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Railways NTPC (Technical Ability) Theory of Machines and Machine Design Question Bank Theory of Machines

  • question_answer
    Two co-axial rotors having moments of inertia \[{{I}_{1}},{{I}_{2}}\] and angular speeds \[{{\omega }_{1}}\] and \[{{\omega }_{2}}\] respectively are engaged together. The loss of energy during engagement is equal to:

    A) \[\frac{{{I}_{1}}{{I}_{2}}{{({{\omega }_{1}}-{{\omega }_{2}})}^{2}}}{2({{I}_{1}}+{{I}_{2}})}\]  

    B) \[\frac{{{I}_{1}}{{I}_{2}}{{(\omega _{1}^{2}-\omega _{2}^{2})}^{2}}}{2({{I}_{1}}+{{I}_{2}})}\]

    C) \[\frac{2{{I}_{1}}{{I}_{2}}{{({{\omega }_{1}}-{{\omega }_{2}})}^{2}}}{({{I}_{1}}+{{I}_{2}})}\]  

    D) \[\frac{{{I}_{1}}\omega _{1}^{2}-{{I}_{2}}\omega _{2}^{2}}{({{I}_{1}}+{{I}_{2}})}\]

    Correct Answer: A

    Solution :

    \[\Delta KE=\frac{1}{2}({{I}_{1}}\omega _{1}^{2}-{{I}_{2}}\omega _{2}^{2})-\frac{1}{2}({{I}_{1}}\omega '_{1}^{2}-{{I}_{2}}\omega '_{2}^{2})\] For momentum balance, \[{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}={{I}_{1}}\omega {{'}_{1}}+{{I}_{2}}\omega {{'}_{2}}\] \[=\omega {{'}_{1}}=\omega {{'}_{2}}=\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{{{I}_{1}}+{{I}_{2}}}\] \[\Delta KE=\frac{{{I}_{1}}{{I}_{2}}}{2({{I}_{1}}+{{I}_{2}})}{{({{\omega }_{1}}-{{\omega }_{2}})}^{2}}\]


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