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Hi! I have a question concerning the properties of centrifugal acceleration and force: Lets say I have a sphere rotating in a circle inside of a spinning cone (it remains the same distance from the center of the cone at all times) Then, lets say you put some thing on that sphere, like a ring (ex. bearings that hold more balls together with a ring) or lets say you simply put a HEAVIER sphere into the cone to rotate alongside the first sphere. Now, my question is: would the heavier sphere rotate at the same distance away from the center as the first sphere (Both spheres are experiencing the same outward ACCELERATION) and would the sphere with the ring still continue to spin at the same distance? In my thinking, I would say yes, because both spheres experience the same acceleration, so they should stay at the same distance away from the center (the outward acceleration and gravity cancel out there - where they are rotating) I would say this because of what we see with gravity: (ex. mass does not matter, ACCELERATION acts the same on any object, but can produce a stronger force. (In the case of the spheres, the heavier sphere produces a larger centrifugal force) Am I right? Please help! Any info will be appreciated! Thanks!
Question Date: 2009-06-21
Answer 1:

This is an interesting question which raises some good points about centrifugal force and acceleration. To make sure were on the same page, lets review the setup of the problem:

There is a cylinder which is oriented so that its circular end is facing upwards in gravity. It spins along its axis with some angular velocity w. The cylinder also has some angle, teta between its axis of rotation and its side.

What will be the distance, r, between the cylinders axis and the center of a sphere which is inside this cylinder?

So thats the setup of the problem. Now lets work it out!

Now, there are three forces to consider which will decide r, the distance of the ball from the cylinders center. These are gravity, the centrifugal force, and the normal force or force the cylinder wall exerts on the ball.

One of Newtons laws says that for every action, there is an equal and opposite reaction. Another way of saying this law is that the forces balance in both the horizontal and vertical direction. Using this law, we can write down two equations.

In the vertical direction,
Fg=Fnsin(teta)Where Fg is the force of gravity and Fn is the normal force (the force the wall exerts on the sphere).

In the horizontal direction,
Where Fc is the centrifugal force.

The sin(teta) and cos(teta) terms come from trigonometry. You might have learned about this in school, but if not, the important thing to recognize is that these are just fractions which tell us how much of the normal force goes in the horizontal direction (sin(teta)) and how much goes in the vertical direction (cos(teta)).

Now we can use another of Newtons laws to figure out Fg and Fc. Newtons second law says Force equals Mass times Acceleration, or F=ma. So we can say

Fg=mg (where g is the acceleration due to gravity)
Fc=mac (where ac is the acceleration due to the centrifugal force)

Now we need to figure out ac. I didnt remember what this was so I looked up on Wikipedia that ac=rw2. So now we know that Fc=mrw2.

Now we can plug this into the force balances and get two equations for the normal force:


Now we can substitute for Fn to set the equations equal:


Now we want to solve for r. You can rearrange this equation to show that:

r=(mg cos)/(mw2 sin(teta))

Now you can look at this equation and see that the mass cancels!

r=(g cos(teta))/(w2 sin(teta))

So it turns out you were right! The mass of the sphere does not matter, and the heavier sphere will rotate at the same distance from the center as the lighter sphere. The mass of a ring should not change this as long as all the assumptions we made to get this equation still hold. Good jobyour intuition was correct! This method of using force balances is a very general method that you can use for many similar problems.

Answer 2:

First, a point of clarification, because it sounds like it's confusing you: there is no such thing as centrifugal force. An object that is revolving in a circle is being accelerated toward the center, not away from it. Remember that if the circle-inducing force were taken away the revolving object would head off tangent to the circle at the point where it was, which will take it farther away from the center. The force that is moving the spheres in an orbit, in this case, is gravity.

Now, onto your question: what you have is a cone that is spinning at a certain speed. This speed is measured as angular velocity, abbreviated w (lower case omega). w is in units of radians/second, so if your cone makes one complete rotation in a second, then it is spinning at 2(pi)r radians/second.

The ball is revolving at velocity v. Its path is a circle of radius r around the apex of the cone. The path that it has to traverse in one revolution therefore is length 2(pi)r. Therefore v = 2(pi)rw.

Now, we want acceleration. a = v2/r = (2(pi)rw)2/r = 4r((pi)w)2.

Notice that there is no mass term in that equation apart from F = ma, but we already know acceleration: it has to be that of gravity. As a result, mass is already accounted for by the time we get to that point: it doesn't matter in determining r, which is what we want, since w also has to be the same if the two balls are spinning in the same cone.

Therefore, you are correct: the two balls will orbit at the same distance.

Answer 3:

Your wording of the problem is a bit strange. Why is the sphere inside the cone? Which way is gravity pointing? Is there gravity? If so, which way is the cone pointing? Is the cone the source of gravity? Is this in a laboratory on the earth and the earth is providing the gravity? What exactly are the forces involved? Is it the contact force between the sphere and the cone that cause the centripetal acceleration? Or is it gravity?

Until you clarify which forces are causing what, it is not possible to answer. Its possible to guess what you mean, but impossible to know for sure.

Physics is fundamentally a very quantitative science. So if you could provide information necessary to quantify the problem that would make the problem much easier to answer. Present it much as you might see a homework problem. Then we can talk a bit more about the solution.

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