题目内容
如图所示,竖直放置的光滑圆环,半径R=20cm,在环上套有一个质量为m的小球,若圆环以w="10" rad/s的角速度转动(取g=10m/s2),则角θ的大小为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241912383643609.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241912383643609.jpg)
A.30° | B.45° | C.60° | D.90° |
C
试题分析:对小球受力分析,受重力,环的支持力,根据几何关系可得:小球转动半径为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824191238488615.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824191238769859.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824191239066602.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824191239315883.png)
带入数据解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824191239549583.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824191239846486.png)
故选C.
点评:要求同学们能结合几何关系解题,注意小球转动半径不是R.
![](http://thumb2018.1010pic.com/images/loading.gif)
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