Trajectory of a projectile on a frictional inclined plane

Abstract

A closed form solution is given for the trajectory of a particle sliding on an inclined plane with Coulomb-type friction. If the inclination of the plane is less than the friction angle, the particle eventually comes to rest and expressions for the location of this point and the duration of the motion are given. If the initial launch is inclined at a small angle with respect to the upward line of greatest slope, the direction of the velocity changes rapidly during the last instants of motion.

Locus of the apices of projectile trajectories under constant drag

“…(iv) As a simple example of friction that depends on the curvature. (v) The Coulomb friction, m = f N, appearing in the description of sliding bodies, has the same description as that presented here for the coplanar motion of a particle on an inclined plane [13].…”

“…In [13] the motion of a particle on an inclined plane under the presence of Coulomb friction is analysed. This friction is constant, and its magnitude is proportional to m a mg cos , μ being the kinetic friction constant and a mg cos the normal to a plane inclined by an angle α.…”

Motion of an inclined cylinder on an inclined plane

“…The cylinder does not stop rotating when it starts sliding but R v y ω < when the cylinder is sliding since a y is then larger than R t d d ω . The ratio R v y ω decreases as θ increases since F y remains approximately constant, so t d d ω remains approximately constant according to equation (5) and a y increases as θ increases, as described by equation (2).…”

“…That is, why does the cylinder roll across the inclined plane in this manner and why does not it slide or roll straight down the incline, given that the gravitational force has a component directed straight down the incline? The motion of a cylinder on an inclined plane is quite different from the motion of a sphere or a rectangular block projected at an angle on an inclined plane [1,2].…”

Spin revolution

“…2c,d. This, however, means that in addition to τ Fs a torque τ R = r × F s emerges [14][15][16], with r being a distance vector pointing from the axis M eq to the contact point (see Fig. 2e).…”