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Moment of inertia. Ball rolling experiment.

  1. Dec 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi,
    I have been asked to find the moment of inertia of a rolling ball. The ball can be any size and radius. I have chosen a solid ball.

    The experiment says that I should roll the ball down a ramp and then measure the time it takes for the ball to roll from the end of the ramp to some fixed distance. I have done this and have the measurements from the experiment but now I do not know how to calculate the moment of inertia?

    2. Relevant equations

    I know that for a stationary solid ball the moment of inertia is (2/5)Mr^2. the experiment talks about using potential, rotational kinetic and translational energy?

    3. The attempt at a solution

    I really am confused about what the moment of inertia actually is!!

    Can anyone point me in the right direction?
    Thank You
    Dominic
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 5, 2012 #2

    Simon Bridge

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    You need to use conservation of energy here ...
    What sort of energy does the ball start with, at the top of the ramp?
    Where does that energy go?

    Write an equation that describes that.

    This should help:
    Moment of inertial plays the same role for rotation that mass plays for translation. Just like mass, it is the same for moving bodies as for stationary ones. You have to do work to start an object spinning - work is change in energy - so the spinning object stores energy.
     
  4. Dec 5, 2012 #3

    ehild

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    From the distance covered and the time you can determine the velocity of the ball. Knowing the velocity, you can apply conservation of energy, as Simon suggested. The ball has both translational kinetic energy and rotational energy. How the rotational energy is related to the moment of inertia?

    ehild
     
  5. Dec 5, 2012 #4

    jedishrfu

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    with the distance and time you can calculate the angular velocity assuming no slipping, right?

    and you can calculate the linear velocity too

    and with these you can calculate the total kinetic energy of the ball using the equation:

    KE= 1/2*m*v^2 + 1/2*I*w^2 for the ball.

    The ball is now in your court...
     
  6. Dec 5, 2012 #5
    Thank you for the replies.
    OK I now have the equation

    Pe = Ke + (1/2)Iω^2

    Which I have rearranged for "I"

    (M(2gh - v^2))/ω^2 = I

    I found the velocity and ω as suggested.
    I have then use this to find I. Is this correct?

    I have tried it with different balls (tennis ball, ball baring and a marble), and have got some wildly varying results! For example the tennis ball I = 3939.9 and the Marble I = 6.89?
     
  7. Dec 5, 2012 #6

    jedishrfu

    Staff: Mentor

    for the velocity did you use the instaneous velocity or the average velocity?
     
  8. Dec 6, 2012 #7
    For the velocity I used D/t.

    For t I used the time the ball was between two points. So I think this must be the average velocity?
     
  9. Dec 6, 2012 #8

    ehild

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    V=D/t is the average velocity, but the object moves with acceleration along the slope. You know the distance travelled and the time, and the ball started from rest: from these, you can calculate the final speed.

    ehild
     
  10. Dec 6, 2012 #9

    Simon Bridge

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    Compare your experimental values with those you calculate. Bear in mind that the marble is a small smooth solid while the tennis ball is large, hairy, and hollow.

    If the balls do not slow much over the distance you timed them, you should be fine with the mean speed. Your equation does call for instantaneous speed at the bottom of the ramp... so, if the test ball does slow a lot you will need to account for that.
     
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