Question

某同學(xué)利用業(yè)余時(shí)間為工廠設(shè)計(jì)了一個(gè)測(cè)定機(jī)器轉(zhuǎn)動(dòng)角速度的裝置如圖10所示,滑桿PN由合金材料制成,電阻不能忽略,PA與電路相連接,電源電動(dòng)勢(shì)為E,內(nèi)阻不計(jì),限流電阻與桿PN的總電阻相等.連接小球的彈簧由絕緣材料制成,彈簧的勁度系數(shù)為k.小球質(zhì)量為m,靜止時(shí)恰好在滑桿PN的中點(diǎn).當(dāng)系統(tǒng)繞OO′軸勻速轉(zhuǎn)動(dòng)時(shí),電壓表的示數(shù)為U,試求此時(shí)系統(tǒng)轉(zhuǎn)動(dòng)的角速度ω.

圖10

Answer 1

解析：設(shè)桿PN電阻為R₀,系統(tǒng)轉(zhuǎn)動(dòng)時(shí)桿PA段的電阻為R

根據(jù)歐姆定律得E=U+IR₀                                                                                 ①

I=                                                                       ②

設(shè)彈簧原長(zhǎng)為L(zhǎng)₀,實(shí)際長(zhǎng)度(反映PN桿接入電路中的長(zhǎng)度)為L(zhǎng),則，所以

L=                                                                   ③

彈簧伸長(zhǎng)量為x=-L₀                                                                              ④

據(jù)胡克定律和牛頓第二定律得kx=mω²L                                        ⑤

由①②③④⑤式解得:ω=.

答案：