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, Fc=Fncos(teta) 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=rw^{2}. So now we know that Fc=mrw^{2}. Now we can plug this into the force balances and get two equations for the normal force: Fn=sin(teta)/(mg) Fn=cos(teta)/(mrw^{2})Now we can substitute for Fn to set the equations equal: sin(teta)/(mg)=cos(teta)/(mrw^{2})Now we want to solve for r. You can rearrange this equation to show that: r=(mg cos)/(mw^{2} sin(teta))Now you can look at this equation and see that the mass cancels! r=(g cos(teta))/(w^{2} 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.
