※ 引述《ghostpig (^ ^)》之铭言：
: 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
: 这题是一般铅直的单摆？但为何算功，题目还会提到 时间呢？ 麻烦高手解答一下
: 不知怎么算><
这题只是功的问题
设最低点位能 = 0
起始总能 = mg[L-Lcos(Pi/6)] + 0 (因为最高点动能=0)
有阻力做负功才会使总能慢慢减少
所以起始总能 + [摩擦力所做的负功] = 0 + 0 位能和动能均为0
所以摩擦力所做的负功 = - 起始总能 = -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
: 这题是求角位移是吗？我的想法是，能否直接从角加速度往回积分，这样就可算角加
: 速
: 和角位移？
: ====================================================================
: 再麻烦各位大大 给予建议 谢谢~~~^ ^
可以
假设题目为α = 4t ^3 - 3t^2
积分α得w
角速度w = t^4 - t^3 + w(0) 最后一项是积分常数
由题目w(0) = 30
5 5
所以在t=3 -> 5 秒之间的角度差 = ∫ wdt = [(1/5)t^5 - (1/4)t^4] + 30t|
3 3
= 500.4 rad
答案D