Ma=2mb=2？ Kg, va = 4.5m/s
Solution: vb=9.0m/s? ②
Let the speed of slider B when it reaches point B be vB, which is obtained by kinetic energy theorem. μmbgL= 12mbv2B？ 12mbv2b③
When entering the circular orbit, let the supporting force of slider B be FN, which is obtained from Newton's second law, FN? mbg=mbv2BR④
By Newton's third law fn =? f’N⑤
The pressure f ′ n of slider B on the track is obtained from ③ ④ ⑤. 59N, vertically downward.
(2) If the small slider B can reach the highest point of the circular trajectory, the speed is vC.
Then by the conservation of mechanical energy,12mbv2b = mbg2r+12mbv2c6.
The solution is vc=3.0m/s⑦.
The speed at which the small piece B can just pass the highest point is V, so mbg=mbv2? R⑧
Solution, v = gr =10m/s pet-name ruby.
Because VC < V, the small slider B can't reach the highest point C of the circular orbit.
Answer: (1) When the small slider B passes through the point B of the circular track, the pressure on the track is 59N, and the direction is vertical downward.
(2) The small slider B cannot reach the highest point C of the circular orbit.