Centripetal Acceleration formula derivation

  • #1
SamitC
36
0
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
 

Attachments

  • Centrepetal.jpg
    Centrepetal.jpg
    23.9 KB · Views: 11,677
Physics news on Phys.org
  • #2
You calculate the distance traveled in Δt. Why? And then you set that distance equal to the change in velocity? Why again?
 
  • Like
Likes SamitC
  • #3
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 ?
 
  • #4
Actually,the assumption that for a small time interval, the change in velocity is equal to the distance traveled is wrong.
 
  • Like
Likes SamitC
  • #5
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.
 
  • Like
Likes SamitC
  • #6
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
 
Back
Top