1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotation and varying friction coefficient

  1. Nov 16, 2012 #1
    1. The problem statement, all variables and given/known data
    (see attachment)


    2. Relevant equations



    3. The attempt at a solution
    I am not able to understand the question and build a scenario in my mind. The question asks the distance travelled when the cylinder starts slipping. I can't think of the situation when the cylinder "slips". I am not sure which equations to start with and i suppose this question involves some integration too.
     

    Attached Files:

  2. jcsd
  3. Nov 16, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    How much friction force is required to prevent slipping?
     
  4. Nov 16, 2012 #3
    I am not sure but do i have to make the equations for torque and the forces to find the frictional force?
     
  5. Nov 16, 2012 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. Apply Newton's 2nd law for rotation and translation.
     
  6. Nov 16, 2012 #5
    Let the mass of cylinder as M, radius R, friction force f, α as angular acceleration, θ as angle of inclination, and a as the linear acceleration, we have
    [tex]fR=\frac{MR^2}{2}α[/tex]
    Since, the cylinder does not slip, the equation α=a/R is applicable, so
    [tex]f=Ma[/tex]
    Now,
    [tex]Mg\sin(\theta)-f=Ma[/tex]
    [tex]Mg\sin(\theta)=2Ma[/tex]
    [tex]a=\frac{g\sin(\theta)}{2}[/tex]
    Therefore, the friction force required to prevent slipping is
    [tex]f=\frac{Mg\sin(\theta)}{2}[/tex]
    Is this correct?
     
  7. Nov 17, 2012 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Good.
    Redo that step and all that follows.

    But you're on the right track. Once you have the correct expression for the friction force, ask yourself what minimum value of μ is required to provide that force.
     
  8. Nov 17, 2012 #7
    Oops, made a small mistake there. :redface:
    The f would be
    [tex]f=\frac{Ma}{2}[/tex]
    Solving using the same method as before, i get
    [tex]f=\frac{Mg\sin(\theta)}{3}[/tex]

    I still can't get the meaning of the problem. Doesn't the cylinder starts slipping the instant it is released or does "slipping" have a different meaning here? :confused:
     
  9. Nov 17, 2012 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Good.


    As long as the surfaces are capable of providing the needed friction force (which you have just calculated), there will be no slipping. But on this surface, the value of μ (and thus the maximum available static friction) decreases with distance down the incline. At some point the surfaces will not be able to provide the needed friction and the cylinder will begin slipping. Find that point.
     
  10. Nov 17, 2012 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The cylinder starts rolling when released if the static friction is enough. Remember the force of static friction ≤ μ FN (FN is the normal force).

    ehild
     
  11. Nov 17, 2012 #10
    Thanks you both for the explanation, i have got the right answer. :smile:

    To find that point, i equate the friction force provided by the surface to the friction force required to prevent slipping.
    [tex]μN=\frac{Mg\sin(\theta)}{3}[/tex]
    Here N is the normal reaction due to the inclined plane and is equal to Mgcos(θ).
    [tex]\frac{2-3x}{\sqrt{3}}Mg\cos(\theta)=\frac{Mg\sin(\theta)}{3}[/tex]
    Solving, i get
    [tex]x=\frac{1}{3}[/tex]

    Thanks once again!
     
  12. Nov 17, 2012 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Good! :approve:
     
  13. Nov 18, 2012 #12
    Here is my attempt at the problem and i am getting an incorrect result ...Kindly check my work and let me know where am i making mistake

    Since at the point of slipping maximum static friction acts

    [itex]f = μN[/itex]

    where [itex]N = Mgcosθ[/itex]

    Thus [itex]f =μMgcosθ[/itex]

    but at the same time ,

    [itex]f = \frac{Ma}{2}[/itex]

    So, [itex]\frac{Ma}{2}=μMgcosθ[/itex]
    [itex]a=2μgcosθ[/itex]

    now θ =60°

    [itex]a=μg[/itex]
    [itex]μ =\frac{2-3x}{\sqrt3}[/itex]

    [itex]a=(\frac{2-3x}{\sqrt3})g[/itex]

    Writing [itex]a=v\frac{dv}{dx}[/itex]

    [itex]v\frac{dv}{dx}=(\frac{2-3x}{\sqrt3})g[/itex]

    [itex]vdv={(\frac{2-3x}{\sqrt3})g}dx[/itex]

    [itex]\int_{0}^{v}vdv=\int_{0}^{x}{(\frac{2-3x}{\sqrt3})g}dx[/itex]

    [itex]\frac{v^2}{2}=\frac{2}{\sqrt3}gx - \frac{3}{2\sqrt3}g{x^2}[/itex]

    Hence ,[itex]v^2=\frac{4}{\sqrt3}gx - \frac{3}{\sqrt3}g{x^2} (1)[/itex]

    Now, we will apply Conservation of energy

    When the cylinder has moved by a distance x along the incline ,then

    Loss in potential energy =Gain in Kinetic energy

    Since rolling without slipping occurs,we can put v=ωR and [itex]I = \frac{MR^2}{2}[/itex]

    [itex]mgxsinθ = \frac{1}{2}Mv^2 + \frac{1}{2}Iω^2[/itex]

    Solving ,we get [itex]v^2=\frac{2}{\sqrt3}gx (2) [/itex]

    Equating (1) and (2) we get x=2/3 which is incorrect.

    Where am i getting it wrong???
     
  14. Nov 18, 2012 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You can not apply conservation of energy.

    ehild
     
  15. Nov 18, 2012 #14
    why cant we apply conservation of energy?? friction is not doing any work since the cylinder rolls without slipping ...energy is definitely conserved till the point cylinder starts to slip
     
  16. Nov 18, 2012 #15

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Well, yes, you are right... The energy is conserved.

    You used the maximum static friction to calculate the speed. a=2Fs/m, but Fs is not μgsosθ during rolling. It reaches that value only at the end when the cylinder starts to slip.
    You can get the acceleration a from the kinematic equations for translation of the CM and rotation about the CM.

    ehild
     
    Last edited: Nov 18, 2012
  17. Nov 18, 2012 #16
    Okay...Then what should I equate 'a' (accelerataion) with ? Kindly let me know how then will i be able to calculate the speed 'v' at distance 'x' .I have showed you the complete working ...
     
  18. Nov 18, 2012 #17

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You do not need v really. But if you want it, find the acceleration first and integrate. What equations do you have for the motion of the cylinder?

    ehild
     
  19. Nov 18, 2012 #18
    For translation motion
    [itex]N=Mgcosθ[/itex]
    [itex]Mgsinθ - f = Ma[/itex]

    For rotational motion

    [itex]fR = Iα[/itex]

    For rolling without slipping [itex]α = \frac{a}{R}[/itex]

    Putting these values , we get

    [itex]a=\frac{2}{3}gsinθ[/itex]

    Integrating , we get [itex]v^2=\frac{2}{\sqrt3}gx[/itex] which is the same result which we get from applying conservation of energy

    What then is the condition mathematically , which we can apply in finding when the cylinder starts to slip ?
     
  20. Nov 18, 2012 #19

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Well, a can be determined from conservation of energy. You know a, then you can find the force of static friction. What is it?

    ehild
     
    Last edited: Nov 18, 2012
  21. Nov 18, 2012 #20
    How can we find a from conservation of energy ???

    Okay...then we are on the same lines as what has been done earlier by Pranav...i.e equating [itex]\frac{Mgsinθ}{3} =μN[/itex] .Isnt it??
     
    Last edited: Nov 18, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rotation and varying friction coefficient
Loading...