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Math Help - flywheel mass and energy

  1. #1
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    flywheel mass and energy

    I do not even know where to start here, any help is much appreciated

    A flywheel (a) with a mass of 100 kg and radius of gyration 1200mm rotates at 150 revs min^-1 (clockwise). The kinetic energy of this flywheel is to be reduced by 20% by impacting it with a second flywheel (b) rotating at 80 revs min^-1 in the opposite direction,such that they have the same (clockwise) angular velocity after impact.

    A) Calculate the required mass of the flywheel (b) if its radius of gyration is 800mm

    B) Calculate the energy lost to the surroundings.

    C) Is the impact elastic? why


    Thanks for looking
    Nic
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  2. #2
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    Quote Originally Posted by ethascar View Post
    I do not even know where to start here, any help is much appreciated

    A flywheel (a) with a mass of 100 kg and radius of gyration 1200mm rotates at 150 revs min^-1 (clockwise). The kinetic energy of this flywheel is to be reduced by 20% by impacting it with a second flywheel (b) rotating at 80 revs min^-1 in the opposite direction,such that they have the same (clockwise) angular velocity after impact.

    A) Calculate the required mass of the flywheel (b) if its radius of gyration is 800mm

    B) Calculate the energy lost to the surroundings.

    C) Is the impact elastic? why


    Thanks for looking
    Nic
    moment of inertia, I = Mk^2 , where k is the radius of gyration.

    the interaction between the two flywheels is analogous to a perfectly inelastic collision in the linear sense ... angular momentum is conserved.

    I_0 \omega_0 = I_f \omega_f

    M_a k_a^2 \omega_{a0} + M_bk_b^2\omega_{b0} = (M_a k_a^2 + M_b k_b^2) \omega_f

    the second equation involves the information given about the loss of rotational kinetic energy ...

    .8(KE_0) = KE_f

    .8\left[\frac{1}{2}(M_ak_a^2)\omega_{a0}^2 + \frac{1}{2}(M_bk_b^2)\omega_{b0}^2\right] = \frac{1}{2}(M_a k_a^2 + M_b k_b^2)\omega_f^2

    using the two equations, you should be able to answer the questions.
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  3. #3
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    Thanks again skeeter for replying to me on this.
    Still can't get my head around it though feel like i am doing this at the moment and just going round and round in circles.
    Cheers
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  4. #4
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    *****The following solution assumes the flywheels are NOT co-axial, and the impact occurs at the outer edges of both flywheels like two gears.*****
    *****If the flywheels are co-axial, then angular momentum conservation would hold (The words "opposite direction" and "same angular velocity indicate this would be the case here and skeeter's solution would be it). That would make the problem pretty simple.

    Notations:
     p= percentage loss of kinetic energy of flywheel (a)
     m_a = mass of flywheel (a)
     m_b = mass of flywheel (b)
     r_a = radius of gyration of flywheel (a)
     r_b = radius of gyration of flywheel (b)
     \omega_a = initial angular velocity of flywheel (a)
     \omega_b = initial angular velocity of flywheel (b)
     \omega_f = final angular velocity of both flywheels
     R_a = outer radius of flywheel (a)
     R_b = outer radius of flywheel (b)
     \bar{f} = average force of impact
     \Delta t = duration of impact

    Assumptions:
    (1) Both flywheels are have the same shape such
     R_a=kr_a, R_b=kr_b
    (2) Impact occurs at the outer edge of both flywheels (the flywheels are NOT co-axial and contact at the edge).

    Solution:
    (1) Required mass of flywheel (b)
    Final angular velocity of both flywheels is easy to obtain,
    \omega_f=\sqrt{1-p}\,\omega_a (1)

    For flywheel (a),
     m_ar_a^2(\omega_a-\omega_f)=\bar{f}R_a\Delta t (2)
    For flywheel (b),
     m_br_b^2(\omega_b-\omega_f)=-\bar{f}R_b\Delta t (3)

    (2) divided by (3) yields
     \frac{m_ar_a^2(\omega_a-\omega_f)}{m_br_b^2(\omega_b-\omega_f)}=-\frac{R_a}{R_b}=-\frac{r_a}{r_b}

    m_b=\frac{r_a}{r_b}\frac{\omega_a-\omega_f}{\omega_f-\omega_b} m_a (4)

    (2) Energy Loss

    Initial kinetic energy
    E_0=\frac{1}{2}m_ar_a^2\omega_a^2+\frac{1}{2}m_br_ b^2\omega_b^2 (5)
    Final kinetic energy
    E_f=\frac{1}{2}m_ar_a^2\omega_f^2+\frac{1}{2}m_br_ b^2\omega_f^2 (6)
    Energy loss
    \Delta E=E_0-E_f (7)

    (3) Elastic?
    Not elastic because there is energy loss.

    Luobo
    Last edited by mr fantastic; September 19th 2009 at 01:57 AM. Reason: Restored original reply
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