A light balloon filled with helium of density | vedclass

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Question

(D) Let $V$ be the volume of the balloon. The buoyant force is $B = \rho_{air} V g$ and the weight of the balloon is $mg = \rho_{He} V g$.The net upwa

Vedclass · Accepted Answer

$T=2\pi \sqrt{(\frac{\rho_{He}}{\rho_{air}-\rho_{He}})\frac{L}{g}}$

Vedclass · Answer

(D) Let $V$ be the volume of the balloon. The buoyant force is $B = ho_{air} V g$ and the weight of the balloon is $mg = ho_{He} V g$.The net upward force at equilibrium is $F_{net} = B - mg = (ho_{air} - ho_{He}) V g$.When the balloon is displaced by a small angle $	heta$, the restoring torque about the pivot point is $	au = -(B - mg) L \sin 	heta \approx -(B - mg) L 	heta$.The moment of inertia of the balloon about the pivot is $I = mL^2 = (ho_{He} V) L^2$.Using $	au = I \alpha$, we have $I \alpha = -(B - mg) L 	heta$, so $\alpha = -\frac{(B - mg) L}{I} 	heta$.This is the equation of simple harmonic motion $\alpha = -\omega^2 	heta$, where $\omega^2 = \frac{(B - mg) L}{I}$.Substituting the values: $\omega^2 = \frac{(ho_{air} - ho_{He}) V g L}{(ho_{He} V) L^2} = \frac{(ho_{air} - ho_{He}) g}{ho_{He} L}$.The period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{ho_{He} L}{(ho_{air} - ho_{He}) g}} = 2\pi \sqrt{(\frac{ho_{He}}{ho_{air} - ho_{He}})\frac{L}{g}}$.

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