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Two objects of masses 1kg and 2kg separated by a distance of 1.2 m are rotating about their centre of mass. Find the moment of inertia of the system. |
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Answer» Solution :When TWO objects of mass `m_(1)` and `m_(2)` are separated by a distance .d. the distance of mass `m_(1)` from the 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))` Hence `I=m_(1)((m_(2)d)/(m_(1)+m_(2)))^(2)+m_(2)((m_(1)d)/(m_(1)+m_(2)))^(2)` On simplification `I=((m_(1)m_(2))/(m_(1)+m_(2)))d^(2)` Here `(m_(1)m_(2))/(m_(1)+m_(2))` is called reduced mass denoted by `mu`. Hence, for a SYSTEM of two particles rotating about their centre of mass moment of inertia `I=mud^(2)` In this problem, `mu=(m_(1)m_(2))/(m_(1)+m_(2))=(1xx2)/(1+2)=(2)/(3)kg` `I=mud^(2)=(2)/(3)(1.2)^(2)=0.96kgm^(2)` |
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