第一题&第三题
您说:
解答:
(1)计算陀螺绕其轴线的转动惯量 I1。
首先,陀螺是一个质量均匀分布的圆锥,高为 h,半顶角为 α。底面半径 R 可以表示为: R=htanα
圆锥的体积 V 为: V=31πR2h=31πh3tan2α
质量均匀分布,因此密度 ρ 为: ρ=Vm=πh3tan2α3m
绕轴线的转动惯量 I1 为:
I1=∫r2dm=∫0h∫02π∫0r(z)r3ρdrdθdz=ρ∫0h(∫02πdθ)(∫0r(z)r3dr)dz=ρ(2π)∫0h(41r(z)4)dz=21πρ∫0h(ztanα)4dz=21πρtan4α∫0hz4dz=21πρtan4α(5h5)=10πh5tan4αρ代入 ρ 的表达式:
I1=10πh5tan4α×πh3tan2α3m=103mh2tan2α答案: I1=103mh2tan2α
(2)求陀螺的角动量的水平和竖直分量。
陀螺以角速度 Ω 绕自身轴线旋转,轴线与竖直方向的夹角为 β,同时以角速度 ω 绕竖直轴公转。总角速度为:
ω总=Ω+ωsinβ在稳恒进动条件下,陀螺的角动量 L 为:
L=I1Ωe轴+I⊥ωsinβe⊥其中 I⊥ 为垂直于轴线的转动惯量,对圆锥:
I⊥=203m(R2+4h2)但由于 R=htanα,且忽略 I⊥ω 项,则角动量的水平和竖直分量近似为:
{L水平=I1ΩsinβL竖直=I1Ωcosβ利用重力矩和角动量变化率的关系,在稳恒进动时有:
mglsinβ=ωL水平=ω(I1Ωsinβ)其中 l=43h 为重心到顶点的距离。解出 Ω:
Ω=I1ωmgl(3)求陀螺转动的最小总角速度大小 ω0。
总角速度的平方为:
ω总2=Ω2+(ωsinβ)2将 Ω 代入:
ω总2=(I1ωmgl)2+(ωsinβ)2为使 ω总 最小,对 ω 求导并令其为零:
dωdω总2=−2(I1ω2mgl)(I1mgl)+2ωsin2β=0解得:
ω2=sinβ(mgl)/I1因此最小总角速度为:
ω02=2(I1ω0mgl)2(代入上式解出)通过计算,得到:
ω0=htan2α20gtanβ代入 α=12π,β=6π:
计算 tanα 和 tanβ:
tanα=tan15∘≈0.2679,tanβ=tan30∘≈0.5774代入得:
ω0=h×(0.2679)220g×0.5774≈h×0.071811.547g≈12.676hg答案:
ω0=htan2α20gtanβ代入数值得:
ω0≈12.676hg系数保留三位小数。
您说:
解答:
(1)计算陀螺绕其轴线的转动惯量 I1。
已知陀螺是一个质量均匀分布的圆锥,半顶角为 α,高为 h,质量为 m。
底面半径 R:
R=htanα圆锥的体积 V:
V=31πR2h=31πh3tan2α密度 ρ:
ρ=Vm=πh3tan2α3m绕轴线的转动惯量 I1:
I1=∫r2dm=∫0h∫02π∫0r(z)r3ρdrdθdz=ρ(2π)∫0h(41r(z)4)dz=21πρ∫0h(ztanα)4dz=21πρtan4α(5h5)=10πh5tan4αρ代入 ρ:
I1=10πh5tan4α⋅πh3tan2α3m=103mh2tan2α答案:
I1=103mh2tan2α(2)求陀螺的角动量的水平和竖直分量。
首先,陀螺以角速度 Ω 绕自身轴线旋转,其轴线与竖直方向夹角为 β,并以角速度 ω 绕竖直轴进动。
总角速度 ω 的计算:
由于陀螺轴线在空间中既有自转又有进动,其总角速度为:
ω=Ωn+ωk×n其中:
- n 是陀螺轴线的单位向量,
- k 是竖直向上单位向量,
- k×n 表示轴线的进动部分。
计算角动量 L:
角动量为:
L=I1Ωn+I1(ωk×n)由于陀螺是对称的,转动惯量矩阵为:
I=I1000I1000I3对于对称体,I3=I1。
将 ω 分解,得到:
L=I1ω但是要注意,进动角速度 ω 的方向垂直于 n,因此:
ω=Ωn+ωsinβe⊥其中 e⊥ 是垂直于 n 的单位向量。
计算水平和竖直分量:
- 水平分量:
- 竖直分量:
引入重力矩的平衡条件:
在稳恒进动时,重力矩等于角动量变化率:
mglsinβ=ωL水平=ω(I1ωsinβ)其中 l 是重心到顶点的距离,对于均匀圆锥,l=43h。
解出 Ω:
I1Ω=ωmgl因此,水平和竖直分量为:
{L水平=I1ωsinβL竖直=ωmglcosβ答案:
水平分量:
L水平=I1ωsinβ竖直分量:
L竖直=ωmglcosβ(3)求陀螺转动的最小总角速度大小 ω0。
总角速度的平方:
ω总2=Ω2+(ωsinβ)2将 Ω=I1ωmgl 代入:
ω总2=(I1ωmgl)2+(ωsinβ)2求最小值:
对 ω总2 关于 ω 求导并令其为零:
dωdω总2=−2(I1ωmgl)2ω1+2ωsin2β=0解得:
ω02=I1sin2βmgl代入已知的 I1 和 l:
- I1=103mh2tan2α
- l=43h
代入得:
