Does mass have an effect on Centripetal Acceleration?

It depends on what you are looking at.

If you are applying a constant centripetal force to objects of different masses, then they will each experience a different centripetal acceleration.

If a bunch of different masses are under going circular motion around a circle of radius $r$ with speed $v$, then they will all be experiencing the same centripetal acceleration (but different centripetal forces).

So "does mass effect centripetal acceleration?" is not specific enough to have an answer. You need to add in what else you are considering, i.e. what you are holding constant and what you are allowing to change as you change the mass.

If you change the mass, then either the acceleration or the centripetal force must change as well (or both). That is what we see from Newton's 2nd law.

The simulator you are using seems to adjust the force. That's why you see no acceleration change. That is just a choice by the software developers or possibly something you can adjust for in the simulation settings. Whether that is a reasonable approach depends on the physical scenario.

• In scenarios such as celestial orbits, it is reasonable since the gravitational force increases proportionally with mass.

• In scenarios such as a game of tetherball, it might not be reasonable since the string force might be limited or constant regardless of mass.

Note that the expression $a=v^2/r$ isn't needed for this analysis.

a =f/m

but if f is proportional to "m" like in gravitation, centripetal force then the m's cancel and you're left with constant acceleration

The first formula "hides" the mass, because it assumes a fixed $v$, which implies that for whatever $m$ you're working with, you're already supplying the required $f$ to bring its velocity up to $v$.

So when the mass changes, it is assumed that the used force also changes, in a way that $v$ is kept the same. And when $v$ is the same (and $r$, which we never changed), then $a=v^2/r$ will always yield the same result.

So, if you are working in a system where the force supplied is fixed, then changing the mass will change the acceleration. This is no different from how a little outboard motor is able to move a small boat better than an ocean liner. Same force, different mass, different acceleration.

However, if you are working in a system where the velocity $v$ is fixed, then changing the mass doesn't matter, because you're already accounting for any changes in mass by also changing the force required to move said mass.

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Maybe it helps to use a different example.

A BMI of over 30 is unhealthy.

Ignoring that BMI is an outdated measure of bodily health, let's take this as a fact.

Also, your BMI is calculated based on your height and weight. So, here's the analogous question as you asked:

Does a change in height have an effect on what BMI values are considered healthy/unhealthy?

And the answer is no. No matter what your height/weight is, whether you're 5ft or 6ft tall, doesn't matter. As long as the ratio between your height/weight is the same (= your BMI), then the healthy/unhealthy label is also the same.

However, now consider this question:

Does a change in my weight have an effect on me being healthy/unhealthy?

And now the answer is yes, because we implicitly assume that when you gain/lose weight, you don't also gain/lose height. Therefore, the ratio between them changes, which means your BMI changes, which means that your bodily health status (healthy/unhealthy) might also change.

In many cases, a change in mass means that acceleration becomes more or less. If the mass of a car is changed, its acceleration will be less. If a ball becomes heavier it will have less speed (and thus acceleration) when thrown. If the mass of an elementary particle is smaller it's easier accelerated (that is, the acceleration increases). A small kid is easier lifted from the ground. A change in mass is normally associated with a change in acceleration.
As far as the force of gravity is concerned, all masses fall down with the same acceleration. There is no one who thinks that if a mass increases it will be harder for the mass to accelerate downwards.
The upward direction though is different. A rocket will accelerate less if it's heavier (compare this to a fire arrow). The reason is that in many cases the force we can apply is limited and the mass varies over a wide range.
So it is expected that if the mass in a circular motion becomes bigger the acceleration becomes less. It is expected that the mass will move slower.
Obviously, this will not happen because if we increase the mass then to move in the same circle with the same velocity, more force is needed. But it feels like that if we increase the mass the object will move slower. If we envision an increasing mass we associate it with smaller velocity. How in the world can the object stay moving with the same velocity if we increase its mass? Doesn't this contradict conservation of energy? Obviously not if we give it the same velocity as a smaller mass object by applying more force for linear acceleration.
That's the point. Circular acceleration doesn't give an object an increase in speed, while linear acceleration does. Circular acceleration only changes the direction of the object's velocity. If we don't apply a force to change the object's linear speed it has to move with a smaller velocity on the circle. This means that the centripetal force will stay the same. if the object maintains the same speed the centripetal acceleration will be the same but a bigger force was needed to give it linear speed.

Here's a more math-y restatement of the same general answer :)

Just from your question, we see that $v^2/r = a = f/m$. If the simulation keeps $v^2/r$ constant ($a_1$ = $a_2$), then when $m$ changes, $f$ should change. Depending how sophisticated the simulation is, it may or may not report the value of $f$.

If you solve for $m$, the result is something like $m=frac{fr}{v^2}$. From this you can see that when $m$ changes, at least one of the other terms will change. Maybe they all do; maybe it's just $f$.