動力學-加速度問題 - 考試
By Faithe
at 2014-01-14T14:22
at 2014-01-14T14:22
Table of Contents
題目:直徑2m,由靜止開始以順時針方向施以3 rad/sec^2 等角加速度3秒鐘,
試求(a)t=0 (b)t=3 (c)t=5秒,輪子底端之速度及加速度。
(a)
v = w*r
v = v0 + a*t
w = w0 + a*t
w = 0 + 3*0 = 0
v = 0
an = (w^2)*r = 0*1 = 0
at = a*r = 3*1 = 3 m/sec^2
(b)
w = 0 + 3*3 = 9 rad/sec
v = w*r = 9 m/sec
an = (w^2)*r = (9^2)*1 = 81 m/sec^2
at = a*r = 3*1 = 3 m/sec^2
(c)
w = 0 + 3*3 = 9 rad/sec
v = w*r = 9 m/sec
an =(w^2)*r = (9^2)*1 = 81 m/sec^2
at = a*r = 0 * 1 = 0
------------------------------------------------------------------------------
w = 角速度
a = 角加速度
v = 速度
an = 法線加速度
at = 切線加速度
------------------------------------------------------------------------------
小弟的問題,為什麼超過3秒後 w5 = w3呢?
為什麼超過3秒後,計算切線加速度時,a角加速度為零呢?
麻煩版上前輩們不吝嗇指導,謝謝!
--
Tags:
考試
All Comments
By Eartha
at 2014-01-16T03:48
at 2014-01-16T03:48
By Ophelia
at 2014-01-17T05:51
at 2014-01-17T05:51
By Kama
at 2014-01-18T08:12
at 2014-01-18T08:12
By Andy
at 2014-01-22T13:11
at 2014-01-22T13:11
By Elvira
at 2014-01-25T00:53
at 2014-01-25T00:53
By Agatha
at 2014-01-27T03:25
at 2014-01-27T03:25
Related Posts
關於總體經濟學的學習方式
By Olivia
at 2014-01-14T12:14
at 2014-01-14T12:14
樂學網生物課程
By Christine
at 2014-01-14T12:13
at 2014-01-14T12:13
補習班的課表
By Joseph
at 2014-01-14T11:31
at 2014-01-14T11:31
有關阿摩的題庫更新
By Gilbert
at 2014-01-14T11:13
at 2014-01-14T11:13
申請國營事業專板
By James
at 2014-01-14T10:00
at 2014-01-14T10:00