Centripetal Acceleration formula derivation

AI Thread Summary
The discussion centers on the derivation of the centripetal acceleration formula, with a participant expressing confusion over the relationship between distance traveled and change in velocity during a small time interval. The key point raised is that equating a change in velocity to distance is incorrect. Instead, using the concept of similar triangles helps clarify the relationship, specifically the ratio of changes in velocity and distance. The clarification provided leads to a better understanding of the derivation process. Ultimately, the participant acknowledges the correction and expresses gratitude for the assistance.
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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