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How does rotational motion influence static friction?

  1. Oct 15, 2015 #1
    I apologize in advance for not exactly adhering to the template, but the question I have here arose from my attempts to solve the following exercises, so please bear with me. (Edit: I also apologize if discussion of a concept belongs in a different forum, as this is not exactly a homework problem, just a confusing aspect of physics that I need to understand in order to solve a homework problem.) In Sears and Zemansky's University Physics, 12th ed., I have come across two problems involving rotational motion that kind of confuse me:

    5.96) Consider a banked, wet roadway, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of 0.25 between the tires and the roadway. The radius of the curve is R = 50 m. (a) If the banking angle is β = 25°, what is the maximum speed the automobile can have before sliding up the banking? (b) What is the minimum speed the automobile can have before sliding down the banking?

    5.119) A small block with mass m is placed inside an inverted cone that is rotating about a vertical axis such that the time for one revolution of the cone is T [figure omitted]. The walls of the cone make an angle β with the vertical. The coefficient of static friction between the block and the cone is μs. If the block is to remain at a constant height h above the apex of the cone, what are the maximum and minimum values of T?

    The only roadblock preventing me from solving these problems is understanding how exactly static friction plays a role in these problems. I figure that, above a certain rotational speed, fs points down the incline in both problems, working in conjunction with the normal force to keep the object in rotational motion, as seen in this free body diagram. But, on the other hand, if the car stops moving in the first problem, or the cone stops rotating in the second, the object starts to slide down the incline, opposed by a frictional force pointing up the incline, as seen in this other free body diagram.

    The only problem I have is, what happens as rotational motion begins? The frictional force before rotational motion starts points in the opposite direction as the frictional force while rotational motion occurs, so what exactly am I missing in the middle that causes the switch to happen in the first place? And, to tie it back to the original questions, how does that shift determine the minimum or maximum rotational velocities necessary for the objects to keep vertical equilibrium?

    This particular concept has been bothering me for about a week and a half, so any help would be greatly appreciated! If it helps, I already understand how rotational motion is maintained at constant speed, it's just when I have to find a range of speed that things fall apart in my head.
     
    Last edited: Oct 15, 2015
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  3. Oct 15, 2015 #2

    Nathanael

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    If there was no friction, then there will only be a single exact speed with which the objects can travel without sliding up or down. The reason there is a range of speeds that will keep the objects level is that the magnitude of static friction can range from 0 to μsN (where N is the normal force with the surface).

    If you assume there's no tangential acceleration, then static friction must act either up or down the slope. If you then write an equation for the net force which describes the objects as not sliding up or down, (then solve for the speed) then you would find a range of solutions for the speed corresponding to the range of values for the static friction force (which can be from -μsN to +μsN).

    Of course there will be two endpoints to the range of speeds, the minimum and maximum. These correspond to when the frictional force has it's maximum value up the slope and down the slope respectively.

    You may imagine the frictional force as starting down the slope, then decreasing to zero as you increase the speed, then growing up the slope as you increase the speed more. In reality, things are slightly more complicated, as there will be tangential acceleration when you speed up the object, which friction must support. (But this is not relevant to the problems.)

    Not sure if this helps, feel free to re-emphasize your confusion if I missed the point. Are you getting the correct answer for these two problems?
     
  4. Oct 16, 2015 #3

    CWatters

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    Ask yourself.. are the tyres skidding over the surface of the road (eg kinetic friction applies) or rolling (static friction applies).

    Correct.

    What changes (if anything) as the wheels begin to rotate? Are the tyres skidding over the surface of the road (eg kinetic friction applies) or rolling (static friction applies).
     
  5. Oct 16, 2015 #4

    haruspex

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    The way I read the OP, the puzzlement had nothing to do with rotation of the wheels. It was rotation of the cone, or the circular motion of the car (viewed as a rotation).
     
  6. Oct 16, 2015 #5
    That is correct. Neither of the problems I posted directly involve rolling friction.

    I tried to imagine that situation, but I think it starts to fall apart for me when I ask myself what provides the change in force in the first place as rotational motion sort of "winds up".
     
  7. Oct 16, 2015 #6

    haruspex

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    Friction acts so as to oppose the relative motion of surfaces in contact. If the rotation is at just the right rate that there is no tendency to relative motion, so there is no frictional force. If the rotation increases a little then there will be a tendency to slide away from the axis. Correspondingly, a small frictional force arises to oppose it. The rotation rate can be increased until the frictional force reaches the limit of static friction.
    Conversely, if the rotation rate is decreased instead then there is a tendency to slide towards the axis, so a frictional force arises to oppose that.
    There is no sudden switch over - it's a smooth transition.
     
  8. Oct 16, 2015 #7
    Okay, so let me see if I understand what's going on. When no rotation is happening, there (obviously) isn't any centripetal acceleration pointed toward the center of the circle. If μs is low enough, the object on the bank may even slide down the ramp, so that Fc can remain at 0 N. Once everything starts rotating, Fc has to increase to begin the circular motion, and the only way for that to occur is by decreasing the magnitude of the friction force pointed up the bank. Eventually, as the rotational speed increases, the force of friction is now pointing in the opposite direction, and, if the friction is static, it's preventing any movement up and down the ramp while its horizontal component contributes to Fc, until a certain rotational speed is achieved where the frictional force exceeds μs⋅n and the object starts sliding up the bank/cone. Does that sound right?
     
  9. Oct 16, 2015 #8

    Nathanael

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    Yep :smile:
     
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