Centripetal acceleration and velocity derivation

AI Thread Summary
Centripetal acceleration is correctly expressed as a = v^2/r, but the derivation presented in the discussion is flawed. The method used calculates average acceleration over a quarter circle rather than instantaneous acceleration, which is essential for proper derivation. The average change in velocity (Δv) was incorrectly identified, leading to an inaccurate formula. For a correct understanding of centripetal acceleration, it is recommended to refer to established physics resources that explain the derivation in detail. Accurate derivation is crucial for grasping the concept of centripetal motion in physics.
3hlang
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in physics, I've been taught that centripetal acceleration is given by v^2/r, but when i try to derive it myself, i get a=2v^2/(pi*r). i said that if the time period is T, after a time of T/4, the velocity would be going in the next direction, i said west to south, and this gives the equation a=4v/T, i then said that T=2*pi*r/v, so a=2v^2/(pi*r). how do you derive the known equation? please help me?
 
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3hlang said:
in physics, I've been taught that centripetal acceleration is given by v^2/r, but when i try to derive it myself, i get a=2v^2/(pi*r). i said that if the time period is T, after a time of T/4, the velocity would be going in the next direction, i said west to south, and this gives the equation a=4v/T, i then said that T=2*pi*r/v, so a=2v^2/(pi*r). how do you derive the known equation? please help me?
Several problems here. First, by taking a quarter of a circle you are only able to calculate the average acceleration during that time interval, aave = Δv/Δt. You want the instantaneous acceleration, not the average. Second, you found the average acceleration for a quarter circle incorrectly; Δv ≠ v.

To learn how the centripetal acceleration formula is derived, read this: http://dev.physicslab.org/Document....e=CircularMotion_CentripetalAcceleration.xml"
 
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