too fast for friction alone?
     
  7. Nov 17, 2015 #6
    The tension of the attached cable...but if this is so, why are you assuming that the car is skidding in the first place? How do I know that the max velocity given is enough to cause it to skid, causing the Tension of the cable to aid in centripetal force?
     
  8. Nov 17, 2015 #7

    CWatters

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    I'm not assuming it will skid. In fact it can't skid unless the rope fails.

    You don't need to know exactly how fast the car is going to write an equation for the tension in the cable and hence its minimum strength. You can always add a note that below a certain speed the minimum strength of the rope is zero.
     
  9. Nov 17, 2015 #8

    CWatters

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    Actually I see they tell you the Max speed. Have you tried calculating if it will skid?
     
  10. Nov 17, 2015 #9
    Okay. I calculated that the max speed is 9.57 m/s which means that it will skid without any additional centripetal forces! So right now I'm conceptualizing that the Force of static friction and the tension in the cable add together keep the car moving at the given max speed, calling it Fc total. So I can find the tension of the cable by finding the Fc total and subtracting the Fstatic friction. If I'm reading this all correctly, does the kinetic friction factor in here at all?
     
  11. Nov 17, 2015 #10

    haruspex

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    That's the correct reasoning.
    Re kinetic friction, that only comes into play once it starts to skid. What will happen to the centripetal force provided by friction if that happens? If the cable is not strong enough to prevent skidding, will it be strong enough to hold a car which is starting to skid?
    (Note also that if we take the cable as being completely inextensible them skidding would imply the cable has already failed.)
     
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