Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotational mechanics question?

  1. Sep 17, 2011 #1
    One more question on rotation.

    first of all see attachment i want to say that if friction is too large then net force is in direction of inclined in upward direction but from common experience we can say that it will move in downward direction so where is it going wrong. why it moves in downward rather than upward.
     

    Attached Files:

  2. jcsd
  3. Sep 17, 2011 #2
    If it goes upward then you are breaking law of conservation of energy.You are increasing the energy(potential) of system by itself without external work being done.
     
  4. Sep 17, 2011 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Your error is in thinking that friction must equal μmgcosθ. But the static friction will be whatever it needs to be to prevent slipping, up to that maximum value.
     
  5. Sep 17, 2011 #4
    I also thinking so.
    frictional force will equal to mgsin(t). Net force in direction of inclined will zero and it will in pure rolling motion. Am i correct here? I am in doubt about this.
     
    Last edited: Sep 17, 2011
  6. Sep 17, 2011 #5
    Hey, but what about energy conservation???
     
  7. Sep 17, 2011 #6
    bhaskar bhai. I have made a solution on the paper.Can you please check this.
    Hope this is correct.


    through the figure.
    torque around IAOR(instantaneous axis of rotation) is T.
    moment of inertia about IAOR is Mk2. think that k is radius of gyration.Since friction is too large so it will static and enough to have pure rolling. a is acceleration linear acceleration of center of mass(of object on incline) and A is angular acceleration.
    now calculation
    T=IA=mgsin(t) R
    A= [itex]\frac{gsin(t)}{2R}[/itex].
    a=[itex]\frac{gsin(t)}{2}[/itex]
    So here we can conclude that frictional force is 0.5*m*g*sin(t).


    tell me if i a wrong somewhere.
     
  8. Sep 17, 2011 #7

    Doc Al

    User Avatar

    Staff: Mentor

    No, you are not correct. You can figure out the friction force by applying Newton's law to both translation and rotation.

    If you call the friction force F, then the net force will be mgsinθ - F. That same force F will also produce a torque on the sphere. By applying Newton's law and the condition for rolling without slipping you can solve for F.
     
  9. Sep 17, 2011 #8

    Doc Al

    User Avatar

    Staff: Mentor

    I don't quite follow what you're doing. See my last post for an outline of how to solve for the friction force.
     
  10. Sep 17, 2011 #9
    Well I didn't understand the significance of 't'. It's okay that mu(coefficient of friction)*g*sin(x) is limiting case and that's what you have to consider. If you can explain me the physical significance of 't'(that's your mistake) then it's okay what you have done(For any middle case).
     
  11. Sep 17, 2011 #10
    friend I am doing the same thing but about the instantaneous of rotation. You may not understanding my way because i skip many step.

    If I do in the way you say then.
    torque about the center of
    Fr=mr2a
    F=mra
    F=mA.
    Now
    mgsin(t)-F=mA
    since mA=F
    so mgsin(t)-F=F
    2F=m*g*sin(t)
    so F=0.5*m*g*sin(t).
    t is angle of inclination of incline, a is angular acceleration, A is linear acceleration.

    Am i correct??
     
  12. Sep 17, 2011 #11
    Sorry didn' read it completely
     
  13. Sep 17, 2011 #12
    t is angle of inclination of inclined plane. In figure it is x. that's Variable mismanagement.
     
  14. Sep 17, 2011 #13
    So am i correct now?
     
  15. Sep 17, 2011 #14
    mA=F is final limiting case.I thought you wanted to calculate for case in between
     
  16. Sep 17, 2011 #15

    Doc Al

    User Avatar

    Staff: Mentor

    So: F = friction; I = mr2 (a cylindrical shell?); a = angular acceleration
    Realize that you are solving for a special case where I = mr2, a cylindrical shell. But given that, you are correct.
     
  17. Sep 17, 2011 #16
    If r is radius of gyration. then will it correct?

    what you mean. I did not understand. can you please explain.
     
  18. Sep 17, 2011 #17

    Doc Al

    User Avatar

    Staff: Mentor

    No. The friction force needed to prevent slipping will depend on the rotational inertia.

    Don't use the same symbol for radius and radius of gyration.
     
  19. Sep 17, 2011 #18
    I can't finally. I have made my tries. I am still thinking that I am correct. This time i will say dear stop this chat and answer me.:cry:(friction and acceleration of object):cry:
     
  20. Sep 17, 2011 #19

    Doc Al

    User Avatar

    Staff: Mentor

    :confused: Answer what? Do you not see where you assumed that I = mr2? If you want to be more general, use I = mk2. But then your answer will be in terms of k and r, which will no longer cancel.

    If you pick the object you want to study, then you can compute its acceleration and the required friction.
     
  21. Sep 17, 2011 #20
    take it a ring of radius R and solve. find friction and acceleration. If you think there is still lag of any variable than assume that but just do it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rotational mechanics question?
  1. Mechanics question (Replies: 1)

  2. Rotational mechanics (Replies: 3)

  3. Rotational mechanics (Replies: 4)

  4. Rotational mechanics (Replies: 5)

Loading...