Rotational Motion Question 11
Question 11 - 30 January - Shift 1
A thin uniform rod of length $2 m$. cross sectional area ’ $A$ ’ and density ’ $d$ ’ 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 then the value of is
$\sqrt{\frac{\alpha E}{Ad}}$
Show Answer
Answer: (3)
Solution:
Formula: Rotational Kinetic Energy
$ \begin{aligned} & (KE) _{\text{Rotational }}=\frac{1}{2} I \omega^{2}=E \\ & E=\frac{1}{2} \frac{m \ell^{2}}{12} \omega^{2} \\ & E=\frac{1}{2} \frac{dA \ell^{3}}{12} \omega^{2} \\ & E=\frac{dA(2)^{3}}{24} \omega^{2} \\ & \sqrt{\frac{3 E}{dA}}=\omega \\ & \alpha=3 \text{ Ans. } \end{aligned} $
