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A thin rod of...

Question

A thin rod of length $4 l$ , mass $4$ m is bent at the points as shown in the figure. What is the moment of inertia of the rod about the axis passing through $O$ and perpendicular to the plane of the paper?

\frac{m l^{2}}{3}

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\frac{10 m l^{2}}{3}

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\frac{m l^{2}}{12}

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\frac{m l^{2}}{24}

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Solution

The correct option is B $\frac{10 m l^{2}}{3}$
Since, the mass of the rod is $4 m$ and the length is $4 l$

so, mass of $A B = B O = O C = C D = m$

and, length of $A B = B O = O C = C D = l$

We know, Moment of inertia of a rod about its end = $\frac{(m l^{2})}{3}$

Moment of inertia of $A B$ about $B (I_{1})$ = $\frac{m l^{2})}{3}$

Moment of inertia of $B O$ about $O (I_{2})$ = $\frac{(m l^{2})}{3}$

Moment of inertia of $O C$ about $O (I_{3})$ = $\frac{(m l^{2})}{3}$

Moment of inertia of $C D$ about $C (I_{4})$ = $\frac{(m l^{2})}{3}$

Now, from parallel axis theorem,

Moment of inertia of $A B$ about $O (I_{5})$ :

$= I_{1} + m l^{2}$ ————(parallel axis theorem)

$= \frac{(m l^{2})}{3} + m l^{2}$

$= \frac{(4 m l^{2})}{3}$

Similarly, Moment of inertia of $C D$ about $O (I_{6})$

$= \frac{(4 m l^{2})}{3}$

So, moment of inertia of the rod about $O$

= $I_{2} + I_{3} + I_{5} + I_{6}$

= $\frac{(m l^{2})}{3} + \frac{(m l^{2})}{3} + \frac{(4 m l^{2})}{3} + \frac{(4 m l^{2})}{3}$

= $\frac{(10 m l^{2})}{3}$

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A thin rod of length 4l, mass 4m is bent at the points as shown in the figure. What is the moment of inertia of the rod about the axis passing through O and perpendicular to the plane of the paper?

A thin rod of length $4 l$ , mass $4$ m is bent at the points as shown in the figure. What is the moment of inertia of the rod about the axis passing through $O$ and perpendicular to the plane of the paper?