6. system of particles and rotational motion — Physics STD 11 Science — Question

let instantaneous axis of rotation is at distance $l$ from end $A .$

$\omega=\frac{v \sin \theta}{l}=\frac{v^{\prime} \cos \theta}{(L-l)}$$...(2)$

from $(1) \&(2) \omega=\frac{v \sin \theta}{l}=\frac{v \cos ^{2} \theta}{\sin \theta(L-l)}$

$l=L \sin ^{2} \theta \ldots(3)$

from $( 2)$$\&(3)$

$\omega=\frac{v \cos e c \theta}{L}$

$\frac{d \omega}{d t}=\frac{v}{L}(-\cos e c \theta . \cot \theta) \times \frac{d \theta}{d t}$

$\frac{d \omega}{d t}=\frac{v}{L}(-\cos e c \theta . \cot \theta) \times \frac{v}{L} \cos e c \theta$

$\alpha=-\frac{v^{2}}{L^{2}} \cos e c^{2} \theta . \cot \theta$

$\alpha=-\frac{v^{2}}{L^{2}} \cos e c^{2} 45^{\circ} \cdot \cot 45^{\circ} \Rightarrow \alpha=-\frac{2 v^{2}}{L^{2}}$