Centripetal Acceleration formula derivation

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Discussion Overview

The discussion revolves around the derivation of the centripetal acceleration formula, focusing on the assumptions and calculations involved in relating distance traveled and changes in velocity over a small time interval. Participants are examining the validity of certain steps in the derivation process.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about their derivation and seeks help in identifying errors.
  • Another participant questions the reasoning behind equating distance traveled in a time interval to the change in velocity.
  • A participant argues that the assumption that a change in velocity can equal distance traveled is incorrect.
  • Some participants suggest using similar triangles to relate changes in velocity and distance, proposing a ratio to derive the formula.
  • A later reply indicates that one participant has understood the need for a ratio in the derivation process.

Areas of Agreement / Disagreement

There is disagreement regarding the validity of equating change in velocity to distance traveled. Some participants challenge this assumption while others attempt to clarify the reasoning behind it. The discussion remains unresolved with multiple competing views on the derivation process.

Contextual Notes

Participants have not reached a consensus on the assumptions made during the derivation, particularly regarding the relationship between change in velocity and distance traveled. The discussion highlights the need for careful consideration of these assumptions in the derivation process.

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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