Very Simple Uniform Circular Motion Derivation

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SUMMARY

The derivation of the centripetal acceleration formula for uniform circular motion is established as \( a = \frac{4\pi^2R}{T^2} \) or equivalently \( a = 4\pi^2Rf^2 \). This derivation stems from the relationship between linear velocity and radius, where the speed \( v \) of a particle moving in a circle of radius \( R \) over a period \( T \) is given by \( v = \frac{2\pi R}{T} \). Substituting this expression into the centripetal acceleration formula \( a = \frac{v^2}{r} \) leads to the desired results.

PREREQUISITES
  • Understanding of uniform circular motion
  • Familiarity with the concepts of period (T) and frequency (f)
  • Knowledge of basic kinematics and acceleration formulas
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of centripetal acceleration in detail
  • Learn about the relationship between period and frequency in circular motion
  • Explore the implications of circular motion in real-world applications
  • Investigate the differences between linear and angular velocity
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for clear derivations of fundamental concepts in kinematics.

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Homework Statement


This is a very easy derivation (I think), I forgot how to do it though.

How do you derive a = \frac{4\pi^2R}{T^2} or a = 4\pi^2Rf^2?

Homework Equations



The Attempt at a Solution


They both mean the same thing since T is Inverse Frequency but I don't understand how it is derived from a = \frac{v^2}{r}. I know it uses the Sphere formula...

Thanks,
 
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Well, if a particle went around a circle of radius R in time T, what would its speed be?
 
Wow, I feel very stupid now lol.

\frac{(\frac{2\pi r}{T})^2}{r}

Thanks,
 

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