Re: [題目] 牛頓力學...
※ 引述《alwyner (Time is money!!)》之銘言：
: ※ [本文轉錄自 Physics 看板]
: 作者: alwyner (Time is money!!) 看板: Physics
: 標題: [題目] 牛頓力學...
: 時間: Sun Apr 27 13:34:07 2008
: 1.阻力R = kv ，質量m的小鋼球從油的表面(y=0)靜止釋放，k為常數，導出在h深處，此
: 球所到達的速度v。
: Ans: h =?
y is negative for downward
v is negative for downward
v = y'
my" = -mg - ky'
set v = y'
=> v' = -g - (k/m)v
=> v = Aexp(-kt/m) - mg/k
v(t = 0) = 0 => A = mg/k
And y' = v = (mg/k)[exp(-kt/m) - 1]
2
=> y = -g(m/k) exp(-kt/m) - mgt/k + C
2
y(t = 0) = 0 => C = g(m/k)
Therefore,
2 2
y = -g(m/k) exp(-kt/m) - mgt/k + g(m/k)
v = (mg/k)[exp(-kt/m) - 1]
The question asks us to find the │v│ given h, but your answer is h = ??
I'm confused.
: 2. ／
: m ／＼／ │
: ＼／ │
: 夾角 θ ／ M │
: ↘ ／ │
: ／ │
: │￣￣ │
: ￣￣￣￣￣￣￣￣￣
: 忽略所有摩擦，求a)M的加速度 b)m相對於M的加速度。
Let A(->) be the acceleration of M, a the component of the acceleration of m
"relative" to M in the x-direction .
Let N' be the normal force between m and M, N between M and floor.
positive direction:upward and rightward
Assume A>0:(->), a>0:(<-)
Momentum conservation in the x-directon plus differentiation about time:
=> MA + m(-a + A) = 0 ---[1]
Force applied in the y-direction:
=> -ma*tanθ = -(m+M)g + N ---[2]
For big block in the y-direction:
=> -N'cosθ - Mg + N = 0 ---[3]
For big block in the x-direction:
=> N'sinθ = MA ---[4]
There is enough information for you to solve A.
----------------------------
From [3] and [4] => N = M(A*cotθ+ g ) ---[5]
2
Substitute [1] and [5] into [2] => A = mgsinθcosθ/(M + m*sinθ)
the acceleration of m relative to M is a*secθ
2
From [1] => a*secθ = (M + m)A*secθ/m = (M + m)g*sinθ/(M + m*sinθ)
Period.
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◆ From: 122.124.100.190
※ 編輯: Honor1984 來自: 122.124.100.190 (05/13 03:13)
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