题目内容
如图所示,一个小球(视为质点)从H=12 m高处,由静止开始通过光滑弧形轨道AB,进入半径R=4 m的竖直圆环,且圆环动摩擦因数处处相等,当到达环顶C时,刚好对轨道压力为零;沿CB圆弧滑下后,进入光滑弧形轨道BD,且到达高度为h的D点时的速度为零,则h之值不可能为(g取10 m/s2,所有高度均相对B点而言)( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005212374911.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005212374911.png)
A.12 m | B.10 m | C.8.5 m | D.7 m |
ABD
试题分析: 小球到达环顶C时,刚好对轨道压力为零,在C点,由重力充当向心力,则根据牛顿第二定律得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005212405697.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825005212421503.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250052124361247.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250052124671044.png)
所以高度范围为8m<h<10m,故选A、B、D。
![](http://thumb2018.1010pic.com/images/loading.gif)
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