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Thread: Finding the curvature of the road, when you have v and m

  1. #1 Finding the curvature of the road, when you have v and m 
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    I have no idea how to tackle this...
    my teacher did not explain anything like this to us, so what might seem
    easy, i havent even been taught it! Please help.

    A 2000kg car is travelling on a banked icy curved road without sliding. The velocity of the car is 30m/s and the road is banked at an angle of 20 degrees.
    The curvature of the road is?

    a) 175m
    b) 204m
    c) 252m
    d) 302m
    e) 375m

    Thank you!! PS: please explain why you're doing certain equations, etc.
    Thank you


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  3. #2  
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    Well, the first thing to do is to draw out a free-body diagram for the situation. In this case, the icy road can, I assume, be taken to mean there is no friction, only momentum and gravity.

    From there, I think you'd need to work out how large of a force is pulling the car inwards, then work out how large of a circle that would create. This would have to match the curve of the road, or the car would go off one side.


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  4. #3  
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    I did draw a free-body diagram. And I believe I am finding Fnx component...is that correct?
    I have no idea how to find it though. I dont know what i am supposed to do with the velocity?
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  5. #4  
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    Fnx component?

    Anyway, the velocity doesn't matter for finding the magnitude of the force pulling the car inward. The velocity plus the acceleration due to that force should give you the radius of the curve though.
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  6. #5  
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    The force pushing the car down the slope is the component of the gravitational force which is directed along the slope. That is mg sin (20 degrees).

    This is counteracted by the component of the centrifugal force in the opposite direction. The centrifugal force is mv^2/r. The component of that pointing up the slope is (mv^2/r)* cos(20 degrees).
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    Forum Bachelors Degree Demen Tolden's Avatar
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    I know Harold, that you know a ton more than I do, but I have been taught that centrifugal (center fleeing) force is only the sensation a person feels while traveling in uniform circular motion because their velocity is always perpendicular to their acceleration. Because of this perpendicularness (made up word) they feel like they are constantly being thrown outside the circle. Centrifugal force does not really exist, however centripetal (center seeking) force exists, but not as a "real force." If you were to draw a free body diagram, centripetal force would not be included because it does not really have an impact on the object, however centrifugal force is a way to interpret a net force.

    Example:
    Forces in an i direction:

    in uniform circular motion:


    Lets say you had a bunch of forces:


    The "real forces" get substituted in on the left side while the ma is adapted to uniform circular motion.

    I also saw that momentum was offered up as a force. I also believe this is not true. Momentum is not a force.

    "The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force." -Newton

    The only two forces in this problem that exist for your purposes will most likely be gravity and the Normal force. Most likely drag force and friction are not used. Because the car is on an incline, you will have to seperate these forces into components like the other guys have done above.
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    Quote Originally Posted by Demen Tolden
    I have been taught that centrifugal (center fleeing) force is only the sensation a person feels while traveling in uniform circular motion because their velocity is always perpendicular to their acceleration.
    Call it centripetal if you like. You get the same answer.
    The banked turn problem is explained here:
    http://en.wikipedia.org/wiki/Centrip...he_banked_turn
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  9. #8  
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    I think that explaination may be more confusing than the one I am about to give. This is your picture of your banked surface, and your object:

    Notice there is not centrifugal or centripetal forces drawn in as they do not act on our object.

    First, to avoid problems that come up later, let us set up an axis where the normal force points strait up. To do this we indicate where our theta angle is, and draw a right triangle to start to seperate the gravitational force into components.



    Now that we have our forces divided into components, we can draw a freebody diagram to show the net forces.


    Now, using the above information, we can write our equations.
    Net forces in j direction:

    Since the Object is not moving in the j direction, it has 0 acceleration.

    Now we substitute our j forces.



