Centripetal acceleration and velocity derivation

[3hlang]

in physics, I've been taught that centripetal acceleration is given by v^2/r, but when i try to derive it myself, i get a=2v^2/(pi*r). i said that if the time period is T, after a time of T/4, the velocity would be going in the next direction, i said west to south, and this gives the equation a=4v/T, i then said that T=2*pi*r/v, so a=2v^2/(pi*r). how do you derive the known equation? please help me?

 

[Doc Al]

in physics, I've been taught that centripetal acceleration is given by v^2/r, but when i try to derive it myself, i get a=2v^2/(pi*r). i said that if the time period is T, after a time of T/4, the velocity would be going in the next direction, i said west to south, and this gives the equation a=4v/T, i then said that T=2*pi*r/v, so a=2v^2/(pi*r). how do you derive the known equation? please help me?

Several problems here. First, by taking a quarter of a circle you are only able to calculate the average acceleration during that time interval, a_ave = Δv/Δt. You want the instantaneous acceleration, not the average. Second, you found the average acceleration for a quarter circle incorrectly; Δv ≠ v.

To learn how the centripetal acceleration formula is derived, read this: http://dev.physicslab.org/Document....e=CircularMotion_CentripetalAcceleration.xml"

 

[astrokreng]

Hello,

Thank you for sharing your thoughts and efforts in deriving the equation for centripetal acceleration. It is always commendable to see students taking an active role in understanding and questioning the concepts they are taught.

The equation for centripetal acceleration, a=v^2/r, is a fundamental equation in physics and has been derived and proven through various methods. The most common method is by using Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass times its acceleration (F=ma).

In the case of an object moving in a circular path, there is a constant force acting towards the center of the circle, called the centripetal force. This force is responsible for keeping the object in its circular motion. Therefore, we can write the equation F=ma as:

Fc = ma

Where Fc is the centripetal force and a is the centripetal acceleration.

Now, we also know that the centripetal force is equal to the product of the mass of the object, its velocity squared, and the radius of the circular path (Fc = mv^2/r). Substituting this into F=ma, we get:

mv^2/r = ma

Solving for a, we get:

a = v^2/r

This is the same equation that you have derived, so your method is correct. However, there are a few things to note:

1. In your derivation, you have assumed that the time period T is equal to the time taken for the object to travel a quarter of the circle (T/4). This is not always the case. The time period is the time taken for the object to complete one full revolution, which is equal to the circumference of the circle divided by the velocity (T=2πr/v). This is why your final equation has a factor of 2 in the denominator.

2. Your method assumes that the velocity changes direction after a quarter of the circle, which is not always the case. In fact, the velocity of the object is constantly changing direction as it moves along the circular path, but its magnitude remains constant.

I hope this explanation helps you understand how the equation for centripetal acceleration is derived. Keep up the good work in questioning and understanding the concepts of physics!

Best regards,
Scientist