- #1
curiouschris
- 147
- 0
Hi all
First of all this is not a homework problem.
Say I have a flywheel of an arbitrary weight and diameter it is fixed in position with a vertical axis and spinning at oh I don't know 1000 revs per min.
Now I bring a second flywheel of the same size and weight which is not spinning at all into contact with the first.
Now my question is how do I calculate the force on the axles of either flywheel.
I am not sure this describes what I am looking for so I will use a clock face analogy to try and make it clearer.
You are looking down at the first flywheel (axle coming out of the screen) and it is spinning clockwise. twelve o'clock is at the top of the screen 6 o'clock at the bottom 3 to the right and 9 to the left.
If the second flywheel came into contact with the first flywheel at the first flywheels 3 o'clock position I imagine the axle of the first flywheel would attempt to move towards the 12 o'clock position, exactly opposite the tangent described at the 3 o'clock position.
With what force would the axle attempt to move? or more precisely how would I calculate that force?
Is that clearer or more muddy?
Please note: the only purpose of the second flywheel is to introduce friction, a rubber block would server the purpose equally well, I am only interested on the forces on the axle of the first flywheel. it just seemed easier to describe that way.
CC
First of all this is not a homework problem.
Say I have a flywheel of an arbitrary weight and diameter it is fixed in position with a vertical axis and spinning at oh I don't know 1000 revs per min.
Now I bring a second flywheel of the same size and weight which is not spinning at all into contact with the first.
Now my question is how do I calculate the force on the axles of either flywheel.
I am not sure this describes what I am looking for so I will use a clock face analogy to try and make it clearer.
You are looking down at the first flywheel (axle coming out of the screen) and it is spinning clockwise. twelve o'clock is at the top of the screen 6 o'clock at the bottom 3 to the right and 9 to the left.
If the second flywheel came into contact with the first flywheel at the first flywheels 3 o'clock position I imagine the axle of the first flywheel would attempt to move towards the 12 o'clock position, exactly opposite the tangent described at the 3 o'clock position.
With what force would the axle attempt to move? or more precisely how would I calculate that force?
Is that clearer or more muddy?
Please note: the only purpose of the second flywheel is to introduce friction, a rubber block would server the purpose equally well, I am only interested on the forces on the axle of the first flywheel. it just seemed easier to describe that way.
CC