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Rolling without slipping on application of impulse

  1. Mar 29, 2012 #1

    I am having some trouble understanding the effects of an impulse on rolling motion, particularly without slipping.

    For eg, if we take a sphere of radius R and mass M, then what is the point h above the centre C at which when an impulse is imparted, will cause pure rolling motion? (i.e., without slipping)

    I have racked my brain two days over this. I think I am seriously missing something.

    plz Help.:cry:
  2. jcsd
  3. Mar 29, 2012 #2


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    hi nerdvana101! :smile:

    if there's rolling without slipping, the centre of rotation is the point of contact, so you can use angular momentum about that point to find the angular velocity

    then do impulse = m∆vc.o.m to find if the linear velocity matches the angular velocity
  4. Mar 29, 2012 #3


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    This problem doesn't make much sense if the sphere is resting on a surface with a high amount of friction, since it will roll with the impulse applied just about anywhere. I assume the idea here is that the surface is frictionless, and the goal is to locate the impulse so that the surface speed due to angular rotation always equals the linear speed as the sphere accelerates. The linear force produces the same linear acceleration no matter where the impulse is located, so only the angular acceleration is dependent on the location of the impulse. The angular acceleration = torque / (angular inertia), and torque = force x radius of the point of appliciation of the impulse. You could think of the sphere as being a yo-yo with a massless hub and a string wrapped around the hub, so that the string tension is how the impulse is imparted at some radius from the center of the sphere's axis of rotation.
  5. Mar 29, 2012 #4
    This is very important in billiards and pool. If the pool cue hits the center of the billiard ball, the ball will slide because no angular momentum is transferred to the ball, except by sliding on the table. The cue has to hit the ball above the midpoint to impart some angular momentum to the ball. The problem is to find the impulse point of impact such that (as tiny tim pointed out) the angular velocity vrot = Rdθ/dt and the linear velocity vlinear = p/m of the ball are equal.
  6. Mar 30, 2012 #5
    Thx tiny-tim. What I forgot to add was the basics of angular momentum -

    L at the point of contact = Icomxω + rcomxmxvcom.

    THX a lot!
  7. Mar 30, 2012 #6
    isn't there any option to select the best answer?
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