Pendulum Conservation of Energy

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SUMMARY

The discussion centers on the conservation of energy in a pendulum system, specifically addressing the relationship between potential energy and kinetic energy. Participants clarify that at maximum displacement, the pendulum has maximum potential energy (mgh), while at the lowest point, the total energy is expressed as the sum of kinetic energy (1/2mv²) and rotational kinetic energy (1/2Iω²). The confusion arises from the misapplication of energy equations, leading to incorrect calculations of velocity. Ultimately, the correct relationship is established as mgΔh = 1/2mv², confirming the conservation of energy principle.

PREREQUISITES
  • Understanding of basic physics concepts such as potential energy and kinetic energy
  • Familiarity with pendulum motion and its parameters (length L, angle θ)
  • Knowledge of rotational dynamics, including moment of inertia (I) and angular velocity (ω)
  • Ability to perform algebraic manipulations and solve equations
NEXT STEPS
  • Study the derivation of the conservation of energy principle in oscillatory systems
  • Learn about the moment of inertia for different shapes and its application in rotational motion
  • Explore the dynamics of pendulum motion using differential equations
  • Investigate the effects of damping and external forces on pendulum motion
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of energy conservation in oscillatory systems.

  • #31
PeroK said:
You need to post a problem. Show your calculations and we can see where you are going wrong. You seem to understand everything so I don't know what it could be!
Hi, so I tried again. I think I'm using the calculator wrong.
When I use √2gΔh = v that still doesn't work out.
But if I use a correct value for v, square it and divide it by 2 I get gΔh. So mgΔh is 1/2mv2 after all. I'm so relieved. Sorry for the trouble. And thank you all for helping me!
 
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  • #32
spsch said:
Hi, so I tried again. I think I'm using the calculator wrong.
When I use √2gΔh = v that still doesn't work out.
But if I use a correct value for v, square it and divide it by 2 I get gΔh. So mgΔh is 1/2mv2 after all. I'm so relieved. Sorry for the trouble. And thank you all for helping me!
Sounds like you need to enter it as √(2gΔh).
 

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