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Direction of friction in rolling motion

  1. May 28, 2012 #1
    I am really confused as to how to determine the direction of friction acting on a rolling object. Could someone help clarify how to determine the direction of friction? ANy help is appreciated ;)
  2. jcsd
  3. May 28, 2012 #2
    The easiest thing to say is that friction opposes motion. May not be much help but it is where you must start
  4. May 28, 2012 #3
    Think of an accelerating car...say increasing speed to the right.....a positive acceleration.

    can you visulaize the direction of friction against the tires??...the 'gripping' between the road and the tire?? the tires are rotating clockwise....so the friction vector must oppose this at the road surface and point to the right....

    If a car is decelerating what happens....as brakes are applied?? the force of friction must again help the tires 'grip'....but this time points towards the left.

    edit: just occurred to me....for steady velocity rolling friction....I could not tell for sure....looked it up:

    Last edited: May 28, 2012
  5. May 28, 2012 #4
    let's consider the problem when a ball initially at rest on a surface with friction is given a horizontal impulse. According to some online sources, the ball will eventually attain a maximum speed after it is given the impulse. However, this makes absolutely no sense to me; isn't the friction opposing the motion, causing the velocity to strictly decrease?
  6. May 28, 2012 #5
    The ball being struck is not an easy problem!!!! depends where it is struck, any spin and friction.
    A great book 'physics of ball games' ny Daish goes into great detail about the motion of (snooker) balls across surfaces.
  7. May 28, 2012 #6

    Ken G

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    The constraint that static friction imposes is that the point on the ball that is in contact with the ground is instantaneously stationary, i.e., has v=0. As said above, the direction of the friction can be either way-- whatever is necessary to achieve v=0. Now this is not a simple constraint, because friction is a force, and v is a velocity, and force does not produce velocity it produces acceleration and torque. But the thing you can "take to the bank" is that the friction will produce a horizontal time dependent force, call it F(t), that is whatever is required to insure the point that touches the ground always has v=0. That constraint is made useful by connecting it to the speed of the ball V and the rotation rate of the ball W, which thus must have the simple connection V=R*W, if R is the radius of the ball. Then you look at whatever other forces are on the ball, and write the free-body diagram for the net force that controls V, and the net torque that controls W, and then F(t) will be whatever it takes to allow V=R*W to prevail at all times. So this will depend on what other forces are in play, and what torque they represent.

    If the problem is described in terms of impulses, instead of F(t), it just means that you are integrating F(t) over time to get an impulse, and you are imagining that the time is very short. The logic is the same-- the impulse of lateral momentum, and the impulse of angular momentum, both have to be specified, and you then assert there is also an impulse of both from the friction, and its value is whatever is required to allow V=R*W to hold after the impulses have been applied. That will suffice to tell you the impulse of lateral momentum coming from friction (since the impulse of angular momentum will just be R times that, with whatever sign is required to make it all work-- which is the answer to your question).

    Note that after the impulses are over, the ball rolls at constant speed, and this will indeed be the maximum speed it attains. There is no static friction at this point, not in the idealized case.
  8. May 29, 2012 #7


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    A ball given a sharp horizontal impulse will almost surely slide to begin with. The peak force during the impulse is likely to overcome static friction. As it slides, the dynamic friction applies a torque, accelerating the ball rotationally and slowing it linearly. When the rotation rate matches the linear speed (so that the point of contact is instantaneously stationary) rolling commences. At this point, there will be far less frictional force. Flexing in the surface tends to slow the ball down. Since that would cause the rotation to outstrip the linear motion, a static frictional force arises acting forwards on the ball. This slows the ball rotationally while helping to maintain its forward motion, keeping the two in synch.

    Perhaps the sources you cite are considering a force applied over some nonzero duration as the impulse, and 'eventually' refers to that duration. Or maybe they're referring to the rotational speed.
  9. May 29, 2012 #8
    If you consider a ball struck with a horizontal blow at the centre no initial rotation is imparted and the ball will begin to slide over the surface with an initial velocity, V.
    A frictional force will come into play at the point of contact
    F = μMg
    where M is the mass of the ball and μ is the coefficient of sliding friction between the ball and the surface. (about 0.2 for snooker, billiard balls).
    This force has 2 effects....first it will produce linear decelleration of the ball = F/M = μg
    second....the moment of the force about the centre of the ball will produce an angular acceleration = torque/moment of inertia. = Fr/I = μMgr/I
    where r = radius of ball and I = moment of inertia of the ball about an axis through the centre. I = (2Mr^2)/5
    angular acceleration = 5μg/2r
    After a time, t, the linear velocity will have decreased to v = V - μgt
    and the angular velocity will have reached a value ω = 5μgt/2r
    The ball will start rolling when v = ωr
    The above equations can be combined to show that atthis point the linear velocity = 5V/7
    and the angular velocity = 5V/7r

