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What is its linear speed at the bottom of the incline?

  1. Nov 23, 2013 #1
    1. The problem statement, all variables and given/known data
    A spherical object with a moment of inertia of 0.584mr2 starts from rest rolling down a 2.25-m high incline. If the sphere is rolling without slipping


    2. Relevant equations
    I the best equation to use for this problem is k(initial)+u(initial)=k(final)+u(final)


    3. The attempt at a solution
    I stretch the equation 1/2mv^2(initial)+1/2Iω^(initial)+mgh(initial)=1/2mv^2(final)+1/2Iω^2(final)+mgh(final).

    since the object started from rest the initial kinetic energy and the final potential energy is zero, which leads me to this equation

    mgh(initial)=1/2mv^2(final)+1/20.584mr^2ω^2(final). As I continue reduce the equations I round up with this.

    gh(initial)=0.792v^2(final)

    v=sqrt(9.81m/s^2)(2.25m)/0.792

    the linear speed I came up with was 5.279m/s.

    Did I do this right?
     
  2. jcsd
  3. Nov 23, 2013 #2

    haruspex

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    The parentheses aren't written correctly there, but the method's fine and I agree with your final answer.
     
  4. Nov 24, 2013 #3
    How would I write the parentheses the correct way?
     
  5. Nov 24, 2013 #4

    haruspex

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    sqrt((9.81m/s^2)(2.25m)/0.792)
     
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