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Rolling Motion (Kinetic Energy)

  1. Nov 5, 2005 #1
    I have a question regarding the Rolling Motion equation.

    [tex](1/2)Mv^2 + (1/2)I\omega^2 + Mgy[/tex]
    (Where M=mass, v=velocity, I=moment of inertia,ω=angular velocity, g=force due to gravity and y = vertical distance)

    The problem I'm having involves the direction of the acceleration due to gravity. I've seen the equation written as...
    [tex](1/2)Mv^2 + (1/2)I\omega^2 = Mgy[/tex]

    Are they assuming acceleration due to gravity is downward? Hence, adding Mgy to both sides to produce the above results?

    I guess the problem I'm having is distinguishing when to make the acceleration due to gravity positive/negative (in Work/Energy related problems).
    Last edited: Nov 5, 2005
  2. jcsd
  3. Nov 5, 2005 #2
    It just depends on where you place the positive y-axis, usually up. The gain of kinetic energy happens at the expense of the loss of potential energy, thus as the ball starts rolling (KE increases, PE decreases) the ball loses potential as it rolls down the ramp. In your first equation take g to be negative, and you shouldn't have a problem getting to the second equation.
  4. Nov 5, 2005 #3
    Okay, do you mind if I ask you another question somewhat related to this question, but sticking to the Kinetic Energy/Potential Energy stuff?
  5. Nov 5, 2005 #4
    Could've already asked :tongue:
  6. Nov 5, 2005 #5
    Ha-ha, I apologize. Well, I can only ask if I can post a small illustration. Is that okay?
  7. Nov 5, 2005 #6
    Could've already done that too..
  8. Nov 5, 2005 #7
    Okay. Here is the illustration & problem discription. I have the solution, I just need an understanding as to why the problem was setup the way it was.
    The problem setup is as follows...

    [Total Change in KE + PE of Box on Incline] - [Total Change in KE + PE of Box hanging from rope] = Work from Friction
    (Keeping in mind that Mass=(3kg + 2kg))

    Is this the correct setup? Why do you need to subtract the box hanging from the rope?
  9. Nov 5, 2005 #8
    Also, I noticed the instructor used +9.8m/s/s as the acceleration due to gravity. Is this correct? I'm assuming that he's using positive upward (from illustration), but why doesn't he have the acceleration negative?
  10. Nov 5, 2005 #9
    Here is the startup equation that relates to this particular problem...
    [tex][(1/2)(5kg)v_{f}^2 + (5kg)(9.8m/s^2)(0.8m)] - [(1/2)(5kg)(2m/s)^2 + (5kg)(9.8m/s^2)(-3m)] = (5N)(3m)(-1) [/tex]

    which was derieved from the equation...

    [tex][(1/2)Mv_{f}^2 - (1/2)Mv_{i}^2 + Mg(h_{f} - h_{i})] - [(1/2)Mv_{f}^2 - (1/2)Mv_{i}^2 + Mg(h_{f} - h_{i})] = Fdcos\theta[/tex]
    (keeping in mind your subtracting the Total KE+PE of the box on the incline from the Total KE+PE of the box on the rope)
    Last edited: Nov 5, 2005
  11. Nov 5, 2005 #10
    If you didn't subtract the change in energy of the box hanging from the rope, then where would the energy that raises [itex]m_1[/itex] come from?

    We should stop editting, we keep confusing each other.
    Last edited: Nov 5, 2005
  12. Nov 5, 2005 #11
    I'm seeing the box on the rope as contributing energy to [itex]m_1[/itex], which could be the problem with my understanding. The way I see it is [itex]m_1[/itex] is moving up the hill because [itex]m_2[/itex] is losing potential energy. Why subtract the differences?
    Last edited: Nov 5, 2005
  13. Nov 5, 2005 #12
    I apologize for all the changes. I'm going through the "Latex" documentation to learn how it works =)
  14. Nov 5, 2005 #13
    Solved on AIM.
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