    I hope this adds to the discussion
    The mathematics comes from 'the physics of ball games' by C B Daish
  10. May 29, 2012 #9
    This likely means at the instantaneous moment after the impulse is applied. I don't understand the use of the word 'eventually'. Seems to me no matter whether the ball is rotating or slipping, once friction forces are acting at the moment after the impulse, the ball would begin to slow immediately.
  11. May 29, 2012 #10


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    Friction in this particular case would not neccessarily be in place. However, a counter force might be the more correct term to use. As a ball rolls on a surface, the ball and the surface will be deformed, causing the ball to roll in a pit that follows the ball. The pit will cause the ball to roll against a slight uphill - even if the ball is moving horizontally. The "friction" in this case does not have to result in heat loss (Which is friction in my understanding of the term). If heat loss is out if the equation, there must be a counterforce rather than friction. It can however be both friction and counterforce - if the ball and/or the surface consists of a soft material that permanently deform as it rolls.

    Correct me if I'm wrong.

    EDIT: Super elastic materials might solely be slowing down due to counter force which is caused by the pit due to the weight of the ball on the surface. A ball of led would roll shorter than a ball made of nitinol or other super elastic materials.

  12. May 29, 2012 #11
    Hi david456103,
    Here is the best way to find the direction of friction force. Its my own style. I think it would really help you. If you ever want to find the direction of friction force on any object, first think that friction is completely absent there. It happens somewhere on the completely wet muddy surface or say on any lubricant surface. Sometimes you might have seen any Tyre slipping on wet wet muddy surface. If you are able to assume this in your mind than think if the complete smooth surface on which the Tyre is slipping instantaneously becomes rough then in what direction friction will act.
    EX. See the attachment I have attached with this reply.
    In this figure assume the van is slipping on a oily surface. The engine of the vehicle makes the rear tyres rotate so the rear tyres will be rotating only not translating in front direction> the bottommost point of the rear tyre "A" will be in contact with the ground and slipping in the back direction. If the friction suddenly comes there, it'll act in the front direction.
    Same thing for the front tyre. There is no engine which rotates the tyre. So on the oily surface it'll only be slipping in the front direction not rotating. The contact point of the front tyre will be "B" and will be translating only in the front direction. If the friction comes suddenly, it'll act in the back direction.

    Attached Files:

  13. May 29, 2012 #12
    @truesearch i think you have the wrong problem in mind, but yeah your thinking process is exactly the one I feel is right.
    Here is the problem I'm concerned about(from MITOCW):
    A spherical billiards ball of uniform density has mass m and radius R and moment of inertia about center of mass of (2/5)mR^2. The ball, initially at rest on a table, is given a sharp horizontal impulse by a cue stick that is held an unknown distance h above the centerline. The coefficient of sliding friction between the ball and the table is μ. You may ignore the friction during the impulse. The ball leaves the cue with a given speed v0 and an unknown angular velocity ω0. Because of its initial rotation, the ball eventually acquires a maximum speed of (9/7)v0(??????). Find h/R.

    The given solution uses angular momentum, which completely makes sense to me, but this problem does not make any conceptual sense to me. How can the ball increase in translational speed when there is friction?
  14. May 29, 2012 #13
    What I can add is where to hit a snooker ball to eliminate sliding (i.e. ball rolls)
    To do this the ball is hit above centre at a height, h, above the table.
    To eliminate sliding the point of impact is known as the 'centre of percussion'
    It can be shown that the centre of percussion is given by (r + I/Mr) where I is the moment of inertia about an axis through the centre of the ball.
    so h = r + I/Mr and I = (2Mr^2)/5 gives h = r +(2Mr^2/5Mr) = 7r/5 or 7x(2r)/10
    so to eliminate sliding the ball needs to be struck at a height above the table equal to
    7/10 of the diameter.
    Again, the analysis is a summary of what is given in 'physics of ball games'
    hope it helps
  15. May 29, 2012 #14

    Ken G

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    In my opinion, the only way this is possible is if the ball is hit so high that it is initially spinning faster than it is rolling. If you add an impulse p to a ball at distance x from the center, the ball will have mv=p, and its angular momentum will be I*w = p*x. Thus v = p/m = I*w/(mx) = 2/5 R2*w/x. That will be less than R*w any time x > 2/5 R. So if you strike it more than 2/5 from its center, assuming the cue doesn't slip off, you can make it spin faster than it is translating, and so it will actually grab and speed up. Too bad you can't do it with conservation of energy-- while it is sliding it will be dissipating energy, so you have to solve for when v = w*R. But v will always exceed the initial value of x > 2/5 R, because the frictional force is forward. It's always kinetic friction too-- by the time static friction kicks in, the ball will roll with constant speed.
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