✔ 最佳答案
直徑1m圓筒滾下一斜坡 = 垂直高度10公尺 + 水平夾角20度
(1) 不計摩擦力.空氣阻力.滾5秒後滾動距離=?
能量不滅: v = 圓周速率 = 圓心速率
m*g*h = m*v^2/2 + I*w^2/2
= m*v^2/2 + (m*r^2/2)*w^2/2
= m*v^2*(1/2 + 1/4)
=> g*h = 3*v^2/4
=> v = √(4g*h/3)
= √(4*9.8*10/3)
= 11.43 m/s
=> L = h/sin20
= 10/0.342
= 29.238 m
=> a = (V^2 - 0)/2L
= 11.43^2/2*29.238
= 2.23 m/s^2
L(t) = 0.5*a*t^2
=> L(5) = 0.5*1.495*25
= 27.93 m
= Answer
(2) 滾動碰到地面需幾秒?
t = √(2L/a)
= √(2*29.238/2.23)
= 5.12"
= Answer
3.不計空氣阻力, 滑動摩擦係數u=0.5, 5秒後滾動距離=?
k = 滾動摩擦係數
≒ 滑動摩擦係數/20
= 0.5/20
= 0.025
能量不滅:
m*g*h = m*v^2/2 + I*w^2/2 + k*m*g*cos20*L
= m*v^2/2 + m*v^2/4 + k*m*g*cos20*L
=> g*h = 3v^2/4 + k*g*L*cos20
=> v^2 = g*(h - k*L*cos20)*4/3
=> v = √[g*(h - k*L*cos20)*4/3]
= √[g*(h - k*L*cos20)*4/3]
= √[9.8*(10 - 0.025*29.238*0.9397)*4/3]
= 11.03 m/s
a = v^2/2L
= 11.03^2/2*29.238
= 2.081 m/s^2
L(t) = 0.5*a*t^2
=> L(5) = 0.5*2.081*25
= 26.013 m
= Answer
4.滾動到地面需幾秒?
t = √(2L/a)
= √(2*29.238/2.081)
= 5.30"
= Answer
Note: 在機械設計裡面.滾珠軸承都是取: k = u/20