题目内容
(10分)如图所示,用细绳一端系着质量为M=0.6kg的物体A静止在水平转盘上,细绳另一端通过转盘中心的光滑小孔O吊着质量为m=0.3kg的小球B,A的重心到O点的距离为0.2m.若A与转盘间的动摩擦因数为0.3,为使小球B保持静止,求转盘绕中心O旋转的角速度ω的取值范围.(取g=10m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620453881313.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620453881313.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162045576352.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162046153305.png)
试题分析:当A欲向外运动时转盘角速度最大
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162046371895.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162046543617.png)
ω1=4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162046153305.png)
当A欲向里运动时转盘角速度最小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162046933888.png)
ω2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162045576352.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162045576352.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162046153305.png)
点评:本题难度中等,此题属于临界问题,因此首先应分析出临界条件,当拉力较大时,判断摩擦力的方向,当拉力较小时,A由离心运动趋势,判断摩擦力的方向,根据牛顿第二定律列公式求解
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目