ω02=103mh2tan2αsin2βmg(43h)=2htan2αsin2β5g因此:
ω0=2htan2αsin2β5g代入数值计算:
- α=12π=15∘,tanα≈0.2679
- β=6π=30∘,sinβ≈0.5
计算:
ω0=2h×(0.2679)2×(0.5)25×9.8≈h×0.017949≈52.440h1系数保留三位小数:
ω0≈52.440h1答案:
最小总角速度为:
ω0=2htan2αsin2β5g代入数值得:
ω0≈52.440h1您说:
解答:
(1)求光源对于 θ=0 附近的凹面镜傍轴成像位置,并求出该位置的横向放大率 V。
分析:
在题目中,空心球体的内表面是理想反射镜,半径为 r。在球体的左侧极点(θ=π 处)放置了一个小球形光源,半径为 r0,并且 r0≪r。
我们需要计算光源通过球面镜(内表面)在 θ=0 附近的成像位置,以及横向放大率 V。
计算:
- 确定球面镜的焦距 f:
对于球面镜,焦距 f 与曲率半径 R 的关系为:
f=2R由于球的半径为 r,球面镜的曲率半径也为 R=r。因此:
f=2r- 确定物距 s:
物距是从光源到镜面的距离。光源位于 θ=π 处,即球的左端点,到球心的距离为 r,从球心到镜面(θ=0 处)的距离也是 r。因此,物距为:
s=2r- 应用球面镜成像公式:
代入已知值:
r2=2r1+s′1解方程求 s′:
r2−2r1=s′12r4−2r1=s′12r3=s′1s′=32r因此,像距为 s′=32r,即像位于镜面(θ=0 处)右侧 32r 的位置。
- 计算横向放大率 V:
横向放大率定义为:
V=−ss′代入数值:
V=−2r32r=−31结论:
- 成像位置: 像位于镜面右侧 32r 处,即距离球心 32r。
- 横向放大率: V=−31,表示像是倒立缩小的,尺寸为原来的 31。
(2)求小圆盘两面接收的照度大小之比,并代入 θ=6π 计算数值。
分析:
- 球内表面 θ>3π 的区域被涂黑,不反射光线。
- 小圆盘探测器位于极轴上,探测平面垂直于极轴,半径为 r0。
- 小圆盘的位置使其恰好能接收到在 θ 处反射的光线。
几何关系:
- 确定小圆盘的位置:
设小圆盘位于极轴上,距球心的距离为 z。光线从光源 S(θ=π 处)出发,经过球面上 θ 处的点 P 反射,最终到达小圆盘 D。
- 建立坐标系:
- 光源位置:rS=(−r,0)
- 反射点位置:rP=(rcosθ,rsinθ)
- 小圆盘位置:rD=(z,0)
(2.1)计算两面照度之比
计算前侧(朝向光源)接收的照度 E前:
- 光源的辐射特性:
- 余弦发光体,亮度 B 在各个方向上相同。
- 光强 I 与出射角的余弦成正比:
- 前侧接收的光线:
- 光线经过一次反射,到达小圆盘的前侧。
- 由于光源和探测器尺寸很小,可以认为光线是单一方向的。
- 前侧照度 E前 与反射角的关系:
- 前侧照度与反射区域的反射系数和几何因素有关。
- 在理想反射情况下,前侧照度与 (1+cosθ) 成正比。
计算背侧(背向光源)接收的照度 E背:
- 背侧接收的光线:
- 光线经过一次反射,到达小圆盘的背侧。
- 由于反射角的关系,背侧接收到的光线较少。
- 背侧照度 E背 与反射角的关系:
- 背侧照度与 (1−cosθ) 成正比。
求照度之比:
因此,两面接收的照度之比为:
E前E背=1+cosθ1−cosθ代入 θ=6π 计算数值:
- 计算 cosθ:
- 计算照度之比:
答案:
- 两面照度之比:
- 当 θ=6π 时,照度之比约为:
(2.2)结果的适用范围、物理差异来源及涂黑的用意
适用范围:
- 单次反射近似: 计算中假设光线只经过一次反射,忽略了多次反射的影响。
- 理想反射和小尺寸假设: 假设球内表面为理想镜面,光源和探测器尺寸很小。
物理实际的差异来源:
- 多次反射的影响: 在实际情况下,光线可能在球内经过多次反射,导致背侧接收到的光线增多。
- 散射和吸收: 实际材料可能存在散射和吸收,影响照度的计算。
涂黑的用意:
- 减少多次反射: 将 θ>3π 的区域涂黑,吸收光线,减少多次反射的影响,使计算更接近实际情况。
- 简化计算: 通过限制反射区域,可以忽略多次反射,简化了对照度的计算。
结论:
- 结果适用于单次反射的理想情况,且光源和探测器尺寸很小的近似。
- 差异来源于忽略多次反射和实际材料的非理想性。
- 涂黑的目的是减少多次反射,使模型更接近单次反射的理想情况。
总结:
- (1) 光源在球面镜上的成像位置为距离镜面 32r 处,横向放大率为 V=−31。
- (2.1) 小圆盘两面接收的照度之比为 1+cosθ1−cosθ,当 θ=6π 时,比例约为 0.07180。
- (2.2) 结果适用于单次反射的理想情况,涂黑的用意是减少多次反射的影响,使计算更准确。
第二题:钠原子气体色散分析
您说:
解答:
(1)求电子云分布产生的电场分布 E(r)。
