Question 432: Centre of Mass and Rotational Motion

Question

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 ______________.

Accepted Answer

Kinetic energy of rod $$E = {1 \over 2}{{m{l^2}} \over {12}}{\omega ^2}$$
or $$\omega  = \sqrt {{{24E} \over {m{l^2}}}}  = \sqrt {{{24E} \over {d 	imes A 	imes {l^3}}}} $$
$$ \Rightarrow \omega  = \sqrt {{{24E} \over {dA{2^3}}}} $$
$$ = \sqrt {{{3E} \over {Ad}}} $$
So, $$\alpha  = 3$$

Centre of Mass and Rotational Motion

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