    Net forces in i direction:

    Since the object is moving in uniform circular motion we can describe its acceleration as:

    If we wanted to substutite we could do the following. Now, we don't necisarily have to. F = ma might be perfectly fine if we had information that gave us acceleration and we were not interested in velocity or the radius of the object's circular path. For this problem though, we want to know this information. By substitution:

    Substituting the net forces in the i direction for F:

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    Quote Originally Posted by Demen Tolden
    You have a sine where my method and the Wikipedia method would have a tangent. That's because you forgot to multiply the centripetal acceleration by the cosine of the angle.(sin/cos=tan)
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  11. #10  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Do you mean that the next step should be:




    This seems less simplified.

    I see that wiki's answer is:


    I hesitate to say it, but I think this may be a case where wiki is wrong. I've seen this problem done many times in the way I have demonstrated it from both my instructer, my physics book, and videos by the author of the physics book. I actually have access to a video that is an excelent demonstration of my method, but I can't seem to be able to get a workable link to it here. At the moment I am trying to work it out wiki's way to figure out what wiki did wrong.

    -------------------------------------------------------------------
    (Later)
    I believe I have figured out why wiki is wrong. You cannot separate the Normal force into two directions unless you are able to represent its resisting force in two directions. If you set up a freebody diagram that splits the normal force into i and j directions, and have your mg only in the j direction, there is nothing to resist the i direction normal force, and normal force cannot exist without an opposing force such as gravity. In order to get the correct normal force, you must find the gravity component that acts in its direction.

    I found this demonstration, though I don't know how the sound is since I don't have sound access at the moment.
    http://www.youtube.com/watch?v=VU0lr5KQ79s
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    I dunno Demen. I googled banked turn and found a few sites that all had the tangent function, not sine in the solution. I don't think they're all wrong.

    The centripetal acceleration is in the direction of the radius of the turn. So you have to find the component that is in the direction parallel to the road surface.
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    Not for nothing, I can see the logic in trying to figure a speed that would at least put you in the ball park, to be able to make small corrections once you get into the turn.

    But lets face it. When you start into the turn, you are going to have to adjust for entry angle, into the turn, to compensate for the sudden change in direction and lack of slope at the start of the turn.

    I would want a few practice runs with chains to get a feel for it.

    But just off the top of my head, if I had no chains. I would start high and fast into the turn, with the extra power to drop into the turn at the end, if I am too slow.

    I could always come off the gas, and aim high, if I am going to fast. It is harder to raise the car even a foot or two on the track, because of the horse power necessary to lift a car, it will naturally slow.

    There is going to be a certain amount of traction necessary no mater what. And a certain amount of friction to compensate for no matter what.



    Sincerely,


    William McCormick
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  14. #13  
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    considering the direction of the weight and angle of its component force, sine or cosine can be used depending on where the angle is measured.

    i can't see where tangent comes from.. it would seem to mean that the weight is paralell to the surface.
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  15. #14  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Hmm, to be fair, I've done a little more looking around and found this in my book:

    "Curved portions of highways are always banked (tilited) to prevent cars from sliding off the highway. When a highway is dry, the frictional force between the tires and the road surface may be enough to prevent sliding. When the highway is wet, however, the frictional force may be negligible, and banking is then essential. Figure 6-13a represents a car of mass m as it moves at a constant radius R = 190 m. (It is a normal car, rather than a race car, which means any vertical force from the passing air is negligible.) If the frictional force from the track is negligible, what bank angle prevents sliding?

    Key Idea: Unlike Sample Problem 6-9, the track is banked so as to tilt the normal force on the car toward the center of the circle (Figure 6-13b). Thus now has a centripetal component of magnitude , directed inward along a radial axis r. We want to find the value of the bank angle such that this centripetal component keeps the car on the circular track without need of friction.

    Radial calculation: As Fig. 6-13b shows (and as you should verify), the angle that force makes with the vertical is equal to the bank angle of the track. Thus, the radial component , is equal to . We can now write Newton's second law for components along the r axis () as

    .
    This will be equation (6-25).

    We cannot solve this equation for the value of because it also contains the unknowns and m.

    Verticle calculations: We next consider the forces and acceleration along the y axis in Fig. 6-13b. The verticle component of the normal force is , the gravitational force on the car has the magnitude mg, and the acceleration of the car along the y axis is zero. Thus we can write Newton's second law for components along the ya axis () as



    from which


    This will be Equation (6-26).

