※ 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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