Re: [其他] 功、轉動共2題~~請教

看板TransPhys作者Honor1984 (希望願望成真)時間10年前 (2013/07/11 04:37)推噓2(2推 0噓 7→)

留言9則, 3人參與討論串2/2 (看更多)

※ 引述《ghostpig (^ ^)》之銘言： : A pendulum consists of a 0.500 kg mass on the end of a light rod 50.0 cm : long. It was set swinging so that the greatest angle the rod makes with : the vertical was 30.0º . After 5 hours and 30 minutes it was seen to come : to rest at its lowest position. How much work was done on the : pendulum by the frictional forces? (Answer in J) : (A) – 1.225 : (B) – 0.825 : (C) + 0.450 : (D) – 0.328 : 答D : 這題是一般鉛直的單擺？但為何算功，題目還會提到時間呢？麻煩高手解答一下 : 不知怎麼算>< 這題只是功的問題設最低點位能 = 0 起始總能 = mg[L-Lcos(Pi/6)] + 0 (因為最高點動能=0) 有阻力做負功才會使總能慢慢減少所以起始總能 + [摩擦力所做的負功] = 0 + 0 位能和動能均為0 所以摩擦力所做的負功 = - 起始總能 = -mg[L-Lcos(Pi/6)] = -0.328 J 答案D : ====================================================================== : If the angular velocity is 30 radians per second at t = 0, find the angle : turned through by the wheel between times t = 3 and t = 5 seconds. : Answer in radians. : (A) 260 : (B) 320 : (C) 380 : (D) 500 : 答D : 這題是求角位移是嗎？我的想法是，能否直接從角加速度往回積分，這樣就可算角加 : 速 : 和角位移？ : ==================================================================== : 再麻煩各位大大給予建議謝謝~~~^ ^ 可以假設題目為α = 4t ^3 - 3t^2 積分α得w 角速度w = t^4 - t^3 + w(0) 最後一項是積分常數由題目w(0) = 30 5 5 所以在t=3 -> 5 秒之間的角度差 = ∫ wdt = [(1/5)t^5 - (1/4)t^4] + 30t| 3 3 = 500.4 rad 答案D -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 128.220.159.5

推

BaBi

07/11 06:17, , 1^F

07/11 06:17, 1^F

→

Honor1984

07/11 06:40, , 2^F

07/11 06:40, 2^F

→

Honor1984

07/11 06:45, , 3^F

07/11 06:45, 3^F