※ 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