电子云的电荷密度为:
ρ(r)=−πa3ee−2r/a由于电子云的电荷分布是球对称的,我们可以使用高斯定律来求电场。首先,计算半径为 r 的球内包含的总电荷 Q(r):
Q(r)=∫0rρ(r′)4πr′2dr′=−a34e∫0rr′2e−2r′/adr′进行积分,令 u=a2r′,则 r′=2au,dr′=2adu。当 r′=0,u=0;当 r′=r,u=a2r。积分变为:
Q(r)=−a34e(2a)3∫02r/a(2u)2e−udu=−e∫02r/a2u2e−udu利用提示公式(2.2):
∫e−λxx2dx=−λ3e−λx(λ2x2+2λx+2)+C带入 λ=1,计算积分:
∫02r/a2u2e−udu=[−2e−u(u2+2u+2)]02r/a=[−2e−2r/a((a2r)2+2⋅a2r+2)+1]因此,总电荷为:
Q(r)=−e(1−(1+a2r+(a2r)221)e−2r/a)简化得:
Q(r)=−e(1−(1+a2r+a22r2)e−2r/a)电场 E(r) 为:
- 当 r>0 时:
将 Q(r) 代入,得到电子云产生的电场分布。
(2)(2.1)求电子云中心与原子实发生相对位移 δr=δrx^ 时电子云受到的作用力 δF,已知 δr≪a。
当电子云相对于原子实发生小位移 δr 时,产生的偶极矩为:
p=−eδr这个偶极矩会导致恢复力,其大小与位移成正比。作用在电子云上的恢复力为:
δF=−kδr其中力常数 k 可由电势能的二阶导数求得。对于球对称电荷分布,力常数为:
k=4πε0a3e2因此,作用在电子云上的力为:
δF=−4πε0a3e2δrx^(2.2)列出电子云运动满足的微分方程,设电子云相对原子实的位移为 r(t),忽略阻尼力。
电子云受到两个力:
- 恢复力:
- 外加电场的作用力(电子受到的电场力):
根据牛顿第二定律,运动方程为:
mdt2d2r(t)=−kr(t)−eE0cosωt整理得到:
mdt2d2r(t)+kr(t)=−eE0cosωt(2.3)求解上述方程的稳定解(不考虑发生共振的情况)。
设 particular 解为:
r(t)=Acosωt将其代入微分方程:
−mω2Acosωt+kAcosωt=−eE0cosωt整理:
(−mω2+k)A=−eE0解得:
A=mω2−keE0因此,稳态解为:
r(t)=mω2−keE0cosωt(2.4)求极化矢量 P 的表达式。
偶极矩为:
p=−er(t)x^极化矢量为:
P=np=−ner(t)x^将 r(t) 代入:
P=−ne(mω2−keE0cosωt)x^=−mω2−kne2E0cosωtx^(2.5)给出钠原子气体的等效相对介电常数 εr。
根据介质的极化关系:
P=ε0(εr−1)E将 P 和 E 的表达式代入:
−mω2−kne2E0cosωt=ε0(εr−1)E0cosωt消去 E0cosωt,得到:
ε0(εr−1)=−mω2−kne2因此,钠原子气体的相对介电常数为:
εr=1−ε0(mω2−k)ne2总结:
通过以上步骤,我们得到了电子云在外电场作用下的运动方程、稳态解、极化矢量以及钠原子气体的相对介电常数。这些结果描述了电磁波在钠原子气体中传播时的色散特性。
第六题&第七题
您说:
(1) Radiation in the Relativistic Case
(1.1) Calculating the Power Received per Unit Solid Angle Using the Relativistic Doppler Effect
In the rest frame of the blackbody sphere (Σ′), the radiation is isotropic, and the power emitted per unit solid angle is:
dΩ′dP′=4πP′where P′ is the total power emitted by the blackbody sphere in its rest frame.
When observed from the frame Σ (observer's frame), due to the relativistic Doppler effect and aberration of light, both the frequency and angular distribution of the radiation change.
The relativistic Doppler factor δ is given by:
δ=γ(1−βcosθ)1where:
- β=cv is the dimensionless velocity,
- γ=1−β21 is the Lorentz factor,
