A rod $\mathrm{PQ}$ of length Lrevolves in a horizontal plane about the axis $Y Y^{\prime}$. The angular…

A rod $\mathrm{PQ}$ of length Lrevolves in a horizontal plane about the axis $Y Y^{\prime}$. The angular velocity of the rod is $\omega$. If $A$ is the area of cross-section of the rod and $\rho$ be its density, its rotational kinetic energy is
  1. $\frac{1}{3} \mathrm{AL}^{3} \rho \omega^{2}$
  2. $\frac{1}{2} \mathrm{AL}^{3} \rho \omega^{2}$
  3. $\frac{1}{24} \mathrm{AL}^{3} \rho \omega^{2}$
  4. $\frac{1}{18} \mathrm{AL}^{3} \rho \omega^{2}$

Solution

If rotation axis is passing through its middle point $\&$ is $\perp$ to its plane, then moment of inertia about $Y Y^{\prime}$ is


$I=\frac{M L^{2}}{12}$ where $M=$ volume $\times$ density $=(L \times A) \times \rho$
so $I=\frac{L^{3} A \rho}{12}$
so rotational $\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{\mathrm{L}^{3} \mathrm{~A} \rho \omega^{2}}{24}$

Asked in: JEE Mains - Rotational Motion - Test 3

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