QUESTION: > dont do homework
I'm 26, not in school.
I'm playing with a yo-yo swinging it around my head thinking that if it was a beach ball of equal weight that the centripetal acceleration would not be the same.
Do you have any insight ?
When you Google centripetal acceleration + density, nothing really shows up.
1. Does centripetal acceleration depend on the density of the ball at the end of the string.
2. If so, what density equation then equals v squared / r
ANSWER: Permit me to make something clear RIGHT FROM THE START.
If it doesn't apply to you, all the better.
Over the years, innumerable people have asked questions of me in a completely disingenuous manner -- they pretend to want to know something, when all they just really want is a lead-in to an argument with me. If this is the game you're trying to play, I DON'T WANT TO JOIN. If you plan to reply to my answer with anything along the lines of, "That's not correct" -- in other words, if you already have an opinion on this question and are just searching for my opinion -- my immediate response will be to click on the button stating, "This is one follow-up too many."
I hope this makes myself clear.
ANYWAY, recall what centripetal acceleration IS: it's the acceleration of an object -- ie, the change in velocity of an object -- that results in an object moving in a constant circular path. If an object ceases to move at a constant speed in a linear path, it is (by definition) accelerating. If the only change in an object's velocity is that it moving in a circular path, then (and ONLY then) it is experiencing a centripetal acceleration.
The size of this acceleration depends on only two variables: the speed of the circular motion and the radius of the circle of its motion. More precisely, the formula is
a(c) = v^2/r
where a(c) is the centripetal acceleration, v is the speed, and r is the radius.
Thus, it doesn't matter what the density of the object is, the only variables are the velocity and the radius. Which is why you won't find much discussion about a(c) and density -- there simply ISN'T a connection.
---------- FOLLOW-UP ----------
QUESTION: Thanks Expert,
In NO way am I wanting to argue. The only thing I desire is what already happened today when I found out (via your wonderful service) that Radians / Second really means just "per second". I couldn't figure out why the radians were cancelled when the moment of inertia was multiplied by the radians per second squared. Do you have any idea how much RELIEF that is?
And so my frustration here still remains;
> the only variables are the velocity and the radius.
My point exactly, the "velocity" would become less with the beachball because of its greater volume (less Density) than the yo-yo (both having equal weight).
Is it because I'm assuming in my mind that the "energy" causing the rotation is kept the same?
I see more resistance against a beachball than a yo-yo therefore less velocity therefore less Centripetal Acceleration.
Where am I going wrong?
I appreciate your help,
> In NO way am I wanting to argue.
Good. That means my statement -- "If it doesn't apply to you, all the better." -- DOES apply to you.
> the "velocity" would become less with the beachball because of its greater volume
Actually, the velocity of both the beach ball and of the yo-yo are completely arbitrary; they can be any velocity you want them to be. You can spin the yo-yo at a frequency of one revolution per second or five revolutions per second. In the latter case, the VELOCITY will be much higher, even if the radius is unchanged. Likewise for the beach ball.
> Is it because I'm assuming in my mind that the "energy" causing the rotation is kept the same?
I wish you would have mentioned that sooner.
The energy you put into the system is that amount of energy is necessary to accomplish the velocity at which you are seeking to spin the object. The higher the velocity you want, the more energy you must input. Even if the two objects are the same mass, there is no way to determine their respective velocities UNLESS you specify that the energies input into each is the same. Then we can begin to make some comparisons.
> I see more resistance against a beachball than a yo-yo therefore less velocity
> therefore less Centripetal Acceleration.
Now you're adding another complication: air resistance. And, unfortunately, it's a complicated complication. There's a reason just about every basic physics question includes the statement, "Ignoring air resistance." That's because the amount air resistance is VERY difficult to calculate. It depends on the shape, surface material, AND velocity of the moving object -- and, to some extent, the SQUARE of the velocity. Aeronautical engineers find they can't just do some calculations and calculate the air resistance of a certain design -- they simply place a scale model of that design in a wind tunnel and just measure the air resistance. For all but the VERY simplest of setups, the calculations are simply too difficult to do.
PERHAPS what you're running into is this: the air resistance of a beach ball is greater than that for a yo-yo. This frictional force will require that, FOR THE SAME AMOUNT OF ENERGY INPUT into the system, the yo-yo will move faster. That's because the air resistance against the beach ball will "eat up" more of the energy you input into the moving object, causing less energy available for kinetic energy. Less kinetic energy available, lower velocity. Lower velocity, lower centripetal acceleration. *IF* the energy you input each of the two systems -- one with a yo-yo, one with a beach ball -- is the same, then the yo-yo will be moving faster.
Thus, the lower density of the beach ball WILL have an effect on the centripetal acceleration, just as you surmise.
> Where am I going wrong?
Actually, you're doing fine. Your only "problem" is that it's pretty much impossible to calculate the exact effect of air resistance. It'd be nice if we could simply say, "The density of the beach ball is one-tenth that of the yo-yo, therefore the air resistance is ten times greater." But the calculations are a lot more complicated than that.
Oh, and also not mentioning that you want to keep the energy input the two systems the same. Unless you make that statement clear, it is impossible to compare the velocities of the two objects.