题目内容
如图7-7-16所示,光滑圆管轨道ABC,其中AB部分平直,BC部分是处于竖直平面的、半径为R的半圆.圆管截面的半径r<<R.有一质量为m、半径比r略小的光滑小球以水平初速度v0从A点射入圆管.问:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204476922182.jpg)
图7-7-16
(1)若要小球能从C端出来,初速度v0需多大?
(2)在小球从C端出来的瞬间,对管壁作用力有哪几种典型情况?初速度v0各应满足什么条件?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241204476922182.jpg)
图7-7-16
(1)若要小球能从C端出来,初速度v0需多大?
(2)在小球从C端出来的瞬间,对管壁作用力有哪几种典型情况?初速度v0各应满足什么条件?
(1)v0>![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447739312.gif)
(2)当v0=
时,小球与细管无相互作用力;
当v0>
时,小球对细管上侧壁有竖直向上的弹力;
当
<v0<
时,小球对下侧壁有竖直向下的压力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447739312.gif)
(2)当v0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447754312.gif)
当v0>
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447754312.gif)
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447739312.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447754312.gif)
小球在细圆管内的运动过程中,因管道光滑,不受摩擦力作用,所受轨道的弹力与运动方向垂直,不做功,只有重力做功,故机械能守恒.
(1)要使小球能从C端出来,则小球到达C点时速度vc应不为零.对小球从A至C运动过程应用机械能守恒定律得
mvc2+mg·2R=
mv02 ①
因vC≠0,有mg·2R<
mv02
故初速度须满足v0>
②
(2)小球经过C点受重力mg和细圆管的弹力FN,根据牛顿第二定律得
FN+mg=
③
由①③式得FN=
-5mg ④
讨论④式可得:
a.若FN=0,则v0=
,这时小球与细管无相互作用力;
b.若FN>0,则v0>
,即当初速度v0>
时,小球受细管上侧壁的竖直向下的弹力(压力),由牛顿第三定律知,小球对细管上侧壁有竖直向上的弹力;
c.FN<0(即小球要受到下侧壁竖直向上的支持力),则结合②式需满足
<v0<
,这时小球对下侧壁有竖直向下的压力.
(1)要使小球能从C端出来,则小球到达C点时速度vc应不为零.对小球从A至C运动过程应用机械能守恒定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447817225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447817225.gif)
因vC≠0,有mg·2R<
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447817225.gif)
故初速度须满足v0>
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447863310.gif)
(2)小球经过C点受重力mg和细圆管的弹力FN,根据牛顿第二定律得
FN+mg=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447863426.gif)
由①③式得FN=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447879422.gif)
讨论④式可得:
a.若FN=0,则v0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447754312.gif)
b.若FN>0,则v0>
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447754312.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447754312.gif)
c.FN<0(即小球要受到下侧壁竖直向上的支持力),则结合②式需满足
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447739312.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120447754312.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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