题目内容
如图所示,半径R=0.40m的光滑半圆环轨道处于竖直平面内,半圆环与粗糙的水平地面相切于圆环的端点A。质量m=0.l0kg的小球与水平地面之间的动摩擦因数为μ=0.3,小球以初速度v0="7.0" m/s在水平地面上向左运动4.0m后,冲上竖直半圆环,最后小球落在C点,取重力加速度g="10" m/s2,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/2014082500273884215171.png)
(1)小球进入圆轨道通过A点时对轨道的压力;
(2)小球经过B点时速度;
(3)A、C间的距离;
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/2014082500273884215171.png)
(1)小球进入圆轨道通过A点时对轨道的压力;
(2)小球经过B点时速度;
(3)A、C间的距离;
(1)FN=7.25N(2)3m/s(3)1.2m
试题分析:(1)在水平面上小球受摩擦力为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002738873894.png)
根据动能定理
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027388891002.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002738904618.png)
在A点,根据向心力公式
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002738920961.png)
得:FN=7.25N
(2)设小球运动到B点时速度为vB,根据动能定理:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027389511043.png)
得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002738967617.png)
设小球到达B点时的最小速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002738982379.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002738998793.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002739029651.png)
由于
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002739045476.png)
(3)小球离开B点后作平抛运动,有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002739060603.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825002739092695.png)
联立得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250027391071113.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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