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Two objects of masses 1 kg and 2 kg separated by a distance of 1.2 m are rotating about their centre of mass. Find the moment of inertia of the system.

Answer» <html><body><p></p>Solution :When two objects of <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> m, and m, are <a href="https://interviewquestions.tuteehub.com/tag/separated-7272382" style="font-weight:bold;" target="_blank" title="Click to know more about SEPARATED">SEPARATED</a> by a distance .d., the distance of mass `m_(1)` from the <br/> centre of mass `=(m_(2)d)/(m_(1)+m_(2))` and the distance of mass `m_(2)` from the centre of mass `=(m_(1)d)/(m_(1)+m_(2))` <br/> Hence `I=m_(1)((m_(2)d)/(m_(1)+m_(2)))+m_(2)((m_(1)d)/(m_(1)+m_(2)))^(2)` <br/> On simplification `I=((m_(1)m_(2))/(m_(1)+m_(2)))d^(2)` <br/> Here `(m_(1)m_(2))/(m_(1)+m_(2))` is called reduced mass denoted by `mu`. Hence, for a system of two <a href="https://interviewquestions.tuteehub.com/tag/particles-1147533" style="font-weight:bold;" target="_blank" title="Click to know more about PARTICLES">PARTICLES</a> rotating about their centre of mass, moment of inertia `I=mud^(2)` <br/> In this <a href="https://interviewquestions.tuteehub.com/tag/problem-25530" style="font-weight:bold;" target="_blank" title="Click to know more about PROBLEM">PROBLEM</a>, `mu=(m_(1)m_(2))/(m_(1)+m_(2))=(1xx2)/(1+2)=(2)/(3)kg` <br/> `I=mud^(2)=(2)/(3)(1.2)^(2)=0.96kgm^(2)`</body></html>


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