※ 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 --

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