题目内容
(10分)如图所示,内半径为R的光滑圆轨道竖直放置,长度比2R稍小的轻质杆两端各固定一个可视为质点的小球A和B,把轻杆水平放入圆形轨道内,若mA=2m、mB=m,重力加速度为g,现由静止释放两球使其沿圆轨道内壁滑动,当轻杆到达竖直位置时,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502527903390.png)
(1)A、B两球的速度大小;
(2)A球对轨道的压力;
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502527903390.png)
(1)A、B两球的速度大小;
(2)A球对轨道的压力;
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150252931567.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150252884641.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150252931567.png)
试题分析:(1)设杆运动到竖直位置时,A、B两球的速度均为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150253102298.png)
AB系统机械能守恒:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502531651154.png)
解方程得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150253243732.png)
(2)在竖直位置时,设杆对B球的弹力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150253461412.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150253523446.png)
对B球
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502536331034.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150253742790.png)
对A球:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241502538671155.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150253929814.png)
由牛顿第三定律,A球对轨道的压力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150253976767.png)
点评:中等难度。在机械能守恒定律的三种表达式中,若应用
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824150254085610.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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