题目内容
在直径1.6m的圆柱体一端截出一圆锥,如下图所示,在看到剖面上,三角形的三边之比为3:4:5, 圆柱体可绕其中心对称轴匀速旋转。将一小木块放置在斜面的中点,它与斜面间动摩擦力因素为0.25,若小木块保持在此位置不动,则圆柱体旋转的角速度应为多大.(
;g取10m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647099462332.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647097271007.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647099462332.jpg)
3.2rad/s≤ω≤5.4rad/s
试题分析:当圆柱体旋转角速度最小为ωmin时,木块有沿斜面向下滑动趋势,木块受最大静摩擦力f的方向沿斜面向上。木块受重力mg、斜面支持力N和静摩擦f,如图所示。木块在水平面内作匀速圆周运动,向心加速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164710133624.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647105703997.png)
根据牛顿第二定律得
水平方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647109601132.png)
竖直方向: Ncosα+fsinα=mg ②
又 f=μN ③
联立①、②、③解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647113181474.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164711662853.png)
当圆柱体旋转角速度最大为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164712208375.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164712707631.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647131125131.png)
由牛顿第二定律得
水平方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647133001095.png)
竖直方向: Ncosα-fsinα=mg ⑤
又 f′=μN′ ⑥
联立④、⑤、⑥解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241647135181448.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824164713861803.png)
圆柱体旋转的角速度ω应为3.2rad/s≤ω≤5.4rad/s
点评:关键是知道当圆柱体旋转角速度最大木块有沿斜面向上滑动趋势,木块受最大静摩擦力f的方向沿斜面向下。当圆柱体旋转角速度最小木块有沿斜面向下滑动趋势,木块受最大静摩擦力f的方向沿斜面向上。
![](http://thumb2018.1010pic.com/images/loading.gif)
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