Question

如圖所示,AB為豎直轉(zhuǎn)軸,細(xì)繩AC和BC的結(jié)點(diǎn)C系一質(zhì)量為m的小球,兩繩能承擔(dān)的最大拉力均為2mg,當(dāng)AC和BC均拉直時(shí)∠ABC=90°,∠ACB=53°,ABC能繞豎直軸AB勻速轉(zhuǎn)動(dòng),因而C球在水平面內(nèi)做勻速圓周運(yùn)動(dòng),求:
 
(1)當(dāng)m的線速度增大時(shí),AC和BC(BC="1" m)哪條繩先斷?
(2)一條繩被拉斷后,m的速率繼續(xù)增加,整個(gè)運(yùn)動(dòng)狀態(tài)會(huì)發(fā)生什么變化?

Answer 1

見解析

(1)當(dāng)小球線速度增大到BC被拉直時(shí),AC線拉力T_AC=1.25mg.當(dāng)球速再增大些時(shí),T_AC不變,BC線拉力隨球速增大而增大,由T_ACcos53°+T_BC=m,可得當(dāng)5.19 m/s時(shí),T_BC=2mg,BC線先斷.
(2)當(dāng)BC線斷后,AC線與豎直方向夾角α因離心運(yùn)動(dòng)而增大,同時(shí)球速因重力而減小.當(dāng)使球速再增大時(shí),角α隨球速增大而增大,根據(jù)T_AC=可知當(dāng)α=60°時(shí),T_AC=2mg,AC也斷.再根據(jù)T_ACsinα=m,可知此時(shí)球速v′=4.95 m/s.