The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If ${{{I_1}} \over {{I_2}}}$ is ${{x{\pi ^2}} \over 3}$, then the value of x will be ____________.
Answer (integer)
8
Solution
<p>${I_1} = {{M{L^2}} \over 3}$ ..... (1)</p>
<p>For ring : ${I_2} = {{M{R^2}} \over 2}$</p>
<p>and $2\pi R = L$</p>
<p>$\Rightarrow {I_2} = {M \over 2}\left( {{{{L^2}} \over {4{\pi ^2}}}} \right)$ ...... (2)</p>
<p>$\Rightarrow {{{I_1}} \over {{I_2}}} = {{8{\pi ^2}} \over 3}$</p>
<p>$\Rightarrow x = 8$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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