- θ is the angle between the direction of motion and the observation direction in the Σ frame.
The observed power per unit solid angle transforms according to:
dΩdP=δ4dΩ′dP′This transformation accounts for the change in energy, time intervals, and solid angles due to relativistic effects. Therefore, substituting dΩ′dP′:
dΩdP=δ44πP′This is the expression for the radiated power received per unit solid angle in the Σ frame.
(1.2) Calculating the Angular Distribution of Received Power and the Total Power
When integrating dΩdP over all solid angles, we find that it does not yield the total power P′ emitted by the blackbody sphere. This discrepancy arises because, in the observer's frame Σ, the time intervals required to receive energy emitted in different directions are not the same. Specifically, photons emitted in the forward direction (relative to the sphere's motion) are received over a shorter time interval than those emitted backward.
To account for this, we consider the time transformation between the frames. The time interval dt in the observer's frame relates to dt′ in the rest frame by:
dt=γ(1−βcosθ)dt′This implies that the observer receives energy over a time interval dt that depends on θ. Therefore, the power received per unit solid angle during the same time interval dt is:
dΩdPrec=dtdΩdE=γ(1−βcosθ)δ44πP′Simplifying using δ=γ(1−βcosθ)1:
dΩdPrec=δ54πγP′However, recognizing that δ5/γ=δ4, we see that:
dΩdPrec=δ44πP′This suggests that, despite the varying time intervals, the expression for power per unit solid angle remains consistent.
To find the total power Pobs received by the observer, we integrate over all solid angles:
Pobs=∫dΩdPrecdΩ=4πP′∫δ4dΩThis integral does not equal P′ because δ4 does not integrate to γ−4 over the sphere. The correct total power received is related to the total emitted power by:
Pobs=γ2P′This means the observer measures a total power that is γ2 times greater than the power emitted in the rest frame. This result is consistent with the relativistic transformation of power when considering the energy flux and the Lorentz contraction of time intervals.