    Combining results: Equation 6-26 also contains the unknowns and m, but note that dividing Eq. 6-25 by Eq. 6-26 neatly eliminates both those unknowns. Doing so, replacing with , and solving for then yield



    12 degrees."

    I apologize for the lack of visuals which this text refers to.
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  16. #15  
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    so the tangent did not come off as part of any force component.

    it was part of the solution.

    ok got it.
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    Quote Originally Posted by Demen Tolden
    Hmm, to be fair, I've done a little more looking around and found this in my book:

    "Curved portions of highways are always banked (tilited) to prevent cars from sliding off the highway. When a highway is dry, the frictional force between the tires and the road surface may be enough to prevent sliding. When the highway is wet, however, the frictional force may be negligible, and banking is then essential. Figure 6-13a represents a car of mass m as it moves at a constant radius R = 190 m. (It is a normal car, rather than a race car, which means any vertical force from the passing air is negligible.) If the frictional force from the track is negligible, what bank angle prevents sliding?

    Key Idea: Unlike Sample Problem 6-9, the track is banked so as to tilt the normal force on the car toward the center of the circle (Figure 6-13b). Thus now has a centripetal component of magnitude , directed inward along a radial axis r. We want to find the value of the bank angle such that this centripetal component keeps the car on the circular track without need of friction.

    Radial calculation: As Fig. 6-13b shows (and as you should verify), the angle that force makes with the vertical is equal to the bank angle of the track. Thus, the radial component , is equal to . We can now write Newton's second law for components along the r axis () as

    .
    This will be equation (6-25).

    We cannot solve this equation for the value of because it also contains the unknowns and m.

    Verticle calculations: We next consider the forces and acceleration along the y axis in Fig. 6-13b. The verticle component of the normal force is , the gravitational force on the car has the magnitude mg, and the acceleration of the car along the y axis is zero. Thus we can write Newton's second law for components along the ya axis () as



    from which


    This will be Equation (6-26).

    Combining results: Equation 6-26 also contains the unknowns and m, but note that dividing Eq. 6-25 by Eq. 6-26 neatly eliminates both those unknowns. Doing so, replacing with , and solving for then yield



    12 degrees."

    I apologize for the lack of visuals which this text refers to.
    What ever speed you come up with, in real life you will need more speed, to keep you from slipping down the track. Due to friction in the tires, due to traction that the tires do have even on icy roads. The ice is not going to be even. The road is not going to be even. Your line through the turn is not going to be exact.

    Radial tires are going to allow the rims to move over the center of the tire, moving the mass of the car compared to the tracking. Either further out on the track or closer to the center of the turn, depending on whether or not you are going to fast or to slow.
    Most cars are not perfectly balanced, for safety. They tend to fish tail when over powered. This unbalance will cause the rear of the car to move outward on the track before the front of the car does. It will also change a perfect tangent path to the curved track, the car will take to a more toed in angle, towards the center of the track through the curve. If there is no traction at all, it will fish tail. I know there has to be some traction.

    You will need to apply power to the rear wheels throughout the turn. Or else you have to plot a dwindling or spiraling downward path through the turn. This power will hurt the traction of the rear wheels. Causing a whole new set of variables.

    The math is just something you take into consideration before you do the tests. Because there is no math for all of these variables, that would not make the mathematician more wrong then right.

    If you have test data already, then you can create ratios and come very close to a real outcome. But you cannot just sit back with a paper and pencil and formulas and come up with a conclusion.

    One night up in Troy New York, I was going about 85-90 miles an hour, with my friend Mark, his girl friend was going to school up there and I drove up.

    I was wearing a "T" shirt, we just came out of a bar, it seemed nice out. But when I hit a banked turn overpass. I hit a patch of ice. The rear of the car instantly moves outward. When I again hit dry road, I luckily was cutting the wheel hard into the fish tail. Or we would have rolled over. I had to just as quickly cut back or we would have gone off the bridge.



    Sincerely,


    William McCormick
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