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Homework Help: Rotational Kinematics

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose a uniform sphere of mass M and radius R is placed on a rough horizontal surface. It rolls without slipping.

    We can apply 2 equations.
    1) considering the torque due to frictional force about the centre of mass,
    here alpha=a/R as the it is in pure rolling motion
    so f=2Ma/5

    2)considering the linear motion,

    why are 1 and 2 contradicting.?
  2. jcsd
  3. Sep 13, 2010 #2
    Well, they're not. Why? f=0 and a=0. No contradiction, right? :biggrin:
    Think of energy conservation law. If the ball rolls without slipping on a HORIZONTAL surface, without any force from us, does the ball accelerate?
  4. Sep 13, 2010 #3
    Ok fine.
    Suppose the ball is moving with uniform velocity (without any acceleration and any sliding motion).
    As you told f=0.
    Then what causes the ball to rotate? there must be some sliding friction right?
  5. Sep 13, 2010 #4
    Let's look at initial conditions. When the ball rolls without sliding, that means the ball has both linear velocity v and angular velocity w=vR. Just like v, w doesn't change.
    The force only causes the CHANGE in v and the torque only causes the CHANGE in w.
    And how does the ball gain v and w? Maybe before you drop it on the ground, you spin it or hit it or so, as long as it gains v and w such that w=vR. But that's out of our concern. We only care what happens when the ball already rolls without sliding.
  6. Sep 13, 2010 #5
    Alright its clear now.

    I got another one.
    A rod is released from vertical position (giving it a negligible small push) on a smooth horizontal surface. Is the direction of normal reaction perpendicular to the ground or along the rod when it is inclined at angle to the ground? Is there any radial acceleration during its motion?. Suppose the ground is rough (such that it does not allow slipping of bottom of the rod). Then will the above answers change?

    P.S.- This doubt is a product of my mind which came while solving problems.
  7. Sep 13, 2010 #6
    What do you think? :wink: (sorry, forum's rule; I cannot answer before you provide some of your thoughts :smile:)

    Okay, that means you're getting better than the book you read / the problem your solved :biggrin:
  8. Sep 13, 2010 #7
    Well, I think the direction of normal force is along the rod and there is centripetal acceleration in both the cases.
  9. Sep 13, 2010 #8
    Why's that?
  10. Sep 13, 2010 #9
    I dont have a clear explanation for it. May be because normal force always acts along the centre of mass of a body.
    Centripetal acceleration is necessary for rotation.
  11. Sep 13, 2010 #10
    1/ About the direction of normal force:
    Actually the normal force ALWAYS acts perpendicular to the contacting surface. Ideally it should be a contacting point, not a surface. But the real situation is never like that. Instead of a rigid rod, you push your thumb on the ground and will see that it's a contacting surface.

    So that's the same with the rod. When the width of the rod is so small compared to its length, you can model the rod as a straight line whose width = 0. But when it comes to normal force, we have to look at it carefully. See the attached picture. If you try to zoom it and zoom it and zoom it, you will see something like that: it's a contacting surface. Therefore the normal force acts perpendicular to the ground / the plane, not along the rod.

    2/ About the centripetal acceleration:
    When you mention centripetal acceleration (C.A.), you MUST refer to C.A. of which point. For example, if it's a rough surface and the lower end of the rod sticks to the ground, then the rod performs rotation about that end. Every point on the rod has its own C.A., except for the lower end: this end doesn't move, so it doesn't have C.A.

    Likewise, in the case that the surface is smooth, the rod falls: the center of mass (COM) of the rod & the lower end of the rod perform motion in a straight line, so these 2 points don't have C.A., while other points perform "weird" motion (their orbits are neither circle nor straight line!), so these points have C.A.

    Attached Files:

  12. Sep 13, 2010 #11
    That cleared the problem.

    Suppose the rod is pivoted on a fixed support, then why do we take the normal force along the rod? it is not perpendicular to the surface here right?

    about centripetal acceleration, consider this problem.

    A thin rod is held resting on a rough ground with its length inclined at an angle (theta) to the horizontal. The coefficeint of friction between rod and ground is (mu). Show that when the rod is let go it will start slipping on the ground if (mu) < 3sin(theta)cos(theta)/(1+3sin^2(theta)

    In this problem, if we consider the limiting case i.e. the rod is just about to slip then do we have to take any centripetal acceleration of centre of mass for calculating its net acceleration (tangential acceleration is also present)?
    Since the centre of mass follows a circular path while falling, I calculated net acceleration of centre of mass including its centripetal acceleration but got wrong answer. (answer considers only tangential acceleration)
    Why is it so?
  13. Sep 13, 2010 #12
    I'm not sure what situation you are referring to, but maybe the attached picture can shed a little light. In the picture, the pivot is lubricated. If the pivot is not so smooth, there is friction, and the resultant force on the rod is no longer along the rod.

    Yes. Just before the rod slides, COM still performs circular motion.

    Why don't you show me your work? :smile:

    Attached Files:

  14. Sep 13, 2010 #13
    Sure dear :smile:

    considering the torque acting about the point of contact,
    mgcos(theta) l/2=ml^2/3 (alpha)
    from here,
    tangential acceleration is 3gcos(theta)/4

    Using conservation of energy,
    mgl/2 (1-cos(theta))=0.5 ml^2/3 w^2
    (taking reference level through the initial position of centre of mass)
    from here i get w^2=3g(1-cos(theta))/l

    centripetal acceleration of the centre of mass=mw^2 l/2

    now the frictional force balances the force due to centre of mass in x-direction
    taking components of tangential and centripetal acceleration along x- direction I got one equation involving N-the normal force which is perpendicular to the ground. ( (mu)N=m(ax))

    then taking components of accelerations along y-direction ,

    solving for N I did not get the right answer.
    Instead if i proceed without considering centripetal acceleration of the centre of mass, I got the correct answer. How?
  15. Sep 13, 2010 #14
    There is one mistake here:
    "mgl/2 (1-cos(theta))=0.5 ml^2/3 w^2 "
    Anyway I also don't get the answer. What is the source of the problem?
    It's midnight at my place so I'll come back to this tomorrow.
  16. Sep 13, 2010 #15
    erm leave it if you don't get. You must be very busy with your first year :smile:
    Source of this problem is my Physics book from Brilliant Tutorials for IIT-JEE (the toughest engineering entrance exam in the world :wink:)
  17. Sep 13, 2010 #16
    What is the mistake there anyway?
  18. Sep 14, 2010 #17
    Ah, I read the question again and now I see where we were wrong. Staying up late is really horrible :uhh:

    The question doesn't mean that you hold and release the rod at the vertical position! The rod is first held at the (theta) position, then released. The question wants us to show that right after being released, the rod will slide if the above expression is satisfied. This is why there is no centripetal acceleration included: right after being released, the rod hasn't gained any speed yet, so centripetal acceleration at that time = 0!

    Eek, how many students can survive, let alone be inspired, when having to face with this exam? The level of this problem is a way too advanced to high school students :uhh:

    It's sin(theta), not cos(theta).
  19. Sep 14, 2010 #18
    Thanks alot.

    Lol. Only 8000 students of a total of 4 lakh survive ::biggrin::
    This is a normal problem. There are much much more tougher problems than this :wink:
  20. Jan 19, 2011 #19
    If the rod hasn't gained any speed yet, why do we have an expression for tangential acceleration? Is it not 0 just after releasing?

  21. Jan 20, 2011 #20
    Eek, you may want to read my first sentence in post 17 :wink:
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