圓周運動 (位移&速度)

2011-03-02 4:19 am
一個物體以每秒(1/3)圈角速度沿半徑為2m的圓移動.....


我想問點樣計出

(1)物體在0.5s內角位移是(兀/3)

(2)物體在0.5s的位移是2m


&想問 向心加速度只是方向轉變?. 量值不變? 平均的加速度是怎樣得出?(因為量值不變) [ 另外..(v ^ 2) / r ] 這一公式的v是線速率..那 速度是怎樣計?..
因是圓型... 移動的只有角度....


請高手指點!!!!!.

回答 (1)

2011-03-02 4:49 am
✔ 最佳答案
(1) angular velocity = 2兀/3 s^-1
angular displacement = (2兀/3) x 0.5 radians = 兀/3 radians

(2) since angular displacement = 兀/3 radians = (兀/3).(180/兀) degrees = 60 degrees
Displacement = 2 x [2.sin(60/2)] m = 2 m

Q: 平均的加速度是怎樣得出?
The average acceleration is given by v^2/R (where v is the speed and R is the radius of the circular motion) in a radial direction towards the centre.

Q: 那 速度是怎樣計?..
If you refer to the instantaneous velocity, it is v with direction tangent to circular path at the point where the object is.
But if you refer to the average velocity, it is the linear displacement over the time taken.
For example, if an object moves with constant speed of 10 m/s around a circle of raduse 2 m. Then the time to complete half a circle is 2.pi.2/10 s = 2.pi/5 s.
The average velocity of the object in moving over a time interval of, say pi/5 s (i.e. half a circle) = diameter/(pi/5) m/s = 4/(pi/5) m/s = 20/pi m/s


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