Circular motion (approximation of centripetal acc.)

[V0ODO0CH1LD]

Say I have a body moving in a circle of radius r with a constant velocity v.

The time it takes the body to go around the circumference once is:

T = 2πr/v

Then the time it takes the body to go around a fourth of the circumference is T/4.

Now, imagine a diagram where when the body is at the leftmost portion of the circumference its velocity vector is pointing straight up, and when it is at the upmost portion of the circumference its velocity vector points entirely to the right. That is the body traveling over a fourth of the circumference.

Now, the magnitude of the acceleration vector required to make the body behave that way over a period of T/4 is (v√2)/(T/4). And (v√2)/(T/4) is an approximation of the centripetal acceleration for this case.

My question is: why is that an approximation? Why if I do the same thing over a smaller change in time I get closer to the centripetal acceleration? A circle is symmetric and the velocity is constant, right?

 

[Redbelly98]

Say I have a body moving in a circle of radius r with a constant velocity v.

The time it takes the body to go around the circumference once is:

T = 2πr/v

Then the time it takes the body to go around a fourth of the circumference is T/4.

Now, imagine a diagram where when the body is at the leftmost portion of the circumference its velocity vector is pointing straight up, and when it is at the upmost portion of the circumference its velocity vector points entirely to the right. That is the body traveling over a fourth of the circumference.

Agreed, so far.

Now, the magnitude of the acceleration vector ...

Not quite. This is the magnitude of the average acceleration, where the average acceleration is defined as
$$\vec{a_{avg}}=\frac{\Delta \vec{v}}{\Delta t}$$

... required to make the body behave that way over a period of T/4 is (v√2)/(T/4). And (v√2)/(T/4) is an approximation of the centripetal acceleration for this case.

My question is: why is that an approximation?

Because it is an average, and the actual instantaneous acceleration changes substantially (because it changes direction) during the time interval.

Why if I do the same thing over a smaller change in time I get closer to the centripetal acceleration?

Because, for a small time interval, the acceleration does not change direction very much, and is more closely approximated by its average over that time interval.

A circle is symmetric and the velocity is constant, right?

No, velocity is a vector and is not constant if its direction changes. Speed is constant, however.

 

[V0ODO0CH1LD]

That helped a lot! Thanks!

One for additional question though. How does all of that get summarised into (ω^2)*r or (v^2)/r?

 

[Redbelly98]

You're welcome

This explains things pretty well:

http://scienceblogs.com/dotphysics/2009/07/basics-centripetal-acceleration/

 

[PJC01]

I would like to clarify that the term "centripetal acceleration" refers to the acceleration that acts towards the center of a circular path, keeping an object in circular motion. This acceleration is always present in circular motion, but its magnitude may vary depending on the speed and radius of the circular path.

In the given scenario, the body is moving in a circular path with a constant velocity, which means that the magnitude of its velocity vector remains the same throughout the motion. However, the direction of the velocity vector changes as the body moves around the circle. This change in direction results in a change in the velocity vector, which in turn leads to an acceleration towards the center of the circle (centripetal acceleration).

The formula T = 2πr/v, which gives the time taken for the body to complete one full revolution, is derived from the relationship between the linear velocity (v) and the angular velocity (ω) of a rotating object. This formula assumes that the angular velocity is constant, which is not always the case.

In reality, the angular velocity of an object moving in a circular path may not be constant. This means that the body may not take exactly T/4 time to cover a quarter of the circumference. As the time interval decreases, the approximation of the centripetal acceleration also improves, because it takes into account the change in angular velocity over a shorter time period.

Additionally, the formula (v√2)/(T/4) is an approximation of the centripetal acceleration because it assumes that the velocity vector changes from pointing straight up to pointing entirely to the right in a linear manner. In reality, the change in velocity vector is not linear, and this leads to a slight difference between the actual and the approximated centripetal acceleration.

In conclusion, the approximation of centripetal acceleration in circular motion is based on simplifying assumptions and may not accurately represent the actual acceleration. As the time interval decreases, the approximation improves, but it can never be exact due to the complex nature of circular motion.