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Rotational motion and an incline

  1. Jan 12, 2014 #1
    Hello, I have a problem that I have not been able to solve

    1. The problem statement, all variables and given/known data
    Here is the problem question:

    A bowling ball rolling at 8 m/s begins to move up an inclined plane. What height does it reach?


    2. Relevant equations

    The equation that I used was the formula relating potential energy to kinetic energy. In the book it showed how the translational equivalent of kinetic energy related to angular kinetic energy. The equivalent of kinetic angular energy is (3/4)mv^2 while potential energy is mgh, with m is mass, g is gravity and h is height.



    3. The attempt at a solution

    take:

    mgh = (3/4)mv^2

    h = (3/4)(v^2/g)


    The value for h I got is 4.90m
     
  2. jcsd
  3. Jan 12, 2014 #2
    Oh and the books answer is 4.57m.
     
  4. Jan 12, 2014 #3
    I think that (3/4)mv^2 is incorrect. Figure it out..
    KE = KE(tranwslational) + KE(rotational)

    KE(translational is (1/2)mv^2 , now go figure the KE(rotational) using the moment of inertia of a solid sphere.
     
  5. Jan 12, 2014 #4
    So the Kinetic energy for any rotating object depends on it's Moment of Inertia? Taking the moment of inertia of a solid sphere and combining it with the value for translational KE, I get (9/10)MV^2 for the KE of a rolling bowling ball and soling for h in mgh = (9/10)mv^2, the value is 5.88m.
     
  6. Jan 13, 2014 #5
    The moment of inertia of a solid sphere about its COM is 0.4MR2

    In general the kinetic energy of a body is nothing but the summation of the kinetic energy of its constituent particles.
    This sum can be broken into two parts for extended bodies-
    (i)The kinetic energy of the COM
    (ii)The rotational kinetic energy about the COM
    This energy depends on the work done by all the forces on the body and its initial kinetic energy.
    (Work Energy Theorem)
     
  7. Jan 13, 2014 #6
    I get .7MR^2 for the combined KE of the rolling sphere.
    I for a sphere is .4mR^2
     
  8. Jan 13, 2014 #7

    haruspex

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    I'm guessing you forgot the 1/2 in (1/2)Iω2.
     
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