有一直徑1公尺的圓筒 從一斜坡(垂直高度10公尺 水平夾角20度的斜坡)滾下 求1.不計摩擦力與空氣阻力 滾5秒後在斜坡滾動距離是多少? 滾動碰到地面需幾秒? 2.不計空氣阻力 但摩擦係數是0.5 筒重100公斤 則滾5秒後在斜坡滾動距離是多少? 滾動碰到地面需幾秒?

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���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