题目内容
如图所示,一绳系一质量为
的球在光滑的小桌面上做匀速圆周运动,绳长
=0.1m,当角速度增大到为ω=20 rad/s时,绳恰好断开,试分析:(忽略空气阻力,g取9.8m/s2)
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(1)绳断时,小球在桌面上运动的速度
;
(2)绳能承受的最大拉力T是多大?
(3)若桌子离地高度
,求小球离开桌子落地时的水平位移
。(结果可保留分数形式)
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(1)绳断时,小球在桌面上运动的速度
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(2)绳能承受的最大拉力T是多大?
(3)若桌子离地高度
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(1)
(2)
(3) x=2/7m
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试题分析:球在桌面上做匀速圆周运动的半径
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(1)绳断时,小球的运动速度
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(2)绳子提供的拉力充当向心力,当绳子断时,此时的向心力
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故
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(3)绳断后,球按原来的速度水平飞离桌面,做平抛运动,运动时间
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由①、②可求得
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点评:本题难度较小,掌握圆周运动的线速度、角速度和向心力公式是关键,平抛运动的运动时间由竖直高度决定
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