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A light rod of length l has two masses m_(1)andm_(2) attached to its two ends. The moment of inertia of the system about and axis perpendicular to the rod and passing through the centre of mass is ……… |
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Answer» `(m_(1)+m_(2))l^(2)`  Position vector of centre of mass `r_(cm)=(m_(1)-r_(1)+m_(2)r_(2))/(m_(1)+m_(2))` `0=(-m_(1)r_(1)+m_(2)r_(2))/(m_(1)+m_(2))` `therefore m_(1)r_(1)=m_(2)r_(2)` `therefore (m_(1))/(m_(2))=(r_(2))/(r_(1))...(1)` Now, `(m_(1)+m_(2))/(m_(2))=(r_(1)+r_(2))/(r_(1))` `therefore (m_(1)+m_(2))/(m_(2))=(l)/(r_(1)) [because r_(1)+r_(2)=l]` `therefore r_(1)=(m_(2)l)/(m_(1)+m_(2)),r_(2)=(m_(1)l)/(m_(1)+m_(2))` `therefore` Moment of inertia about an axis passing through centre of mass and perpendicular to it for this arrangement `I_(cm)=m_(1)r_(1)^(2)+m_(2)r_(2)^(2)` `=m_(1)XX(m_(2)^(2)l^(2))/((m_(1)+m_(2))^(2))+m_(2)xx(m_(2)^(2)l^(2))/((m_(1)+m_(2))^(2))` `=m_(1)m_(2)l^(2)[(m_(2)+m_(1))/((m_(1)+m_(2))^(2))]` `=(m_(1)m_(2)l^(2))/(m_(1)+m_(2))` NOTE : `(m_(1)m_(2))/(m_(1)+m_(2))mu` is known as REDUCE mass `therefore I=mur^(2)` |
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