Deriving a formula for max. speed of a simple pendulum bob

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SUMMARY

The maximum speed V_max of a simple pendulum bob can be derived using the conservation of energy principle. The formula is established as V_max = √{2gl(1 - cos(Θ)}, where g is the acceleration due to gravity (9.8 m/s²), l is the length of the pendulum, and Θ is the maximum angle of swing. The derivation involves equating gravitational potential energy (E = mgh) at the maximum height with kinetic energy (E = 0.5mv²) at the lowest point of the swing. This relationship confirms the derived formula is correct.

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  • Understanding of basic physics concepts, particularly energy conservation.
  • Familiarity with pendulum motion and angular displacement.
  • Knowledge of trigonometric functions, specifically cosine.
  • Ability to manipulate algebraic equations involving square roots.
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  • Study the principles of energy conservation in mechanical systems.
  • Learn about the dynamics of pendulum motion and its equations of motion.
  • Explore the effects of varying the length of the pendulum on its period and speed.
  • Investigate the role of damping forces in real-world pendulum systems.
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Students studying physics, educators teaching mechanics, and anyone interested in the mathematical modeling of pendulum dynamics.

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Homework Statement



Derive a formula for the maximum speed V_max of a simple pendulum bob in terms of g, the length l and the maximum angle of swing Θ_max. (Hint: Use the fact that the same amount of energy at the top of the arc is the same as at the bottom of the arc.)

The picture depicts a string tied from the floor to the ceiling, and then hanging down from the ceiling most of the way down, which is swinging. It starts at the left hand amplitude. The distance from the left hand amplitude to the floor is labeled h = l-lcosΘ ; the distance to the ceiling from that point is labeled as lcosΘ. The angle is Θ and l is the distance of the string from the ceiling to where it hangs.

Homework Equations



Θ = Θ_max * cos(εt+Φ)
ε = √(k/m) = √(g/l)
g = 9.8 m/s^2
Φ = arctan{-v(0)/[ε*x(0)]}
E = 0.5mv^2
E = mgh

The Attempt at a Solution



I know that the answer, according to the book, is going to be v_max = √{2gl[1-cos(Θ)]} but I'm not sure if I got there correctly. The top of the pendulum would have the same energy as the bottom, so I suppose the E = 0.5mv^2 would mean that velocity is the max velocity. Where the pendulum swings, E = mgh, since it's at the amplitude?

Maybe if I set E = mgh = E = 0.5mv^2 and solve for v_max that will help.
mgh = 0.5mv^2
2gh = v^2
v_max = √(2gh)
v_max = √[2g(l-lcosΘ)]
v_max = √[2gl - 2glcosΘ]
v_max = √[2gl(1-cosΘ)]

^^^Does that all seem correct? Thank you!
 
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Yes, that's right, though your reasoning seemed uncertain (it was more or less correct though). Rather than distributing the 2g and then factoring, why not just factor the l right away?
 
Mindscrape said:
Yes, that's right, though your reasoning seemed uncertain (it was more or less correct though). Rather than distributing the 2g and then factoring, why not just factor the l right away?

Ok, awesome, thank you! And why distribute first...? I suppose it's a habit, I try to distribute everything first and then see what I can factor out to make sure nothing gets "lost."
 

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