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Two Problems on Rotation of a Rigid Body

  1. Dec 5, 2006 #1
    the first
    and the second:
     
  2. jcsd
  3. Dec 5, 2006 #2

    radou

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    5.0 m/s cannot be angular velocity, look at the units. It is the translatoral velocity of the sphere's center of mass, I assume.
     
  4. Dec 5, 2006 #3
    that was a mistake on my part. I meant to put velocity, not angular velocity.
     
  5. Dec 5, 2006 #4

    radou

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    Ok, now all you have to do is use energy conservation. Note only that the sphere has translatoral and rotational kinetic energy.
     
  6. Dec 5, 2006 #5
    I'm a bit unsure as to how to set up the formula, still.
     
  7. Dec 5, 2006 #6

    radou

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    Do you know how to use conservation of energy? The kinetic energy (translatoral + rotational) of the sphere must equal the potential energy of the sphere at the highest point on the incline (since the sphere's kinetic energy equals zero at that point).
     
  8. Dec 5, 2006 #7
    I've used Conservation of Energy before, but I haven't encountered the translatoral (at least the term) or rotational energy formulas yet.
     
  9. Dec 5, 2006 #8

    radou

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    [tex]E_{K,T}=\frac{1}{2}mv^2[/tex], and
    [tex]E_{K,R}=\frac{1}{2}I \omega^2[/tex].
     
  10. Dec 5, 2006 #9
    okay, now I"ve gotten it to

    (25/2)m+(1/3)mr^2(omega)^2 = 9.8mh

    am I on the right track?
     
  11. Dec 5, 2006 #10

    radou

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    Yes, but note only that the moment of inertia of the sphere must be calculated with respect to the bottom point, since the translational velocity at that point equals zero, so it is the center of rotation. This link may be a useufl reference: "http://www.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_4_3.html" [Broken]. I hope you'll get everything right now.
     
    Last edited by a moderator: Apr 22, 2017 at 2:26 PM
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