Question

（2013綿陽(yáng)一診）如圖所示，傳送帶的兩個(gè)輪子半徑均為r=0.2m,兩個(gè)輪子最高點(diǎn)A、B在同一水平面 內(nèi)，A、B間距離L=5m，半徑R=0.4的固定、豎直光滑圓軌道與傳送帶相切于B點(diǎn)，C 點(diǎn)是圓軌道的最高點(diǎn)。質(zhì)量m=0.1kg的小滑塊與傳送帶之間的動(dòng)摩擦因數(shù)μ=0.4。重力加速 度 g=10m/s²。求：

(1)傳送帶靜止不動(dòng)，小滑塊以水平速度v₀滑上傳送帶，并能夠運(yùn)動(dòng)到C點(diǎn)，v₀至少多大？

(2)當(dāng)傳送帶的輪子以ω=10rad/s的角速度勻速轉(zhuǎn)動(dòng)時(shí)，將小滑塊無(wú)初速地放到傳送帶 上的A點(diǎn)，小滑塊從A點(diǎn)運(yùn)動(dòng)到B點(diǎn)的時(shí)間t是多少？

(3) 傳送帶的輪子以不同的角速度勻速轉(zhuǎn)動(dòng)，將小滑塊無(wú)初速地放到  傳送帶上的A點(diǎn)，小滑塊運(yùn)動(dòng)到C點(diǎn)時(shí)，對(duì)圓軌道的壓力大小不同，最大壓力F_m是多大？

Answer 1

解：（1）設(shè)小滑塊能夠運(yùn)動(dòng)到C點(diǎn)，在C點(diǎn)的速度至少為v，則

mg=m················································································· (2分)

mv²－mv₀²=－2mgR-μmgL  ············································· (2分)

解得v₀=2m/s  ··································································· (1分)

（3）輪子轉(zhuǎn)動(dòng)的角速度越大，即傳送帶運(yùn)動(dòng)的速度越大，小滑塊在傳送帶上加速的時(shí)間越長(zhǎng)，達(dá)到B點(diǎn)的速度越大，到C點(diǎn)時(shí)對(duì)圓軌道的壓力就越大。

小滑塊在傳送帶上一直加速，達(dá)到B點(diǎn)的速度最大，設(shè)為v_Bm，對(duì)應(yīng)到達(dá)C點(diǎn)時(shí)的速度為v_cm，圓軌道對(duì)小滑塊的作用力為F，則

v_Bm²=2aL  ··············································································· (2分)

mv_Cm²－mv_Bm²=－2mgR   ················································ (2分)

mg+F=m  ········································································ (1分)

F_m=F   ···················································································· (1分)

解得F_m=5N  ············································································ (1分)