Centripetal Acceleration formula derivation

In summary, the conversation discusses the calculation of distance traveled and its relation to the change in velocity. The assumption that for a small time interval, the change in velocity is equal to the distance traveled is incorrect. The correct formula can be derived using similar triangles and the ratio between the change in velocity and the velocity.
  • #1
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: 11,637
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
 

1. What is the formula for centripetal acceleration?

The formula for centripetal acceleration is given by a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular motion.

2. How is the centripetal acceleration formula derived?

The centripetal acceleration formula is derived from the equation for acceleration, a = ∆v/∆t, and the definition of centripetal acceleration, a = v^2/r. By combining these equations, we can derive the formula a = v^2/r.

3. What is the significance of the centripetal acceleration formula?

The centripetal acceleration formula is significant because it allows us to calculate the acceleration of an object moving in a circular path. This is important in many scientific fields, such as physics, engineering, and astronomy.

4. Can the centripetal acceleration formula be used for any type of circular motion?

Yes, the centripetal acceleration formula can be used for any type of circular motion, as long as the motion is uniform and the radius of the circle is known.

5. How does the centripetal acceleration formula relate to Newton's laws of motion?

The centripetal acceleration formula is related to Newton's laws of motion in that it is a specific application of Newton's second law, F = ma. In circular motion, the centripetal force is the force that causes the object to move in a circular path, and the centripetal acceleration is the result of this force acting on the object.

Suggested for: Centripetal Acceleration formula derivation

Back
Top