题目内容
一细绳与水桶相连,水桶中装有水,水桶与细绳一起在竖直平面内做圆周运动,如图所示,水的质量m=0.5kg,水的重心到转轴的距离
=50cm。(取g=10m/s2,不计空气阻力)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241924413632296.jpg)
(1)若在最高点水不流出来,求桶的最小速率;
(2)若在最高点水桶的速率v=3m/s,求水对桶底的压力大小。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824192440317281.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241924413632296.jpg)
(1)若在最高点水不流出来,求桶的最小速率;
(2)若在最高点水桶的速率v=3m/s,求水对桶底的压力大小。
(1)2.42m/s(2)4N
试题分析:⑴水在最高点恰好不流出来,说明水的重力恰好提供其做圆周运动所需的向心力,
mg=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824192441378424.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824192442221430.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824192443032322.png)
⑵设桶底对水有一向下的压力FN,则:
FN+mg=m
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824192441378424.png)
根据牛顿第三定律可知水对桶底的压力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824192444795707.png)
点评:本题应用牛顿第二定律破解水流星节目成功的奥秘,关键在于分析受力情况,确定向心力的来源.
![](http://thumb2018.1010pic.com/images/loading.gif)
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