Question

如圖4－3－13所示為某種魚餌自動(dòng)投放器中的投餌管裝置示意圖，其下半部AB是一長(zhǎng)為2R的豎直細(xì)管，上半部BC是半徑為R的四分之一圓弧彎管，管口沿水平方向，AB管內(nèi)有一原長(zhǎng)為R、下端固定的輕質(zhì)彈簧．投餌時(shí)，每次總將彈簧長(zhǎng)度壓縮到0.5R后鎖定，在彈簧上端放置一粒魚餌，解除鎖定，彈簧可將魚餌彈射出去．設(shè)質(zhì)量為m的魚餌到達(dá)管口C時(shí)，對(duì)管壁的作用力恰好為零．不計(jì)魚餌在運(yùn)動(dòng)過(guò)程中的機(jī)械能損失，且鎖定和解除鎖定時(shí)，均不改變彈簧的彈性勢(shì)能．已知重力加速度為g.求：

(1)質(zhì)量為m的魚餌到達(dá)管口C時(shí)的速度大小v₁；

(2)彈簧壓縮到0.5R時(shí)的彈性勢(shì)能E_p；

(3)已知地面與水面相距1.5R，若使該投餌管繞AB管的中軸線OO′在90°角的范圍內(nèi)來(lái)回緩慢轉(zhuǎn)動(dòng)，每次彈射時(shí)只放置一粒魚餌，魚餌的質(zhì)量在m到m之間變化，且均能落到水面．持續(xù)投放足夠長(zhǎng)時(shí)間后，魚餌能夠落到水面的最大面積S是多少？

Answer 1

解析　(1)質(zhì)量為m的魚餌到達(dá)管口C時(shí)做圓周運(yùn)動(dòng)的向心力完全由重力提供，

則mg＝m                                                                                                   ①

由①式解得

v₁＝                                                                                                                ②

(2)彈簧的彈性勢(shì)能全部轉(zhuǎn)化為魚餌的機(jī)械能，

由機(jī)械能守恒定律有

E_p＝mg(1.5R＋R)＋mv₁²                                                                                                                                                                 ③

由②③式解得

E_p＝3mgR                                                                                                          ④

(3)不考慮因緩慢轉(zhuǎn)動(dòng)裝置對(duì)魚餌速度大小的影響，

質(zhì)量為m的魚餌離開管口C后做平拋運(yùn)動(dòng)，

設(shè)經(jīng)過(guò)t時(shí)間落到水面上，

離OO′的水平距離為x₁，

由平拋運(yùn)動(dòng)規(guī)律有

4．5R＝gt²                                                                                                                                                                                                        ⑤

x₁＝v₁t＋R                                                                                                                ⑥

由⑤⑥式解得x₁＝4R                                                                                         ⑦

當(dāng)魚餌的質(zhì)量為m時(shí)，

設(shè)其到達(dá)管口C時(shí)速度大小為v₂，

由機(jī)械能守恒定律有

E_p＝mg(1.5R＋R)＋v₂²                                                                                                                                                     ⑧

由④⑧式解得v₂＝2                                                                                         ⑨

質(zhì)量為m的魚餌落到水面上時(shí)，

設(shè)離OO′的水平距離為x₂，則

x₂＝v₂t＋R                                                                                                                ⑩

由⑤⑨⑩式解得x₂＝7R

魚餌能夠落到水面的最大面積

S＝(πx₂²－πx₁²)＝πR²(或8.25πR²)．

答案　(1)

(2)3mgR

(3)πR²(或8.25πR²)