※ 引述《ib61632003 (北北)》之銘言：
: 小弟有三題普物想請教
: 希望能有強者解答，要有計算過程和解釋
: 1.A uniform thin wire is bent into a semicircle of radius r. Determine
: the coordinates of its center of mass with recpect to an origin of coordinates
: at the center of the ''full'' circle.
因為the thin wire is uniform
根據點對稱性質..可知圓線質心在幾何中心處..
將圓座標設為幾何中心處..
令質塊δm位置向量與半圓中段位置向量的夾角為θ..設線密度為α..
根據面對稱性質..半圓質心會位於幾何中心與半圓中段之間的位置..
則質心距與幾何中心位置距離為..
∫(-π/2→π/2) (ρ*r*δθ)*(r*cosθ) / ρ*π*r
= 2*R0/π
: 2.A centrifuge rotor is accelerated from rest to 2000 rpm in 30s.
: (a) What is its average angular acceleration? (b) Through how many
: revolutions has the centrifuge rotor turned during its acceleration period,
: assuming constant angular acceleration?
2000 rpm = (200*π/3) rad/s
α(average) = (200*π/3)/30 = (20*π/9) rad/s^2
根據angular acceleration is constant
設旋轉數為N..則
N = 2000*(30/60)*(1/2) = 500
: 3.A bullet of a mass m moving with velocity v strikes and becomes
: embedded at the edge of a cylinder of mass M and radius R0. The cylinder,
: initially at rest ,begins to rotate about its symmetry axis,which remains
: fixed in position.Assuming no frictional torque,what is the angular velocity
: of the cylinder after this collision?What is the change in kinetic energy?
假設圓柱是均勻的..設體密度為ρ..高度為h..
設symmetry axis為旋轉軸之轉動慣量 I(cyl)..則
I(cyl) = ∫(0→R0) (2*π*r*δr)*ρ*h*r^2
= ρ*π*R0^4*h/2
= M*R0^2/2............................(a)
設angular velocity為ω
根據角動量守恆
m*v*R0 = [I(cyl) + m*R0^2]*ω
將(a)代入上式解得
ω = m*v*/{[(M/2)+m]*Ro}
設the change in knetic energy為ΔE
則ΔE = (1/2)*m*v^2 - (1/2)*[I(cyl) + m*R0^2]*ω^2
= (1/2)*m*v^2*{1-m/[(M/2)+m]}
: 感謝耐心閱讀
: 謝謝回答
: 微薄P幣以表心意
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※ 編輯: kramnik 來自: 122.116.104.102 (01/16 08:30)