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Galileo's inclined plane

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    I need a little help here.

    2. What could be the absolute largest acceleration we could obtain in this experiment? (just look at the equation you derived in lab).


    We are working with an inclined plane set at15.86 degrees.
    The ball is solid and weighs 67 grams
    The diameter of the ball is 2.53 cm

    I really have no idea what I am doing!

    after traveling .80 meters the ball passed through a gate and took 0.0145 seconds to enter and exit. This gave me a velocity of 3.045m/s2

    2. Relevant equations
    (We were given a=kgsin(theta))


    3. The attempt at a solution

    I can solve for K, but am unsure what K actually is. I know that it is a factor of inertia, but how do I solve for the question above?

    Here is what I have done.
    >
    > My ball is 2.53cm in diameter.
    > Its mass is 67 g
    > my angle is 15.86
    > acceleration is 1.90 m/s2
    >
    > a = k g sin theta
    > 1.90 = k (9.8) (0.2733)
    > 1.90 = k (2.678)
    > divide both sides by 2.678
    > k = 0.709
    >
    > I = k m r2
    > I= (0.709) (0.067 kg) (1.6x10^-4 m)
    > I = 7.6x10^-5
    >
    >
    > I also found:
    >
    > Solid sphere:
    >>
    >> I = 2/5 m r2
    >>
    >> where m = mass of sphere (lbm, kg)
    >
    > r = radius in sphere (ft, m)
    >
    > I = 2/5 (0.067 kg) (1.6x10^-4 m)
    > I = 4.288x10^-6
     
  2. jcsd
  3. Sep 26, 2011 #2
    A diagram is a must here!
    Why not use resultant force = mass x linear acc for linear motion down the plane
    and torque = moment of inertia x angular acc for rotational motion.
     
  4. Sep 26, 2011 #3
    We have not covered inertia or angular acceleration.

    Force = mass * acceleration I get, The rotational motion I don't understand.

    I found the inertia equations on the internet, and am not sure what I am looking at.

    Thanks for the help!
     
  5. Sep 26, 2011 #4

    gneill

    User Avatar

    Staff: Mentor

    I suppose that the k is there to account for effects of rotational inertia. The idea is that some of the gravitational potential energy that gets converted to kinetic energy as the ball falls through a given height (thus accelerating the ball as it falls) will be sidetracked into rotational energy in the ball. The details of what might go into deriving a theoretical value for k are not important at this juncture; For all intents and purposes here, the gravitational constant is 'kg', a single variable, for this particular setup.

    Looking at your given equation [itex]a = kg sin(\theta)[/itex], if 'kg' is of fixed value then the largest acceleration would occur when the [itex] sin(\theta)[/itex] term is maximum. When does that occur? Is it a reasonable extrapolation?
     
  6. Sep 26, 2011 #5
    If I set the angle to 90 Degrees, Sin(90) equals 1. My max acceleration would then be g. It makes sense to me, but I am still confused about k. I am sitting in the physics help room now and all of the graduate assistants are also confused.
     
  7. Sep 26, 2011 #6

    gneill

    User Avatar

    Staff: Mentor

    The maximum acceleration would be g if the ball no longer touched the ramp and therefore did not rotate as it fell. Until the ball loses contact (or the force of static friction is no longer sufficient to prevent the ball from sliding rather than rolling), the extrapolated maximum would be k*g.
     
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