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Rotational energy

  1. Mar 13, 2005 #1
    initially, there is a cylinder with a moment of inertia [tex]I_1[/tex] and angular velocity [tex]\omega_i[/tex] A second cylinder that has a moment of inertia of [tex]I_2[/tex] and is not rotatiing drops onto the first cylinder show here. There is friction between the surfaces and the two objects reach the same angular speed of [tex]\omega_f[/tex]

    I need to show that the kinetic energy decreases in this interaction and also calcualte the ratio of the final rotational energy to the initial rotational energy.

    [tex].5I_i \omega _i^2 = .5I_f \omega _f^2[/tex]
    [tex]I_1 \omega _i^2 = (I_1+I_2) \omega _f^2[/tex]
    ratio of initial to final:
    [tex]\frac{\omega _i^2}{\omega _f^2} = \frac{I_1+I_2}{I_1}[/tex]

    how do I show the decrease in rotational energy?
    Last edited: Mar 13, 2005
  2. jcsd
  3. Mar 13, 2005 #2


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    This is wrong.
    You have conservation of angular momentum, not conservation of kinetic energy.
    Last edited: Mar 13, 2005
  4. Mar 13, 2005 #3

    Andrew Mason

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    You can't do it with your equation. Your equation says that energy is conserved.

    This is analagous to an inelastic collision where the two objects collide and remain together after the collision.

    Since there is no external torque on the system, we know that angular momentum is conserved. So:

    [tex]I_1\omega_i = (I_1 + I_2)\omega_f[/tex]

    [tex]\omega_f = \frac{I_1\omega_i}{I_1 + I_2}[/tex]

    The initial energy is:

    [tex]E_i = \frac{1}{2}I_1\omega_i^2[/tex]

    [tex]E_f = \frac{1}{2}(I_1 + I_2)\omega_f^2[/tex]

    [tex]E_f = \frac{1}{2}(I_1 + I_2)\frac{I_1^2\omega_i^2}{(I_1 + I_2)^2}[/tex]

    [tex]E_f = \frac{1}{2}\frac{I_1^2\omega_i^2}{(I_1 + I_2)}[/tex]

    comparing the two energies:

    [tex]\frac{E_f}{E_i} = \frac{\frac{I_1^2\omega_i^2}{(I_1 + I_2)}}{I_1\omega_i^2}[/tex]

    [tex]\frac{E_f}{E_i} = \frac{I_1}{(I_1 + I_2)}[/tex]

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