# Marble spinning in steel tube

1. Feb 25, 2008

### Nightrider55

1. The problem statement, all variables and given/known data
A 10g steel marble is spun so that it rolls at 150rpm around the inside of a vertically oriented steel tube. The tube is 12 cm in diameter. Assume that the rolling friction is small enough for the marble to maintain 150 rpm for several seconds. During this time, will the marble spin in a horizontal circle, at constant height, or will it spiral down the inside of the tube.

Known

150rpm
12cm diameter
coefficient of Static friction for steel on steel (dry)=.80

2. Relevant equations

A=V^2/r
Fsmax=(mew)s x n

3. The attempt at a solution

I solved for the velocity which came out to be .3pi m/s for the ball and then got stuck from there. I solved for the centripetal acceleration but dont know if that would apply in this situation. I know that I need to find the normal force so I can figure out the static max, and in order for the ball to remain horizontal it must be under the static max right? Any insight that would aim me in the right direction would be very much appreciated.

Last edited: Feb 25, 2008
2. Feb 25, 2008

### Dick

If you know the centripetal acceleration, then you know the normal force, right? If you know the normal force, then you know the frictional force. Now just ask if the frictional force is greater than mg.

3. Feb 25, 2008

### Nightrider55

Is the normal force directly related to the centripetal acceleration? Dont you have to factor mass into the normal force? Wouldn't the free body of the ball have normal force perpendicular to the wall with gravity straight down and static friction straight up? Then how would you calculate the normal force?

4. Feb 26, 2008

### Dick

Yes, the free body diagram has a normal force perpendicular to the wall with gravity straight down and friction straight up. The normal force is m times the centripetal acceleration. If you assume the ball has negligible diameter so you can ignore torques I think you can assume the ball would have to start sliding before it can move down the tube.

5. Feb 26, 2008

### Nightrider55

Ok I calculated the normal force which came out to be 1.45 N then I calculated Fsmax which is (mew)s x normal force so .80 x 1.450---> 1.16 N. This means that the force of static friction has to be below this value in order for the ball to remain horizontal and not spiral downwards. Then does this mean that the static friction would be equal to gravity x mass, making it a larger value than the static max, or is there something that I am missing because
I'm not too sure about my method.

6. Feb 26, 2008

### Dick

I'm getting a centripetal acceleration of 14.8m/sec^2. Normal force is m times that. Friction is 0.8 times that. mg is m times 9.8m/sec^2. So I'm getting static friction a bit larger than mg. Can you check your numbers?

7. Feb 26, 2008

### Nightrider55

I get 14.8 m/s^2 for the centripetal acceleration after using a=V^2/r then normal force is m times that which is 14.8 x .01 kg= .148044 N. Then friction is .8 times that which is .1184 N, for the Fsmax. Then Fs is calculated by m x g which is .098 N right? So this is below the max so the ball remains horizontal which the rpm is at 150. Did I do it right?

8. Feb 26, 2008

### Dick

That's the way I would answer it. Hope I'm not missing something.