Simple Harmonic Motion of Inverted Simple Pendulum with a Helium Balloon

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SUMMARY

The discussion centers on the analysis of an inverted simple pendulum system consisting of a helium-filled balloon with a density of 0.175 kg/m³, attached to a string of length 3.35 m. The motion of the balloon, when displaced from equilibrium, is confirmed to be simple harmonic. The period of the motion is derived using the formula T = 2π√(L/g), where g is the acceleration due to gravity, resulting in a calculated acceleration of 62.44 m/s². The participant successfully applies the relevant equations after initial confusion, demonstrating the importance of understanding the dynamics of buoyancy and pendulum motion.

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  • Understanding of simple harmonic motion principles
  • Familiarity with buoyancy concepts and density calculations
  • Knowledge of pendulum dynamics and related equations
  • Basic algebra for manipulating equations and solving for variables
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  • Learn about buoyancy forces and their effects on motion in fluids
  • Explore advanced pendulum dynamics, including damping effects
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to enhance their understanding of simple harmonic motion and buoyancy effects in fluid dynamics.

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Homework Statement


A light balloon filled with helium of density 0.175 kg/m3 is tied to a light string of length L = 3.35 m. The string is tied to the ground, forming an "inverted" simple pendulum (Fig. P13.63a). If the balloon is displaced slightly from equilibrium, as in Figure P13.63b, show that the motion is simple harmonic (do this on paper. Your instructor may ask you to turn in this work), and determine the period of the motion. Take the density of air to be 1.29 kg/m3. (Hint: Use an analogy with the simple pendulum discussed in the text, and see Chapter 9.)



Homework Equations


Ft = [-(density of air - density of He)Vg)/L]s
T= 2pi*sq.root of L/g

The Attempt at a Solution


I tried to play around with the formulas, however, I was unsuccessful. I'm not sure if those formulas are relevant to the question. Any suggestion on how to start solving or any hints?
 

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Hint: If the balloon were released, what would its acceleration be? I assume you are supposed to neglect the resistance of the air to the motion of the balloon. You can't have the balloon floating without the air, but you can neglect the resistance effect.
 
The acceleration would be
a = ((d(air) - d(gas))V - m)g/(m + d(gas)V)
So...
I neglected mass but it says a light balloon...
a = (1.29-0.175)(9.8)V/(0.175V)
a=62.44 m/s

Hmmm... then I use the period formula and...
GOT IT!
thanks OlderDan =D
 

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