A light balloon filled with helium of density | vedclass

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(C) When the balloon is slightly displaced horizontally,the horizontal component of the buoyant force produces a torque about the end of the string at

Vedclass · Accepted Answer

the balloon undergoes simple harmonic motion with period $2 \pi \sqrt{\left(\frac{\rho_{He}}{\rho_{air}-\rho_{He}}\right) \frac{l}{g}}$

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(C) When the balloon is slightly displaced horizontally,the horizontal component of the buoyant force produces a torque about the end of the string attached to the ground. This torque produces sideways oscillations of the balloon.The restoring torque is given by:$	au_1 = F_b \sin 	heta 	imes l = V(ho_{air} - ho_{He})g l \sin 	heta$For small angular displacements,$\sin 	heta \approx 	heta$,so:$	au_1 = V(ho_{air} - ho_{He})g l 	heta$The inertial torque on the balloon is:$	au_2 = I \alpha = (m l^2) \alpha = (V ho_{He}) l^2 \alpha$Equating the torques (considering the restoring nature):$V ho_{He} l^2 \alpha = -V(ho_{air} - ho_{He}) g l 	heta$$\alpha = -\left(\frac{ho_{air} - ho_{He}}{ho_{He}}ight) \frac{g}{l} 	heta$Comparing with the standard $SHM$ equation $\alpha = -\omega^2 	heta$,we get:$\omega^2 = \left(\frac{ho_{air} - ho_{He}}{ho_{He}}ight) \frac{g}{l}$The time period $T$ is:$T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\left(\frac{ho_{He}}{ho_{air} - ho_{He}}ight) \frac{l}{g}}$

$A$ light balloon filled with helium of density $\rho_{He}$ is tied to a long light string of length $l$ and the string is attached to the ground. If the balloon is displaced slightly in the horizontal direction from the equilibrium and released,then:

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