A thin uniform rod of length $2 \mathrm{~m}$, cross sectional area '$A$' and density '$\mathrm{d}$' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$. If value of $\omega$ in terms of its rotational kinetic energy $E$ is $\sqrt{\frac{\alpha E}{A d}}$ then value of $\alpha$ is ______________.
Answer (integer)
3
Solution
<p>Kinetic energy of rod $E = {1 \over 2}{{m{l^2}} \over {12}}{\omega ^2}$</p>
<p>or $$\omega = \sqrt {{{24E} \over {m{l^2}}}} = \sqrt {{{24E} \over {d \times A \times {l^3}}}} $$</p>
<p>$\Rightarrow \omega = \sqrt {{{24E} \over {dA{2^3}}}}$</p>
<p>$= \sqrt {{{3E} \over {Ad}}}$</p>
<p>So, $\alpha = 3$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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