As shown in the figure,a particle A of mass 2 | vedclass

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(A) The center of mass of the system is at a distance $r_A = \frac{m \cdot L}{2m + m} = \frac{L}{3}$ from particle $A$ and $r_B = \frac{2m \cdot L}{2m

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$qE + qE	heta_0^2$

Vedclass · Answer

(A) The center of mass of the system is at a distance $r_A = \frac{m \cdot L}{2m + m} = \frac{L}{3}$ from particle $A$ and $r_B = \frac{2m \cdot L}{2m + m} = \frac{2L}{3}$ from particle $B$.The moment of inertia about the center of mass is $I = (2m)r_A^2 + m r_B^2 = 2m(\frac{L}{3})^2 + m(\frac{2L}{3})^2 = \frac{2mL^2}{9} + \frac{4mL^2}{9} = \frac{6mL^2}{9} = \frac{2mL^2}{3}$.The restoring torque for a small angle $	heta$ is $	au = -(qE \cdot r_A \sin 	heta + qE \cdot r_B \sin 	heta) = -qE L \sin 	heta \approx -qE L 	heta$.Using $	au = I \alpha$,we get $-qE L 	heta = \frac{2mL^2}{3} \alpha$,so $\alpha = -\frac{3qE}{2mL} 	heta$.This represents simple harmonic motion with $\omega^2 = \frac{3qE}{2mL}$.The angular velocity at $	heta = 0$ is $\omega_{max} = \omega 	heta_0 = 	heta_0 \sqrt{\frac{3qE}{2mL}}$.The tension $T$ in the rod provides the centripetal force for the particles. For particle $B$,$T - qE = m \omega_{max}^2 r_B = m (	heta_0^2 \frac{3qE}{2mL}) (\frac{2L}{3}) = qE 	heta_0^2$.Thus,$T = qE + qE 	heta_0^2$.

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