I am using an online simulation for a lab concerning Centripetal Acceleration. When I change the mass the graph indicates that the magnitude of the acceleration is constant. According to the Centripetal Acceleration formula: $a=v^2/r$, this is true because no mass is present in the relationship. However, when I use Newton's Second Law of Motion, $a=f/m$, I can see that the mass and the acceleration are inversely proportional. Both of these ideas are found when I look them up online, now I am a bit confused on which one might be more valid.
[Physics] Does mass have an effect on Centripetal Acceleration
accelerationcentripetal-forcenewtonian-mechanics
Related Solutions
The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and always aligned with the X and Y axes respectively, while the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ rotate with an angular velocity of $\omega=\|\dot{\theta}\|$ and point in the directions of increasing radius and angle (respectively). The included graphic below shows the two basis vector pairs overlaid on top of one another.
The position vector of the object is obviously defined as:
$\vec{p}(t)=x\hat{i}+y\hat{j}=rcos(\theta)\hat{i}+rsin(\theta)\hat{j}$,
with
$\|\vec{p}(t)\|=\sqrt{(rcos{\theta})^2+(rsin{\theta})^2}=\sqrt{r^2(sin^2(\theta)+cos^2(\theta))}=r\sqrt{(1)}=r$
Less obviously, it can be shown that the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ can be expressed solely in terms of the Cartesian unit vectors $\hat{i}$ and $\hat{j}$ and the angular position $\theta$ as,
$\boxed{\hat{e}_r=cos(\theta)\hat{i}+sin(\theta)\hat{j}}$ and $\boxed{\hat{e}_\theta=-sin(\theta)\hat{i}+cos(\theta)\hat{j}}$.
These two equations are extremely important, as they will be the key to expressing the Cartesian acceleration in polar coordinates, of which one of the terms will be our desired $v^2/r=\omega^2r$ centripetal acceleration. Moving forward, the vector acceleration of the object in Cartesian coordinates is simply
$\vec{a}(t)=\frac{d^2}{dt^2}\left[\vec{p}(t)\right]=\ddot{x}\hat{i}+\ddot{y}\hat{j}$.
Starting with $x=rcos(\theta)$ and $y=rsin(\theta)$ and differentiating once, we have
$\boxed{\dot{x}=\dot{r}cos(\theta)-r\dot{\theta}sin(\theta)}$ and $\boxed{\dot{y}=\dot{r}sin(\theta)+r\dot{\theta}cos(\theta)}$.
Differentiating again, we will have
$\ddot{x}=\ddot{r}cos(\theta)-\dot{r}\dot{\theta}sin(\theta)-\dot{r}\dot{\theta}sin(\theta)-r\frac{d}{dt}\left[\dot{\theta}sin(\theta)\right]$
$=\ddot{r}cos(\theta)-2\dot{r}\dot{\theta}sin(\theta)-r\left[\ddot{\theta}sin(\theta)+{\dot{\theta}}^2cos(\theta)\right]$, such that
$\boxed{\ddot{x}=(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))}$.
Similarly, the y acceleration $\ddot{y}$ becomes
$\ddot{y}=\ddot{r}sin(\theta)+\dot{r}\dot{\theta}cos(\theta)+\dot{r}\dot{\theta}cos(\theta)+r\frac{d}{dt}\left[\dot{\theta}cos(\theta)\right]$
$=\ddot{r}sin(\theta)+2\dot{r}\dot{\theta}cos(\theta)+r\left[\ddot{\theta}cos(\theta)-{\dot{\theta}}^2sin(\theta)\right]$, such that
$\boxed{\ddot{y}=(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})cos(\theta)}$.
Now, we must plug these scalar derivatives into our formulation for the vector acceleration. In Cartesian coordinates, this is
$\vec{a}(t)=\ddot{x}\hat{i}+\ddot{y}\hat{j}=\{(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))\}\hat{i}+\{(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(cos(\theta))\}\hat{j}$
which can be rearranged into the following form:
$\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\{cos(\theta)\hat{i}+sin(\theta)\hat{j}\}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\{-sin(\theta)\hat{i}+cos(\theta)\hat{j}\}$
But as we have already seen, this is simply equal to
$\boxed{\boxed{\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\hat{e}_r+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{e}_\theta}}$
As we can now appreciate from carrying out the full derivation, there are actually two components each to both the radial and tangential accelerations. The $\ddot{r}$ term is straightforwardly equal to the second derivative of the position vector magnitude. The second term, $r\dot{\theta}^2$, is our long sought-after centripetal acceleration $r\dot{\theta}^2=\omega^2r=v^2/r$, and (as expected) it points in the negative radial direction. The tangential terms are perhaps a bit less intuitive. The $r\ddot{\theta}$ term is the acceleration that occurs whenever the radius and angular acceleration $\ddot{\theta}$ are both non-zero (imagine the tangential acceleration of a turbine blade of a jet engine as the engine spools up). The final term $2\dot{r}\dot{\theta}$ is what's commonly known as the Coriolis acceleration, and it occurs whenever the radius and angle change simultaneously. It arises because, for a given angular velocity, the arc length travelled every second increases with radius (tangential velocity increases with radius). Thus, an object with a given angular velocity will have different tangential velocities at different local radii of rotation. If the radius changes with time ($\dot{r}\not=0$) and the angular velocity $\dot{\theta}$ is not equal to zero, then the tangential velocity will change with time, which is by definition a tangential acceleration.
This is a common misinterpretation of Newton's third law, often stated as "to every action, there's an equal and opposite reaction." As you surmise, "action" and "reaction" refer to forces. However, they refer to forces acting on different things. Otherwise, nothing could accelerate, ever: if every force were always canceled out by an equal and opposite force, no force could ever do anything. Instead, forces occur between objects--say car and road, to take your example. The road exerts an inward force on the car, which, you're right, is the centripetal force. The equal and opposite force is exerted by the car, on the road. The two forces are acting on different things, so they do not cancel. This second force (the force exerted by the car on the road) is sometimes referred to as the "reactive centrifugal force," which is confusing, because it's different from the more common meaning of centrifugal force.
Best Answer
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.