题目内容
如图所示,竖直放置的光滑圆轨道被固定在水平地面上,半径r=0.4m,在最低点处有一小球(其半径远小于轨道半径r)。现给小球以水平向右的初速度v0,g取10m/s
,如果要使小球运动过程中不脱离圆轨道运动, v0的大小满足的条件可表示为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241456011054017.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145600731242.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241456011054017.jpg)
A.v0≥0 | B.v0≥4m/s | C.v0≥2![]() | D.v0≤2![]() |
CD
试题分析:当小球以初速度v0沿圆环轨道运动到水平位置,即上升r高度后,速度为零,则小球会沿轨道下滑下来,不会脱离轨道,此时根据机械能守恒可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145601355789.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145601402669.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145601761691.png)
当小球能做完整的圆周运动,此时的临界条件是小球在最高点的向心力只有重力充当,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145602104693.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145602197957.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824145602322720.png)
点评:本题的关键是找出小球不脱离轨道的临界条件,然后运用机械能守恒定律解题
![](http://thumb2018.1010pic.com/images/loading.gif)
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