Centripetal Acceleration formula derivation

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SUMMARY

The discussion focuses on the derivation of the centripetal acceleration formula, specifically addressing a common misconception regarding the relationship between change in velocity and distance traveled during a small time interval (Δt). Participants clarify that the assumption equating a change in velocity to distance is incorrect. Instead, they emphasize the use of similar triangles, represented by the equation Δv/v = Δx/r, to accurately derive the formula for centripetal acceleration.

PREREQUISITES
  • Understanding of basic physics concepts, particularly motion and acceleration.
  • Familiarity with the concept of centripetal acceleration.
  • Knowledge of similar triangles and their application in physics.
  • Basic algebra skills for manipulating equations.
NEXT STEPS
  • Study the derivation of the centripetal acceleration formula in detail.
  • Learn about the application of similar triangles in physics problems.
  • Explore the relationship between velocity, distance, and time in motion equations.
  • Investigate other types of acceleration and their formulas in circular motion.
USEFUL FOR

Students of physics, educators teaching motion concepts, and anyone interested in understanding the principles of centripetal acceleration and its derivation.

SamitC
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Hello,
This is a very basic question. I am sure I am doing something wrong in the derivation as shown in the picture. But I am not able to find out where I am doing it wrong. It would be very helpful if you can pls. let me know what I am doing wrong here.
Thanks
 

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You calculate the distance traveled in Δt. Why? And then you set that distance equal to the change in velocity? Why again?
 
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Doc Al said:
You calculate the distance traveled in Δt. Why? And then you set that distance equal to the change in velocity? Why again?
Because in Δt, the velocity changes from v0 to v. And for small change in time this can be equal to the distance traveled because of the triangle similarities. That is what the assumption is behind the effort. Can you pls. elaborate where you find something wrong or missing ?
 
Actually,the assumption that for a small time interval, the change in velocity is equal to the distance traveled is wrong.
 
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SamitC said:
Because in Δt, the velocity changes from v0 to v. And for small change in time this can be equal to the distance traveled because of the triangle similarities. That is what the assumption is behind the effort. Can you pls. elaborate where you find something wrong or missing ?
A change in velocity cannot equal a distance. You can use similar triangles: ##\frac{\Delta v}{v} = \frac{\Delta x}{r}##. Use that and you can easily derive the formula.
 
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Doc Al said:
A change in velocity cannot equal a distance. You can use similar triangles: ##\frac{\Delta v}{v} = \frac{\Delta x}{r}##. Use that and you can easily derive the formula.
I got it now. It has to be the ratio...Thanks a lot
 

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