動力學-加速度問題 - 考試

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
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w = 角速度

a = 角加速度

v = 速度

an = 法線加速度

at = 切線加速度
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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！

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