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Am I solving this correctly? Acceleration of a car glidin down a curve

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    ZiDFHm7.gif
    A car is resting on a curved path in point A as seen in the picture. The mass of the car is known.
    The angle θ and the radius of the curved path are known.
    The kinetic coefficient of friction is known.

    The car starts gliding backwards on the curved path. What is the acceleration of the car when it starts gliding from point A?
    State both the magnitude and the direction of the acceleration.

    3. The attempt at a solution
    I start by drawing the free body diagram. Since the acceleration of a circular motion is directed towards the center, then I already know the direction of the acceleration. I also draw the normal force (n) the gravity force (mg) and the kinetic friction force (fk):
    AP8PiBd.gif

    I rotate the coordinatesystem with θ degrees:
    bmaKUuC.gif

    I now use Newton's second law on the horizontal forces only (since there is no acceleration vertically).

    [itex]\uparrow \sum F_y=ma_{rad}[/itex]
    [itex]n-mg \cdot cos(\theta)=ma_{rad}[/itex]
    [itex]a_{rad}=\frac{n}{m} \cdot g \cdot cos(\theta)[/itex]

    But I don't know the normal force? :(
     
  2. jcsd
  3. Nov 14, 2013 #2
    This is just a guess but is it possible to use Newton's first law on the vertical forces:
    [itex]f_k-mg \cdot sin(\theta)=0[/itex]

    Inserting the definition of the friction force because we know the coefficient of friction:

    [itex]\mu_k \cdot n - mg \cdot sin(\theta)=0[/itex]

    Now solving for the normal force yields:
    [itex]n=\frac{mg \cdot sin(\theta)}{\mu_k}[/itex]

    And then inserting this expression for the normal force into the expression of the acceleration? What do you guys think?
     
  4. Nov 14, 2013 #3

    haruspex

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    That's only true for motion at constant speed. Since it will get faster, there is also tangential acceleration.
     
  5. Nov 14, 2013 #4
    Oh, that's true. So are all my calculations wrong? :(
     
  6. Nov 14, 2013 #5

    haruspex

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    'fraid so.
     
  7. Nov 14, 2013 #6
    Do you have any suggestions to some other formulas I can use? I've really tried to think this through and that was the best I could come up with.
     
  8. Nov 14, 2013 #7

    haruspex

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    The initial state is at rest. What is the usual expression for centripetal acceleration?
    What equation can you write for the tangential forces?
     
  9. Nov 14, 2013 #8
    This is the centripetal acceleration:
    [itex]a_{rad}=\frac{v^2}{R}[/itex]

    I've thought about this expression a lot but couldn't use it in any way. R is known, however if I assume that the velocity is zero because it's initially at rest, then that would make the radial acceleration zero?

    For a non-uniform motion, I know that the tangential acceleration is the rate of change in speed:
    [itex]a_{tan}=\frac{d|\vec{v}|}{dt}[/itex]

    But here I don't have any info about the time or again the speed.
     
  10. Nov 14, 2013 #9

    haruspex

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    Exactly.
    But you know the forces. Now that you know there is no radial acceleration, you can write down the usual ƩF=ma equations for the radial and tangential directions and deduce the tangential acceleration.
    (It's remarkable how often the first response needed to a post is the advice to write out those equations.)
     
  11. Nov 15, 2013 #10
    So for the radial acceleration:
    [itex]\sum F=m\cdot 0[/itex]

    But what forces do I use here? I have to use the forces that work on the direction of the radial acceleration but if this radial acceleration is 0, how do I determine if I need the forces in the x or y direction?

    Tangential direction:
    [itex]\sum F=m\frac{d|\vec{v}|}{dt}[/itex]

    I know that the radial direction is perpendicular on the velocity and the tangential acceleration is parallel to the velocity. But how do I determine what forces that need to be applied for the radial and tangential accelerations? :)
     
  12. Nov 15, 2013 #11

    haruspex

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    You want the forces in the radial direction, regardless of whether there is radial acceleration.
    Again, it's the forces in the tangential direction. Look at your FBD. Which forces have components in that direction, and what are those component values?
     
  13. Nov 15, 2013 #12
    Okay, so for the radial acceleration, I have the forces:
    [itex]n-mg\cdot cos(\theta)=0[/itex]

    And for the tangential acceleration, I have:
    [itex]mg\cdot sin(\theta)-f_k=m\frac{|\vec{v}|}{dt}[/itex]

    But what do I do about the [itex]\frac{|\vec{v}|}{dt}[/itex]? I don't have any velocity?

    Or perhaps you want me to write it this way instead:
    [itex]mg\cdot sin(\theta)-f_k=m\cdot a_{tan}[/itex]

    So it is [itex]a_{tan}[/itex] that causes the initial acceleration when the car starts to move, correct?

    If that is the case, I can isolate [itex]a_{tan}[/itex] and find an expression for it and use the formula for the radial acceleration and find an expression for the normal force which I can use to determine the friction force, right?
     
  14. Nov 15, 2013 #13

    haruspex

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    Just because the velocity is zero initially doesn't mean the rate of change of velocity is zero at that point.
    Same thing.
    Right.
    One way to think about this problem is that the car doesn't 'know' the track is going to curve. The acceleration and velocity at t=0 are exactly as though it were a straight slope.
     
  15. Nov 15, 2013 #14
    Cool, that makes sense. :)
    So if I were to find the direction of the acceleration, that would be easy since the velocity is zero initially that means we have zero radial acceleration and so in order to find the vector acceleration from the radial and tangential acceleration we get:

    [itex]\vec{a}=\sqrt{a_{rad}^2+a_{tan}^2}=\sqrt{0+a_{tan}^2}=a_{tan}[/itex]

    So our acceleration is equal to the tangential acceleration and so they have the same direction, which would be [itex]180\deg+\theta[/itex]

    ZmA6FZW.gif
     
  16. Nov 15, 2013 #15

    haruspex

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    Yes.
     
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