The speed of small parts and conveyor belt is * * *, and the time used is: t = v0a = 55s =1s.
The displacement of motion is △ x = V202A = 522× 5 = 2.5m < (L? x)=6m
Therefore, after reaching the same speed as the conveyor belt, the small piece moves to O at a uniform speed of v0=5m/s, and then rushes onto the smooth circular track to just reach N point, so there are:
mg=2mc2NyN
According to the law of conservation of mechanical energy:12mv20 = mgyn+12mv2n.
Solution: yN= 1m
(2) Set the small object on the conveyor belt at the coordinate x 1, and if it can just reach the m point on the right side of the center of the circle, it is obtained by the conservation of energy:
μmg(L-x 1)=mgyM
Substituting the data, the result is: x 1=7.5? m
μmg(L-x2)= 12mgyN
Substituting the data, the result is: x2=7? m
If we can just reach the point m on the left side of the center of the circle, we can know from (1) that x3=5.5? m
Therefore, the position coordinate range of small pieces placed on the conveyor belt is:
7m≤x≤7.5m，0 ≤ x ≤ 5.5m。
Answer: (1) The ordinate of point N is1m; ;
(2) The abscissa range of the position where a small object can move along a smooth arc orbit (the small object always moves along the arc orbit without derailing) is 0≤x≤5.5m, 7m ≤ x < 7.5m 。