A radius r of the wheel to the horizontal speed v in the horizontal plane for a linear motion, set the water plane for the x-axis. There is a circular deceleration zone with a radius of curvature R and a maximum point from the ground. When the wheel passes through the deceleration zone, the vertical speed immediately becomes zero. Find the wheel with the arc contact with the moment the wheel by the external force, as well as the wheel after the arc speed. (Assuming no friction, gravity acceleration g)
According to individual humble opinion, the subject of the ball is not out of the buffer zone, so along the point of contact point to the center of the speed V1 = 0, and the collision moment by the buffer force along the V2 direction is 0, so V2 kinetic energy is not lost, V1 = 0, that is, the loss of kinetic energy. Next with the above answer.
There is also a situation, V is large enough, when the wheel into the deceleration with the vertex, the speed is greater than the root under the g * r, then leave the deceleration zone, do flat throwing motion. At this time, the final speed is the highest point of the speed. The kinetic energy theorem is easy to use.
The key question of this question is that the friction is not considered, so that after the wheel and the slowdown belt contact, the wheel by the deceleration zone on his power has been directed to the direction of the wheel center, this way, this force will not work, Can only change the direction of movement of the wheel, and can not change its speed. The speed of the collision instantaneous wheel is then calculated. V is decomposed into a straight line along the center line V1 and the same centerline V2, by the radius and high can be obtained SIN1, you can get V2. When the wheel to the same place, decomposition V2 (divided into horizontal and vertical), find V3, V3 is the speed required. As for the contact force, if you can find, then by my strength, not out, consult experts
