The correct formula for centripetal acceleration is a = v²/r, where v is the tangential velocity and r is the radius of the circular path. The discussion highlights a common misunderstanding in deriving this formula, where the average acceleration during a quarter circle was incorrectly calculated as a = 2v²/(πr). The key distinction is that average acceleration differs from instantaneous acceleration, which is what the centripetal acceleration formula represents. For accurate derivation, one must focus on instantaneous changes in velocity rather than average changes over time intervals.
PREREQUISITESStudents of physics, educators teaching circular motion, and anyone interested in understanding the dynamics of objects in circular paths.
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.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?