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(B) The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its centre and inclined at an angle $	heta$ to th

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(B) The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its centre and inclined at an angle $	heta$ to the rod is given by $I_{centre} = \frac{ML^2}{12} \sin^2 	heta$.From the graph,the maximum value of $I$ is $0.6 \ kg-m^2$ at $	heta = \frac{\pi}{2}$.Substituting these values: $0.6 = \frac{ML^2}{12} \sin^2(\frac{\pi}{2}) = \frac{ML^2}{12} \Rightarrow ML^2 = 7.2 \ kg-m^2$.The moment of inertia of the rod about an axis passing through one of its ends and inclined at an angle $	heta$ to the rod is $I_{end} = \frac{ML^2}{3} \sin^2 	heta$.For $	heta = \frac{\pi}{3}$,we have $I_{end} = \frac{ML^2}{3} \sin^2(\frac{\pi}{3}) = \frac{ML^2}{3} 	imes (\frac{\sqrt{3}}{2})^2 = \frac{ML^2}{3} 	imes \frac{3}{4} = \frac{ML^2}{4}$.Substituting $ML^2 = 7.2$,we get $I_{end} = \frac{7.2}{4} = 1.8 \ kg-m^2$.

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