题目内容
如图所示,竖直平面内的3/4圆弧形光滑轨道,轨道半径为R,A端与圆心等高,AD为水平面,B点在圆心的正下方,一小球m自A点正上方由静止释放,自由下落至A点进入轨道,小球巧好能够通过最高点C,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242241083382415.png)
(1)小球到B点时的速度vB;
(2)释放点距A的竖直高度h;
(3)落点D与A的水平距离s。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242241083382415.png)
(1)小球到B点时的速度vB;
(2)释放点距A的竖直高度h;
(3)落点D与A的水平距离s。
(1)
(2)
(3)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224112971637.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224110116600.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224111692624.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224112971637.png)
试题分析:(1)小球恰好通过最高点C点,即C点轨道弹力等于0,根据径向合力提供向心力,在最高点满足
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224114281742.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224115592547.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242241174011033.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224110116600.png)
(2)从释放点到B点,只有重力做功,由动能定理得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242241200841084.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224111692624.png)
(3)物体离开轨道最高点C点后将做 平抛运动,落在AD水平面上,那么平抛的高度就是R,竖直方向自由落体运动,则有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224121317641.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224122627658.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224123891677.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824224112971637.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目