一人騎腳踏車以10m/s的速度前進

用戶37276

2013-01-26 7:59 am

1.一人騎腳踏車以10m/s的速度前進當其進入水平彎路後車申傾斜一角度若彎路之曲率半徑為20米則車身與鉛直方向所成之角度為西搭若g=10m/s平方 tan西搭為何

2.承上題車輪與路面摩擦係數最小須為多少

更新1:

1.mgtan西搭=m*v平方R tan西搭=V平方/gR =10平方/10*20 =1/2 2.f=mgtan西搭<=usmg us>=tan西搭=1/2 u最小為1/2

回答 (1)

天同

2013-01-26 8:25 am

✔ 最佳答案

1. Let R be the normal reaction, and Ff be the frictional force on the bicycle tyre
For vertical equilibrium, R = mg where m is the mass of the bicycler and its rider

Take moment about the centre of mass of the bicycle and its rider
R.[h.sin(theta)] = Ff[h.cos(theta)]
where h is the height of the centre of mass above ground.
hence, mg.sin(theta) = Ff.cos(theta)
But Ff provides for the centripetal force for the bicycle to turn the corner
hence, Ff = m(10^2)/20

Therefore, mg.sin(theta) = [m(10^2)/20].cos(theta)
tan(theta) = 10^2/20g = 0.5
(theta) = 26.6 degrees

2. Ff = u.R where u is the coefficient of friction
Hence, u.R = m(10^2)/20
u.(mg) = m(10^2)/20
u = 10^2/20g = 0.5

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