# Cylinder rolling down a ramp

If we had a cylinder rolling down a ramp, with scalar friction coefficient us and kinetic friction coefficient uc, and we assume for example that the inclination of the ramp is enough to make the cylinder reach the acceleration needed to exceed the scalar friction of rolling. It means that the scalar velocity of the cylinder is not equal to ωR anymore (where ω is the angular velocity). So how can we calculate the angular and scalar velocity now? Or in other words, the parcel of the kinetic energy directed to translational and rotational energy?

Is it possible to calculate the minimum inclination of the ramp that satisfies the statement above?

If needed use other variables like h for the height of the cylinder, g for gravity, etc...

Related Other Physics Topics News on Phys.org
haruspex
Homework Helper
Gold Member
If we had a cylinder rolling down a ramp, with scalar friction coefficient us and kinetic friction coefficient uc, and we assume for example that the inclination of the ramp is enough to make the cylinder reach the acceleration needed to exceed the scalar friction of rolling.
I assume you mean that the slope is sufficient that the forces overcome static friction. If the slope is constant, it's not a question of the cylinder reaching such a condition after some time; it'll happen straight away or not at all.
Suppose cylinder has mass m, moment of inertia I (you didn't say whether it was hollow or solid, etc.) linear acceleration a, radius r. The normal force is N, the frictional force F. Slope has angle θ to vertical.
Forces parallel to the slope: ma = mg cos θ - F
Perpr. to slope: N = mg sin θ
Angular accn = a/r
Moments about cylinder's centre: Fr = Ia/r

At rolling limit: F = usN = us mg sin θ
Fr2 = us mgr2 sin θ = Ia = mgr2 cos θ - mar2
a = mgr2 cos θ /(I + mr2)
tan θ = I /((I + mr2)us)

Once it has started to slip, F reduces to ucN, so the acceleration will increase a bit.