# Rolling body dynamics

Consider a cylinder of radius R and mass m, which was under pure rolling initially. This cylinder collides and goes over a, say, ledge of height R/4, where at the corner of the ledge, there was friction which is why the cylinder was able to climb up. We have to find the final angular velocity of the cylinder. I did this using conservation of angular momentum about corner, but was unable to do this using energy conservation

I also wanted to know that when the cylinder comes into contact of the ledge,:

• What will be the magnitude and direction of the normal reaction force on the cylinder by the ledge? Will it be equal to mg? Or equal to the change of linear momentum?
• What will be the magnitude of friction? I know it will be = friction coefficient*normal reaction, but what will it be opposite to? Friction is helping the cylinder climb up, but it will not be acting vertically upwards. Then what will be its direction?

Also, will mechanical energy be conserved? My reasoning is no, it will not be conserved, because of friction, a non-conservative force. Again, is energy ever conserved in a rolling body if friction is helping it roll?

static friction does not dissipate energy because there is no displacement at the point in which the force is being made, assuming the cylinder climbs without slipping. The static friction force will be vertical, because the cylinder's edge tends to move downward at the vertical surface, so the friction will be opposite to this potential motion and point upwards. It is difficult to imagine this with a sharp edge, but imagine it is actually rounded and smooth, even at a very small scale).

The vertical force will be larger than mg, otherwise the cylinder will not move up. You can calculate the magnitude of the friction force because you know that it is accelerating up at an acceleration equal to $$R\alpha$$

• Also, I wanted to know how to calculate the final angular velocity? I have tried it both by conserving angular momentum and energy. But answers here are different.