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Help needed -- Analysis of torque and force on turning vehicles

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-11-16 at 4.55.10 PM.png
    Screen Shot 2016-11-16 at 4.55.36 PM.png

    2. Relevant equations
    Torque and Newton's 2nd Law Equations

    3. The attempt at a solution
    Apart from drawing a free-body diagram, I couldn't do much else. I can't seem to find the source of torque causing the car to rotate clockwise about its pivot (i defined it as the right tyre). I read from somewhere that we could take the centre of mass as a pivot as well but i'm not sure if that's right or how to go about analysing forces about that pivot.
     
  2. jcsd
  3. Nov 16, 2016 #2

    Simon Bridge

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    You don't need to argue about torque exactly - you need to compare to the motorcycle case. Why does the motorcyclist need to lean?
     
  4. Nov 16, 2016 #3

    jbriggs444

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    When you drew up the free body diagram, did you assume the inertial frame of reference where the car is accelerating leftward? Or the rotating frame where it is stationary? I am guessing that you chose the former.

    In that case, how does the leftward acceleration of the car affect its angular momentum?
     
  5. Nov 16, 2016 #4
    I'm sorry but I don't quite get what your first question means, I only know of 1 kind of FBA (the one also used in banked curve and pulley problems?) if it helps. If I took the right wheel to be the pivot, wouldn't friction produce no torque/angular momentum about it (due to line of action going through the pivot)? Or is my concept of the problem completely wrong? Thanks for your help.
     
  6. Nov 16, 2016 #5

    jbriggs444

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    What I am getting at is the roster of forces that are included on your free body diagram and whether an acceleration is expected to result.

    Is friction the only horizontal force acting?
    Is angular momentum changing over time?

    To answer these questions, you have to specify a reference frame. Hence the question: What reference frame are you using?
     
  7. Nov 16, 2016 #6
    If I'm understanding the question correctly, i was using the non-inertial reference frame of the vehicles.
     
  8. Nov 16, 2016 #7

    jbriggs444

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    That is a valid choice. The problem allows you an alternative: "You may take the perspective of the inertial frame of the earth or the non-inertial reference frame of the rider plus motorcycle".

    Now, given that you are using the non-inertial reference frame, are there any forces that arise due to that choice?
     
  9. Nov 16, 2016 #8
    Here's the best I could come up with, though i have a feeling it's missing something.

    PFphoto.jpeg
     
  10. Nov 16, 2016 #9

    jbriggs444

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    Both drawings feature a "a <- rad" notation. What does that denote? You have adopted the non-inertial frame in which the motorcycle is stationary. What is its acceleration in this frame?
     
  11. Nov 16, 2016 #10
    It was supposed to be radial acceleration which was caused by friction, is this right?
     
  12. Nov 16, 2016 #11

    jbriggs444

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    In the non-inertial frame, the motorcycle and rider are stationary. They are not moving. They are not accelerating.

    There is, however, a centrifugal force...
     
  13. Nov 17, 2016 #12
    could you explain what this FBD would look like? I read in my textbook that centrifugal force was fictional and was not to be included in a FBD though
     
  14. Nov 17, 2016 #13

    jbriggs444

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    If the textbook said that, the textbook was wrong. The minute you decide to use the non-inertial frame, the centrifugal force becomes valid for use.

    Edit: Let me expand a bit on that...

    In the non-inertial frame, the driver plus motorcycle are stationary. Not moving. Not accelerating. Yet if we add up all of the real forces on them, there is a net leftward force. In order to preserve the usefulness of Newton's second law (f=ma) in this environment, the net force needs to be zero. The centrifugal force fills this requirement. In this case, it balances out the real forces so that the net is zero.

    If you use a non-inertial frame and leave the centrifugal force out of your analysis then you will have objects moving around for no reason. Or unbalanced torques with no resulting rotation.

    If you prefer not to use the centrifugal force that is fine. But then you should do your analysis in the inertial frame where the driver plus motorcycle are accelerating leftward. In this frame there is no centrifugal force. There is a net leftward force and a net leftward acceleration. There is also a net torque. But there is also a changing angular momentum if you look for it.
     
    Last edited: Nov 17, 2016
  15. Nov 18, 2016 #14
    Turns out I misread the text, it says N's Laws can't be applied in non-inertial frames. I apologize but I'm new to non-intertial reference frames. Could you show me an example equation of the real and fictional forces on the motorcycle? (Is it something like fs = mv2/r?) I have also read about how we can shift the normal and frictional forces to the vehicle's center of mass as long as we take into account the torques they exert on the centre of mass (??) but I'm not sure how that works.
     
  16. Nov 18, 2016 #15

    jbriggs444

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    Newton's laws CAN be applied in non-inertial frames. The catch is that you have to invent "fictitious" inertial forces to make them work. In an accelerating frame, a object that is subject to no "real" forces is seen to accelerate. So we invent a force that accounts for the acceleration.

    For objects at rest in a rotating reference frame, this invented force is called the "centrifugal force". It is proportional to the square of the rotation rate of the reference frame and directly proportional to the distance from the axis of rotation. ##F_c = m \omega^2 r##.

    The formula ##F_c = \frac{mv^2}{r}## also works, but is somewhat unsatisfactory since, from the viewpoint of the rotating reference frame, the object is stationary and its velocity is zero. The v in this formula is the velocity in the inertial frame. Mixing reference frames invites confusion.

    For moving objects or for reference frames that do not rotate at a fixed rate there are other fictitious forces needed (Coriolis and Euler). But we need not examine these. In the case at hand, we do not even need to know the formula for centrifugal force. All we need to know is that it is a rightward force that is proportional to mass and (for our purposes) can be considered to act at the center of mass.

    NOTE: Newton's first and second laws can be preserved in accelerating frames by the use of fictitious inertial forces. But Newton's third law only applies to the real forces, not the fictitious ones. Fictitious forces have no third-law partners.
     
  17. Nov 18, 2016 #16
    Thank you, I believe i understand it slightly better now. Could you also explain how the shifting of forces to the CM of the vehicles work?
     
  18. Nov 18, 2016 #17

    jbriggs444

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    I do not know what you mean by shifting forces to the CM. If I were to take a guess...

    If one has a force applied to an object at one point, it is equivalent to the same force applied to the object at another point along with a torque that compensates for the shift. The torque can be computed as the vector cross product of the force times the displacement of the original point of application from the modified point of application. [Or, equivalently in two dimensions, as the ordinary product of the force times the component of the offset that is perpendicular to the force with the appropriate sign applied to the result]
     
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