Rotational Motion

JEE Main level, numerical answer, medium

156 of 280

Question 156 of 280

A thin uniform rod of length 2m2m, cross sectional area AA and density dd is rotated about an axis passing through the centre and perpendicular to its length with angular velocity ω\omega. If value of ω\omega in terms of its rotational kinetic energy EE is αEAd\sqrt{\dfrac{\alpha E}{Ad}}, then the value of α\alpha is ______.

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Answer: 3

Rotational kinetic energy is given by E=12Iω2E=\dfrac{1}{2}I{\omega }^{2}, here, II is moment of inertia of rod about an axis through its centre and perpendicular to it.
So, E=12ml212ω2E=\dfrac{1}{2}\dfrac{m{l}^{2}}{12}{\omega }^{2}
⇒E=12dAl312ω2(∵mass=density×volume)\Rightarrow E=\dfrac{1}{2}\dfrac{dA{l}^{3}}{12}{\omega }^{2}\left(∵mass=density\times volume\right)
⇒E=dA(2)324ω2\Rightarrow E=\dfrac{dA(2{)}^{3}}{24}{\omega }^{2}
⇒ω=3EdA\Rightarrow \omega =\sqrt{\dfrac{3E}{dA}}
Thus, the value of α=3\alpha =3.

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