長25cm之輕繩 - Ans.fyi 參考答案

長25cm之輕繩

用戶37276

2013-01-22 9:28 am

長25cm之輕繩上端固定於天花板下端懸掛6kg之物體若物體沿水平作等速率圓週運動 (西撘=53度) 則

1.張力為何

2.物體所受之向心力量值為何

3.速率為何

4.週期為何

5.若西搭改成37度則週期為何

更新1:

答案是1.2.3.4 1.Tcos西搭=mg T=mg/cos西搭 =6*10/cos53度 =100N 2.Tsin西搭=100*4/5 =80N 3.100=m*v平方/R v平方=80*R/m =8/3 v=2根號6/3 4.T=2拍*根號h/g =2拍*根號0.15/10 =根號6/10*拍

回答 (1)

天同

2013-01-22 4:19 pm

✔ 最佳答案

1. I suppose angle (theta) is the angle the string makes with the vertical.
Let T be the tnesion in string.
T.cos(53) = 6g
T = 60/cos(53) N = 99.7 N (take g = 10 m/s^2)

2. Centripetal force = T.sin(53) = 99.7 x sin(53) N = 79.62 N

3. Centripetal acceleration = 79.62/6 m/s^2 = 13.27 m/s^2
Radius of cicular motion = 0.25.sin(53) m = 0.2 m
Let v be the speed of the 6 kg mass
v^2/0.2 = 13.27
v = 1.629 m/s

4. Period = 2.pi x 0.2/1.629 s = 0.771 s

5. Repeat the above calculation with angle (theta) = 37 degrees.

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