(A) The rod is pivoted at one end. The moment of inertia of the rod about the pivot is $I = \frac{mL^2}{3}$.The gravitational force $mg$ acts at the c

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(A) The rod is pivoted at one end. The moment of inertia of the rod about the pivot is $I = \frac{mL^2}{3}$.The gravitational force $mg$ acts at the center of mass of the rod,which is at a distance $L/2$ from the pivot.The torque $\tau$ about the pivot due to gravity is $\tau = mg \cdot (L/2) \sin \theta$.Using the rotational analog of Newton's second law,$\tau = I \alpha$,where $\alpha$ is the angular acceleration:$mg \frac{L}{2} \sin \theta = \left( \frac{mL^2}{3} \right) \alpha$Solving for $\alpha$:$\alpha = \frac{mg(L/2) \sin \theta}{mL^2/3} = \frac{mgL/2}{mL^2/3} \sin \theta = \frac{3g}{2L} \sin \theta$.Thus,the angular acceleration is $\frac{3g \sin \theta}{2L}$.

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

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$A$ thin uniform rod of mass $m$ and length $L$ is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is [consider negligible friction at the pivot] ($g=$ acceleration due to gravity).

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A
$\frac{3g \sin \theta}{2L}$
B
$\frac{3g \cos \theta}{2L}$
C
$\frac{2g \sin \theta}{3L}$
D
$\frac{2g \cos \theta}{3L}$

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

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