How does a screw roll down an inclined plane? I was thinking about how various objects would slide down on an inclined plane, and I just couldn't figure this problem out.
So let's say I have this screw or cone on its side, on an inclined plane. If friction exists, what would the motion of the screw be as it slides down the inclined plane?
There is no initial velocity and there is no air resistance either. The only forces acting on the object would be gravity, normal force, and friction.
 A: It would certainly depend on the inclination angle, the dimensions of the screw, the mass of the screw, and the coefficients of static & kinetic friction. I'll give a qualitative answer, and maybe somebody can provide a more quantitative one. The point, however, is to illustrate that a full quantitative answer will be very very complex, probably best left to numerical simulations.
For a small but finite inclination angle, the screw will roll and rotate about a pivot point, pendulating back and forth until it is stopped by rolling friction. At no point during its motion will its acceleration parallel to the surface (centripedal acceleration plus acceleration due to gravity) overcome the force of friction.
For a moderate inclination angle, the screw will again begin to roll and rotate about a pivot point, but because the angle is steeper it will roll and rotate more quickly. Due to this increased centripetal acceleration, plus the increased tangential gravitational acceleration due to the steeper angle, at some point the net tangential acceleration will overcome the force of static friction. Thus the screw will start to slide outwards and downward. If the inclination angle isn't too steep, it will probably come to a stop.
If the inclination angle is steep, it will probably not come to a stop, and slide all the way to the bottom. The steeper the angle, the earlier the screw will begin sliding down.
