题目内容
(10分)如图所示,AB是一段位于竖直平面内的弧形轨道,高度为h,末端B处的切线沿水平方向。一个质量为m的小物体P(可视为质点)从轨道顶端处A点由静止释放,滑到B点时以水平速度v飞出,落在水平地面的C点,其轨迹如图中虚线BC所示。已知P落地时相对于B点的水平位移OC=l,重力加速度为g,不计空气阻力的作用。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242206318184264.png)
(1)请计算P在弧形轨道上滑行的过程中克服摩擦力所做的功;
(2)现于轨道下方紧贴B点安装一水平传送带,传送带右端E轮正上方与B点相距。先将驱动轮锁定,传送带处于静止状态。使P仍从A点处由静止释放,它离开B点后先在传送带上滑行,然后从传送带右端水平飞出,恰好仍落在地面上C点,其轨迹如图中虚线EC所示。若将驱动轮的锁定解除,并使驱动轮以角速度ω顺时针匀速转动,再使P仍从A点处由静止释放,最后P的落地点是D点(图中未画出)。已知驱动轮的半径为r,传送带与驱动轮之间不打滑,且传送带的厚度忽略不计。求:
①小物块P与传送带之间的动摩擦因数;
②若驱动轮以不同的角速度匀速转动,可得到与角速度ω对应的OD值,讨论OD的可能值与ω的对应关系。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242206318184264.png)
(1)请计算P在弧形轨道上滑行的过程中克服摩擦力所做的功;
(2)现于轨道下方紧贴B点安装一水平传送带,传送带右端E轮正上方与B点相距。先将驱动轮锁定,传送带处于静止状态。使P仍从A点处由静止释放,它离开B点后先在传送带上滑行,然后从传送带右端水平飞出,恰好仍落在地面上C点,其轨迹如图中虚线EC所示。若将驱动轮的锁定解除,并使驱动轮以角速度ω顺时针匀速转动,再使P仍从A点处由静止释放,最后P的落地点是D点(图中未画出)。已知驱动轮的半径为r,传送带与驱动轮之间不打滑,且传送带的厚度忽略不计。求:
①小物块P与传送带之间的动摩擦因数;
②若驱动轮以不同的角速度匀速转动,可得到与角速度ω对应的OD值,讨论OD的可能值与ω的对应关系。
(1)
;
(2)①
;
②当0﹤ω﹤
时,OD=l;当ω﹥
时,OD=(![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220637652351.png)
;当
﹤ω﹤
时,OD=
。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242206333001009.png)
(2)①
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220634252732.png)
②当0﹤ω﹤
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220635437424.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220636529504.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220637652351.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220638838514.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220640008424.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220636529504.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220642551687.png)
试题分析:(1)小物块从A到B的过程中有重力和摩擦力对其做功,
故根据动能定理有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242206433621015.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408242206333001009.png)
(2)① 需要先求出物块在E点的速度;
没有安装传送带时,小物块从B到C的过程,做平抛运动,水平方向有l=vt;
安装传送带后,小物块从E到C的过程沿水平方向有l/2=vEt,二者的竖直高度相等,落下时所用的时间相等,故联立以上两个方程,解得vE=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220644953400.png)
设小物块与传送带之间的动摩擦因数为μ,小物块从B到E的过程,根据动能定理有
-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220646061970.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220634252732.png)
②(a)当传送带的速度0﹤v带=ωr﹤vE,
即0﹤ω﹤
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220635437424.png)
(b)如果传送带的速度较快,物体在传送带上一直加速而未与传送带共速,则物体的加速度始终为a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220649586392.png)
根据运动学公式有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220650834743.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220652285492.png)
当传送带的速度v带=ωr﹥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220653330482.png)
即ω﹥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220636529504.png)
则OD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220654953440.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220656076513.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220637652351.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220638838514.png)
(c)当传送带的速度vE﹤v带=ωr﹤vmax,
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220640008424.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220636529504.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220654953440.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824220642551687.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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