Find the conditions for the static coef. for non-slip conditions

  1. Jun 28, 2012 #1
    1. The problem statement, all variables and given/known data

    It's a cilinder of mass M and radius R rolling without slipping and I'm asked to find the maximum value of the static friction coefficient for the cilinder to roll without slipping.

    2. Relevant equations
    Non-slip conditions:
    [tex] \displaystyle v=\omega R[/tex]
    [tex] \displaystyle a=\alpha R[/tex]

    [tex] \displaystyle \tau =I\alpha [/tex]

    3. The attempt at a solution
    Net force in axis X (my axis is along with the inclined plane):

    [tex] \displaystyle -mg\sin \theta -{{\mu }_{s}}N=ma[/tex]

    Equation in axis y will give me the normal force:

    [tex] \displaystyle N=mg\cos \theta [/tex]
    [tex] \displaystyle -mg\sin \theta -{{\mu }_{s}}mg\cos \theta =ma[/tex]
    Solving for u:

    [tex] \displaystyle {{\mu }_{s}}=\frac{-g\sin \theta -a}{g\cos \theta }=\frac{-g\sin \theta -\alpha R}{g\cos \theta }[/tex]

    So I need to find out the angular aceleration:

    [tex] \displaystyle \tau =I\alpha [/tex]
    [tex] \displaystyle -{{\mu }_{s}}mg\cos \theta \cdot R=\frac{1}{2}m{{R}^{2}}\cdot \alpha [/tex]
    [tex] \displaystyle \alpha =\frac{-2g{{\mu }_{s}}\cos \theta }{R}[/tex]

    Replacing alpha in the equation I get that:
    [tex] \displaystyle {{\mu }_{s}}\le \tan \theta [/tex]

    But the option marked as correct is:
    [tex] \displaystyle {{\mu }_{s}}\ge \frac{\tan \theta }{3}[/tex]

    any help?


    Attached Files:

  2. jcsd
  3. Jun 28, 2012 #2
    Check your I
  4. Jun 28, 2012 #3
    What's wrong out the inertia of the cilinder? Isn't it 1/2*MR^2? Thank you!
  5. Jun 28, 2012 #4
    Hello Hernaner28,
    You have implicitly assumed the limiting friction case in your rolling situation (fixing fs=μN )
    First of all treat friction as an unknown independent of the normal reaction of the cylinder on the inclined plane to get an expression for friction in terms of Mass of object,the inclination of the plane ,The radius and the Moment of inertia.Now the maximum friction that your incline can offer at the given N is ____________?(answer) So your calculated friction must not exceed _________?
    Try to answer them and using the expression that comes as an interpretation calculate the condition again.I pretty much think this gives the actual answer as marked.Does this help?

  6. Jun 28, 2012 #5
    But what was wrong about what I did? I worked with the limiting friction case because that would be the maximum coefficient of friction... that's why I finally get it has to be "less or equal".

  7. Jun 28, 2012 #6
    Hello again!
    Less than or equal to what?The maximum value isnt it?There should be a separate equation for the general expression of friction isnt it?Here is an advised algorithm .Find the general expression for friction for the non limiting case (i.e. not involving the usage of N ;just treat f as an unknown fully and just apply the pure rolling rolling condition with Gravity and friction ;get an expression and compare it with the maximum value of friction sought).Now let us see -What if the maximum value of friction is less than this required value of friction? There will be inevitable slipping right?
    By the way the expression for friction that comes from the general case will give you the general expression for friction for a pure rolling body on an inclined plane.
    Does this help?
    Last edited: Jun 28, 2012
  8. Jun 28, 2012 #7
    Less or equal to tan(theta) as I finnaly worked out. I understand your thinking but I don't understand what is the difference between mine and yours, and what's the mistake about mine..

  9. Jun 28, 2012 #8
    I think you have by mistake yielded your derivations to a≥Rα.Contradictory to pure rolling isn't it?(The last step quoted)

  10. Jun 28, 2012 #9
    Ah yes, you're right. So what is what I got? Because it is not incorrect but maybe it is not what I was asked
  11. Jun 28, 2012 #10

    ma =mgsinθ-friction
  12. Jun 28, 2012 #11
    No, both forces point the same direction. The cilinder rolls downward so friction force does an opposite torque (downward).

    But Yukoel, what was what I got then?
  13. Jun 28, 2012 #12
    How the cyclinder roll counter clockwise?
    Where the torque comes from?
    The mgsinθ acts through the center of the cylinder.
  14. Jun 28, 2012 #13
    Ah yes you're right :D ! THANKS! Anyway, let's assume the last result is fine, then, what's the meaning of what I got? Is it the maximum coefficient of friction for the cilinder not to slip?
  15. Jun 28, 2012 #14
    I don't know why but now I get the correct answer... even with my first thinking (nothing was wrong about it). I think it was because of the direction of the force which azizlwl corrected.

  16. Jun 28, 2012 #15
    I get:

    [tex] \displaystyle F=\frac{mg\sin \theta }{3}[/tex]

    The friction F force is:

    [tex] \displaystyle {{\mu }_{s}}mg\cos \theta =\frac{mg\sin \theta }{3}[/tex]

    [tex] \displaystyle {{\mu }_{s}}=\frac{\tan \theta }{3}[/tex]

    For the limit case... so that would be the minimum value, or maximum?
  17. Jun 28, 2012 #16
    The exact answer is μ≥ (2/7)tanθ
    The maximum ratio friction/N is μs.
  18. Jun 28, 2012 #17
    Hmm... I am not so sure but it is a kind of slipping on the limiting case (as azizlwl points out there are sign mistakes in your equations).
    Let me get to point on signs here
    You can take friction arbitrarily let us suppose.For pure rolling the point of contact has to be at rest. Suppose the gravity and friction are in the same direction and look at the bottom .It will be accelerated.This is why friction has to be on the opposite of gravity.
    This however has nothing to do with limiting cases.As a matter fact it is quite comfortable to take things down the incline as positive.
  19. Jun 28, 2012 #18
    Yukoel, I got the correct answer now:

    [tex] \displaystyle {{\mu }_{s}}=\frac{\tan \theta }{3}[/tex]

    But that is a equality, not inequality, now I have to decide the sign of the inequality. That value is the minimum or maximum value for the cilinder not to slip?
  20. Jun 28, 2012 #19
    Hello azizlwl
    I think you have mistaken cylinder as a sphere (I=(2/5)M*(r^2)
    However for cylinders (I=(1/2)M(r^2) )
    I am pretty much sure that the answer sought as third of the value of tan(θ) is correct for the cylinder. And (5/2) times your answer your answer is correct for the sphere.
    Last edited: Jun 28, 2012
  21. Jun 28, 2012 #20
    Yes you are right, I made a mistake there. Thanks.
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