A thin rod of length L is placed vertically on a frictionless horizontal floor and released with a n

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6.System of Particles and Rotational Motion

hard

$A$ thin rod of length $L$ is placed vertically on a frictionless horizontal floor and released with a negligible push to allow it to fall. At any moment, the rod makes an angle $\theta$ with the vertical. If the center of mass has acceleration $= A$, and the rod an angular acceleration $= \alpha$ at initial moment, then

A

$A= (L\alpha ).sin\theta$

B

$A/2 = (L\alpha ).sin\theta$

C

$2A = (L\alpha ).sin\theta$

D

$A = L\alpha$

Solution

$\therefore a_{c m}=A=\frac{L}{2} \alpha \sin \theta$

$\therefore \mathrm{A}=\frac{\mathrm{L}}{2} \alpha \sin \theta$

$\therefore 2 \mathrm{A}=\mathrm{L} \alpha \sin \theta$

Standard 11

Physics

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