Question

（2013绵阳一诊）如图所示，传送带的两个轮子半径均为r=0.2m,两个轮子最高点A、B在同一水平面 内，A、B间距离L=5m，半径R=0.4的固定、竖直光滑圆轨道与传送带相切于B点，C 点是圆轨道的最高点。质量m=0.1kg的小滑块与传送带之间的动摩擦因数μ=0.4。重力加速 度 g=10m/s²。求：

(1)传送带静止不动，小滑块以水平速度v₀滑上传送带，并能够运动到C点，v₀至少多大？

(2)当传送带的轮子以ω=10rad/s的角速度匀速转动时，将小滑块无初速地放到传送带 上的A点，小滑块从A点运动到B点的时间t是多少？

(3) 传送带的轮子以不同的角速度匀速转动，将小滑块无初速地放到  传送带上的A点，小滑块运动到C点时，对圆轨道的压力大小不同，最大压力F_m是多大？

Answer 1

解：（1）设小滑块能够运动到C点，在C点的速度至少为v，则

mg=m················································································· (2分)

mv²－mv₀²=－2mgR-μmgL  ············································· (2分)

解得v₀=2m/s  ··································································· (1分)

（3）轮子转动的角速度越大，即传送带运动的速度越大，小滑块在传送带上加速的时间越长，达到B点的速度越大，到C点时对圆轨道的压力就越大。

小滑块在传送带上一直加速，达到B点的速度最大，设为v_Bm，对应到达C点时的速度为v_cm，圆轨道对小滑块的作用力为F，则

v_Bm²=2aL  ··············································································· (2分)

mv_Cm²－mv_Bm²=－2mgR   ················································ (2分)

mg+F=m  ········································································ (1分)

F_m=F   ···················································································· (1分)

解得F_m=5N  ············································································ (1分)