物理圓周運動

物理圓周運動

goodgood

2012-03-26 6:47 pm

一個重量為5N的方塊放在1個水平轉盤上, 方塊和轉盤之間可產生的最大摩擦力為2N 假如轉盤以每秒6周轉動,在方塊沒有打滑的情況下與轉盤中心最大大可相距多少?

一個錐擺中一顆質量為0.4kg的小珠沿著一個水平圓至路徑轉動繩子的的長度為L=1m 與垂直方向成角◎ 繩子可抵反的最大張力為Tamx=10N
原本是有圖的但沒法上載上來 SORRY

本人物理很WEAK 請列出詳細步驟說明如有不便敬請原涼

回答 (1)

天同

2012-03-26 10:16 pm

✔ 最佳答案

1. 6 revolution per second = 6 x2.pi radians/s = 37.7 radians/s
Centripetal force (向心力) needed = (5/g).R.(37.7^2)
where g is the acceleration due to gravity, taken to be 10 m/s2
R is the maximum radial distance from the centre.

The centripetal force is provided by friction, hence
(5/g).R.(37.7^2) = 2
solve for R gives R = 2.81 x 10^-3 m = 0.28 cm

2. What do you like to calculate?

2012-03-26 19:49:28 補充：
2. 10.sin(◎) = (0.4).R.w^2
where R is the radius of the circle and w is the angular velocity (角速度)
but R = 1 x sin(◎)
hence, 10.sin(◎) = (0.4).sin(◎).w^2
w = square-root[10/0.4] s^-1 = 5 s^-1

2012-03-26 19:52:45 補充：
1. Your equation: 2=5(r)37.7^2 is wrong
It should be 2=(5/g).(r)37.7^2
since 5 N is the weight, (5/g) is the mass of the cube
r = (2g/5)/(37.7)^2 m = 2.81x10^-3 m

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物理 圓周運動

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