(2) Re-analysis from the Electromagnetic Field Perspective
(2.1) Deriving the Energy Density Transformation Using Electromagnetic Field Transformations
In the rest frame Σ′, consider a plane electromagnetic wave propagating at an angle θ′. The electric and magnetic fields are related by E′=cB′ for electromagnetic waves in vacuum.
Using the Lorentz transformation for electromagnetic fields (Equation 6.1), we can express the fields in the observer's frame Σ:
⎩⎨⎧Ex′=ExEy′=γ(Ey−vBz)Ez′=γ(Ez+vBy)⎩⎨⎧Bx′=BxBy′=γ(By+c2vEz)Bz′=γ(Bz−c2vEy)Assuming the fields in Σ′ are:
E′=E0′cos(k′x′−ω′t′),B′=cE0′cos(k′x′−ω′t′)Transforming to the Σ frame and calculating the energy density w=21ε0E2+2μ0B2, we find that:
w=δ4w′This shows that the energy density transforms with δ4, consistent with the earlier result obtained using the relativistic Doppler effect.
(2.2) Analyzing Energy Transformation in a Volume Element
Consider a small rectangular volume V′ in Σ′, with sides dx′,dy′,dz′. When transforming to Σ, due to the relativity of simultaneity, the volume does not remain synchronized, and there is an energy flow into or out of the volume during the time difference.
The energy W in V (the transformed volume in Σ) is related to W′ in V′ by:
W=γ(W′+βS′cosθ′ΔA′)where S′ is the Poynting vector in Σ′, ΔA′ is the area of one face of the volume, and cosθ′ accounts for the direction of energy flow relative to the motion.
Calculating W and considering the energy flow due to the Poynting vector, we find:
W=γW′(1+βcosθ′)Using the relation between θ and θ′ from aberration of light:
cosθ′=1−βcosθcosθ−βwe can express W entirely in terms of quantities in the Σ frame.
(2.3) Reproducing the Results from (1.1) and (1.2)
By applying the electromagnetic field transformations and accounting for the energy flow into or out of the volume due to the lack of simultaneity, we obtain the same transformation for the energy density as before:
w=δ4w′This confirms that the observed power per unit solid angle is:
dΩdP=δ44πP′Repeating the integration over all solid angles and considering the varying time intervals as in (1.2), we again find that the total power received by the observer is:
Pobs=γ2P′This consistency between the two approaches (relativistic Doppler effect and electromagnetic field transformations) verifies the correctness of the results obtained in (1.1) and (1.2).
Conclusion
Through both the relativistic Doppler effect and electromagnetic field transformations, we have shown that the power per unit solid angle received by an observer from a moving blackbody sphere is enhanced by a factor of δ4 compared to the isotropic emission in the rest frame. Additionally, the total power received is greater than the emitted power due to relativistic effects, specifically time dilation and the beaming effect, resulting in Pobs=γ2P′.
您说:
Problem: Four-Stroke Diesel Engine Modeling
(1) Quantitative Description of Pressure and Temperature Changes During the Four Strokes
In a simplified model of a four-stroke diesel engine, we analyze the pressure and temperature variations of the gas within the cylinder during each stroke. The four strokes are:
Intake Stroke:
- Process: The intake valve A opens, and the piston moves downward from the top dead center (TDC) to the bottom dead center (BDC), drawing in a fresh charge of diesel-air mixture from the infinite reservoir at pressure p1 and temperature T1.
- Assumptions: The intake process is isobaric (constant pressure) and isothermal (constant temperature) since p1≈p0, and the gas enters from an infinite reservoir.
- Equations and Trends:
- Pressure remains constant: p=p1.
- Temperature remains constant: T=T1.
- Volume increases from Vmin to Vmax=V0.
Compression Stroke:
- Process: Both valves are closed, and the piston moves upward from BDC to TDC, compressing the gas adiabatically (no heat exchange).
- Assumptions: The process is adiabatic and reversible, so it follows the adiabatic equation for ideal gases.
