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Friction and Total Energy

  1. Dec 3, 2011 #1
    My group was assigned to make a lab about potential/kinetic energy and present the lab to the class. The lab we designed was to roll a ball down an incline, which then made a loop upwards like a rollercoaster, and calculate how high we need to set the ball on the incline in order that the ball completely make the loop without dropping.

    "without friction" we used the following formula, which worked perfectly for the ball that had little friction:

    Kinetic Energy Final + Potential Energy Final + Rotational energy (of a sphere) = KE Initial + PE Initial

    For Velocity final we used √(g*(radius of the circle)

    The slick ball was off by 3 cm (62 cm theory, 65cm experiment), but the rougher ball was off by an entire 20 cm. So we decided to try and find the energy lost due to friction. Here's where things go downhill.

    To find the coefficient of rolling friction we set up another experiment where we just used a stopwatch and see how fast each ball rolled down the incline part of the track, tilted to be almost flat so we could actually measure the time right. I'll use some hard numbers here since this might be where the problem is.

    Avg time for "White ball": 2.4s
    Avg time for "yellow ball": 2.9s
    Horizontal length: 70cm
    Vertical length: 2.5cm
    sinθ: 2.05 degrees

    To figure out the coefficient we just used ma = mg - μmgcosθ.

    For "a" we used some kinematics. Vfinal^2 = Vinitial^2 +2ad. Vinitial is 0, leaving

    a = Vfinal^2/2d

    And Vfinal is just 2Vavg

    So we got

    μ(White ball) = .011
    μ(Yellow ball) = .019

    Which makes sense since rolling friction is pretty small. Unfortunately, I didn't really know what to do at this point. What I tried was alter the original equation like this:

    KE(final) + PE(final) +RE(final) + Friction(incline) + Friction(semi-circle) = PE(inital) + KE(Final)

    giving this: (H = diameter of circle, h = height of ball at start)

    .5mv^2 + mgH + .5(Iω^2) + (μmghcosθ)/sinθ + .5μmg(Pi)(H/2) = mgh

    I got the (μmghcosθ)/sinθ by μ(Normal force)(distance), where distance = h/sinθ

    I got the .5μmg(Pi)(H/2) also by μ(Normal force)(distance). distance is (Pi)(H/2), and normal force I just made as a rough average of 0 at the top and μmg at the bottom, yielding .5μmg. Not perfect I know, but certainly not a big difference.

    Solving for h... only changed the original 62cm by a few centimeters for each ball, nowhere near 20 cm for the yellow ball. The error must lie somewhere in the two frictions I used, if I'm even supposed to due frictional energy this way.

    Anyways, sorry if this is a mess, but does anyone know where I went wrong? Was the frction incorrectly added to the energy equation? Wrong numbers? Other factors? If any more info is needed, I can change this. We can probably can get a decent grade without getting the friction since we're designing our own lab, but this is really bugging me :/
    Last edited: Dec 3, 2011
  2. jcsd
  3. Dec 3, 2011 #2
    What is the weight of each ball? I'm wondering if you'll need to consider air drag on the ball. The drag due to the air is a function of the velocity squared. It is a separate issue from rolling friction.
  4. Dec 3, 2011 #3
    Well the white ball was 42.33 g and the yellow was 9.69 g, and the white ball had a much bigger radius. We sort of glossed over drag, so I'm not sure what information is needed. However, I thought friction was generally a lot stronger than air resistance anyways.
  5. Dec 3, 2011 #4
    Well, you can check it out. Look up the drag coefficient for a sphere. The force on the sphere due to air drag is:

    F = rho*A*Cd*V^2/(2*g)

    rho is air density in pounds/ft^3
    A is area pi*d*d/4 in ft^2
    Cd is drag coefficient - look it up for a sphere
    V is velocity in ft/sec
    g is gravitational acceleration in ft/sec^2

    Units on F are pounds force.
    Last edited: Dec 3, 2011
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