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What slows down a rolling sphere?

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data
    My teacher said that static friction can't slow down a sphere, when I during his office hours, and he gave he said that they don't do work on a rolling sphere..... instead rolling friction is what slows down a sphere. Can someone explain to me how rolling friction works and why static friction can't slow down a sphere (when its rolling without slipping).

    2. Relevant equations

    uhhh... [itex] F_{fs} = \mu _k F_n [/itex]

    3. The attempt at a solution
    Just a conceptual question.
     
  2. jcsd
  3. Jun 11, 2015 #2

    Simon Bridge

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    See this: https://www.lhup.edu/~dsimanek/scenario/rolling.htm
    ... it starts out without friction - but bear with it.

    Note: if the static friction could slow the ball down, then it must do work on the ball ... work is force times the distance the ball moves in the direction of the force. So now you can figure out the work done by the static friction. Or just use logic: if the static friction acted on the motion of the ball, then it would not be static would it?
     
    Last edited: Jun 11, 2015
  4. Jun 11, 2015 #3
    then how does static friction cause a person to move? its the static friction between the feet and the ground.... but that causes a displacement correct?
     
  5. Jun 11, 2015 #4
    No - static friction is defined as a resistive force between two surfaces that are initially at rest wrt and prevents motion between them.
     
  6. Jun 11, 2015 #5

    haruspex

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    It is generally more correct to say that a rolling ball is slowed by 'rolling resistance'. That may have some small contribution from atmospheric drag, but is mostly to do with deformation of surfaces. If you could examine the ball and ground in great detail you would find that they get compressed at the leading edge of contact and rebound at the trailing edge. Since it is not perfectly elastic, the rebound force is weaker than the initial compression force, so there is a net torque opposing the motion of the ball.

    That said, it is not true to say that static friction cannot slow a moving object. Consider instead a cart with tyres that are perfectly elastic but there is some friction in the axles. The only external force on the cart with a horizontal component is static friction, so it must be what slows the cart down, and the integral of this wrt distance will equal the KE lost. But at the same time as the cart is doing work against this friction, that friction is in turn doing work turning the wheels and overcoming axle friction. Thus, there is no net work done on the ground.
    Note that if there were no friction between tyres and ground the cart would not slow down.

    Edit: going back to the rolling ball case, will there be a static friction force, and which way will it act?
     
  7. Jun 11, 2015 #6

    jbriggs444

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    The force of static friction does no work in the frame of reference in which the two involved surfaces are at rest. That's not the same thing as saying that it cannot be involved in causing a person to move. There are at least two details that come in.

    First, there is the distinction between [real] work and center of mass work. If you look at the motion of the sole of the person's shoe, that sole is not moving with respect to the ground. No work can have been done on it. If you look at the motion of the person's center of mass, it is moving with respect to the ground. Work is being done on it. "Real" work is the force applied across an interface multiplied by the [parallel component of] the displacement of the material upon which that force is applied -- the force of ground on shoe leather multiplied by the displacement of the shoe leather. "Center of mass" work is the force applied to an object multiplied by the [parallel component of] the displacement of the center of mass of the object. For point objects the two are identical. For rotating or non-rigid objects, the two can be different. Human legs are non-rigid. The muscles in the legs can be seen as the source of the real work that is done while walking.

    In the case of a sphere rolling on a stationary surface, static friction cannot do [real] work. However, it can do either positive or negative center of mass work.

    Second, then there is the choice of frame of reference. The work done by a force will vary depending on your choice of reference frame. For instance, if you throw a ball forward in a train, the work that your hand does on the ball will be larger as judged by an observer on the ground than as judged by an observer on the train. Work is force times distance moved. The distance moved from starting coordinate to ending coordinate depends on the motion of the coordinate system. A measured quantity that does not depend on coordinate system is called "invariant". Neither work nor energy are invariant measures.

    In the case of a sphere rolling around on the bed of a truck, static friction can do work as judged using the ground as a reference frame.
     
  8. Jun 11, 2015 #7
    ok, so what's the force that does work on a person that moves (that causes his displacement).
     
  9. Jun 11, 2015 #8

    jbriggs444

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    Leg muscles.

    Edit: Looking for a "cause" in a physics equation is not generally a good idea. An equation asserts a relationship between quantities. It says nothing about which quantity is cause and which is effect. You asked for a force that does work on a person that moves. The forces at the attachment points of the leg muscles to the person's body are examples of such forces. The net work done by all such forces (if you add them up) comes out positive. If you want to drill down, the contracting muscle fibers exert such forces at their end points. The sum of those over all the muscle fibers also comes out positive.
     
    Last edited: Jun 11, 2015
  10. Jun 11, 2015 #9
    what would you call that in a Free body diagram? an applied force?
     
  11. Jun 11, 2015 #10

    jbriggs444

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    Looking at muscle forces in the context of a free body diagram implies that you are splitting the body up and considering its various (semi-)rigid components separately.

    For instance, you could account for the gastrocnemius at its attachment points on the lower leg near the knee and on the back of the ankle. You want to have at least two free body diagrams. One for the lower leg and one for the foot. The foot would have external forces from ankle joint, gastrocnemius and the ground. The force from the gastrocnemius would be an external force.
     
  12. Jun 11, 2015 #11

    Simon Bridge

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  13. Jun 11, 2015 #12
    sorry i started reading it during class, then went back to copying notes. ill read it tonight.
     
  14. Jun 12, 2015 #13

    Simon Bridge

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    ... reading other work during class is usually frowned upon yes ;)
    The article answers most of your question.
     
  15. Jun 12, 2015 #14
    oh dont worry.... it was just a math class ;)
     
    Last edited: Jun 12, 2015
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