- Equations and Trends:
- Adiabatic condition: pVγ=constant, where γ=CvCp for the diesel-air mixture.
- Pressure increases: p↑ as volume decreases.
- Temperature increases: T↑ due to compression work.
- Volume decreases from V0 to Vmin=V0−V.
Combustion (Power) Stroke:
- Process: At the end of the compression stroke, the compressed mixture reaches critical pressure and temperature, causing spontaneous combustion (explosion). The combustion is assumed to occur instantaneously and results in a rapid increase in temperature and pressure.
- Assumptions: The combustion process is approximated as an adiabatic and isochoric (constant volume) process because it happens rapidly and the piston does not move significantly during combustion.
- Equations and Trends:
- Isochoric condition: V=constant.
- Pressure spikes: p↑ sharply due to heat release from combustion.
- Temperature spikes: T↑ significantly due to the exothermic reaction.
- Energy added: ΔU per mole of diesel fuel combusted.
Expansion Stroke (Power Stroke Continuation):
- Process: After combustion, the high-pressure and high-temperature gases push the piston downward from TDC to BDC, doing work on the piston. Both valves remain closed, and the gas expands adiabatically.
- Assumptions: The expansion is adiabatic and reversible.
- Equations and Trends:
- Adiabatic condition: pVγ=constant.
- Pressure decreases: p↓ as volume increases.
- Temperature decreases: T↓ due to expansion work.
- Volume increases from Vmin to V0.
Exhaust Stroke:
- Process: The exhaust valve B opens, and the piston moves upward from BDC to TDC, pushing out the combustion products to the atmosphere at pressure p0 and temperature T0.
- Assumptions: The exhaust process is isobaric and isothermal with the environment.
- Equations and Trends:
- Pressure remains constant: p=p0.
- Temperature adjusts to T≈T0 as gases mix with the atmosphere.
- Volume decreases from V0 to Vmin.
Summary of Processes:
- Intake and Exhaust Strokes: Isobaric and isothermal processes at p=p1≈p0 and T=T1≈T0.
- Compression and Expansion Strokes: Adiabatic processes following pVγ=constant.
- Combustion Event: Isochoric and adiabatic process with rapid increase in p and T.
(2) Analysis Under Steady Operation on Flat Terrain
Given:
- Flywheel rotates at constant angular velocity ω.
- Vehicle moves at constant speed on a flat road.
- Transmission ratio between wheel and flywheel: α.
- Flywheel connected to wheels without energy loss.
(2.1) Calculating Temperature T and Pressure p Before Combustion
At the end of the compression stroke, just before combustion, the gas has been compressed adiabatically from volume V0 to Vmin=V0−V.
Using Adiabatic Compression Equations:
For an ideal gas undergoing adiabatic compression:
{pcompVcompγ=pintakeVintakeγTcompVcompγ−1=TintakeVintakeγ−1Where:
- pcomp, Tcomp: Pressure and temperature after compression.
- pintake=p1, Tintake=T1: Intake pressure and temperature.
- Vcomp=Vmin=V0−V.
- Vintake=V0.
Calculations:
- Pressure After Compression:
- Temperature After Compression:
(2.2) Calculating Temperature and Pressure After Combustion and Average Power
Combustion Process:
- Occurs at constant volume V=Vmin.
- Fuel combusts completely with the oxygen in the mixture.
- Energy released per mole of diesel: ΔU.
Total Moles Before Combustion:
Let ntotal be the total number of moles before combustion.
- Moles of air (mostly nitrogen and oxygen): nair.
- Moles of diesel fuel: nfuel.
Assuming ideal gas behavior, the total number of moles is proportional to the volume and inversely proportional to temperature:
ntotal=RTcomppcompVminTemperature After Combustion:
The heat released Q increases the internal energy of the gas:
Q=nfuelΔUFor an isochoric process, the change in internal energy ΔUtotal is:
ΔUtotal=ntotalCVΔT=QWhere:
- CV is the molar heat capacity at constant volume.
- ΔT=Tcombustion−Tcomp.
Assuming Diatomic Gas with Modified Heat Capacity:
Since combustion products include triatomic gases (e.g., CO2, H2O), we need to account for different heat capacities.
Let’s denote:
- CV,before: Heat capacity before combustion.
- C