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Work and forces

  1. Jun 15, 2010 #1
    1. The problem statement, all variables and given/known data
    A question I had in an exam today, just wanted to see whether it was right:
    A ball is rolling down the inside of a concave surface. The radius of the ball and the curvature of the surface is such that there is no slide. Is work done by a) the friction force b) the Normal reaction force and c) the weight of the ball.

    2. Relevant equations



    3. The attempt at a solution
    Frictional force does no work, as the ball is rolling.
    The Weight of the ball does positive work (mg sin theta)
    The Normal reaction force does negative work (-mg sin theta)

    Does that sound right?
     
  2. jcsd
  3. Jun 15, 2010 #2

    collinsmark

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    Hello tomwilliam,
    So far so good. :approve:
    I like the general idea, but the equation you gave has units of force, not energy (i.e. not work). :frown: You'll have to redo that equation. What part of the ball's movement is parallel to the direction of its weight (whether it is rolling or not)?
    You should rethink the normal reaction force.

    Work is defined as the path integral

    [tex] W = \int _P \vec F \cdot d \vec l [/tex]

    where [itex] \vec F [/itex] is the force at any given position, and [itex] d \vec l [/itex] is the differential length of the movement. Notice there is a dot product involved! Even if your class is not calculus based, you'll still need to use an equation that involves a dot product.

    What can you say about the direction of the normal force compared to the direction of the movement at any given point on the surface?

    (btw, if you consider angular energy due to the rotation of the ball, in addition to the ball's translational energy, it becomes an interesting problem to contemplate. You kind of have to do that at least a little for the first part because if the ball wasn't rolling, and instead sliding, friction would be doing work. But in the other two parts, since the question is not asking for any final velocities or final ratios of translational and rotational energy, simply treat energy as energy -- simply consider the direction of the ball's overall movement compared to the direction of the forces involved.)
     
    Last edited: Jun 15, 2010
  4. Jun 15, 2010 #3

    Doc Al

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    That's good.
    OK.
    Is there a displacement in the direction of the normal force? (And why do you think mg sin theta is the reaction force?)
     
  5. Jun 15, 2010 #4
    Sorry, yes, I realise that the equations I gave were for force, not work. Of course as the ball passes through a height h then the work can be calculated. The question only asked me to state whether or not work was done.

    The second part of the question (given the radius and mass of the ball, and height h) was to calculate the final speed. Of course you can use mgh as the initial gravitational energy, taking the bottom of the surface as the zero of potential energy, which you then have to equate to the total kinetic energy, rotational plus translational. I knew the moment of inertia so could calculate translational kinetic energy but didn't know how to find out the angular speed, so couldn't finish the question.
     
  6. Jun 15, 2010 #5

    collinsmark

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    Yes, I understand.

    But you'll still need to rethink part b), the part about the normal reaction force. You claimed that it did negative work, but that's not quite right.
    Very nice. That's the way to do it. :approve:
    The trick is to realize that a point on the outer perimeter of the ball is moving due to rotation. And that rotational speed must be the same speed that the center of the ball is moving. Combining the two makes that point (on the outer perimeter of the ball) have a velocity of 0 at the instant the point touches the surface. If it wasn't that way, the ball wouldn't be rolling, it would be sliding! This provides a tidy relationship between |v| and omega.
     
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