Question

Figure shows the variation of the moment of inertia of a uniform rod, about an axis passing through its centre and inclined at an angle $\theta$ to the length. The moment of inertia of the rod about an axis passing through one of its ends and making an angle $\theta =\frac{\pi }{3}$ is ( Take the figure to be a part of sine curve)

• A. $0.45kg{m}^2$
• B. $1.8kg{m}^2$
• C. $2.4kg{m}^2$
• D. $1.5kg{m}^2$

Answer 1

The correct option is B $1.8kg{m}^2$

Moment of inertia of rod about an axis passing through centre and making an angle $\theta$ is given by

$I=\frac{M{L}^2}{12}si{n}^2\theta$
Now, from graph maximum $I=0.6$ when $\theta ={90}^{\circ }$

$\Rightarrow 0.6=\frac{M{L}^2}{12}sin\left(\frac{\pi }{2}\right)\Rightarrow M{L}^2=7.2$

Moment of inertia of rod about an axis passing through centre and making an angle $\theta$ is given by
$I=\frac{M{L}^2}{3}si{n}^2\theta$
at $\theta =\frac{\pi }{3},I=\frac{M{L}^2}{3}\left(\frac{3}{4}\right)=\frac{M{L}^2}{4}=\frac{7.2}{4}=1.8kg{m}^2$