課程名稱︰力學下
課程性質︰物理系必帶
課程教師︰龐寧寧（陳義裕代課）
開課學院：理學院
開課系所︰物理學系
考試日期（年月日）︰2019年4月8日
考試時限（分鐘）：180分鐘
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題 :
Mechanics II, Spring 2019
Instructor: Yih-Yuh Chen
Midterm
This exam is due 6:30 pn, Monday, April 8, 2019.
為方便顯示，x'表示dx/dt、ex表示x單位向量、dbar為義裕老師上課的notation
1.(40 points) Euler-Lagrange equation reads
d ∂L ∂L
─(──)=──.
dt ∂q' ∂q
In polar coordinates, the Lagrangian L for a central filed problem with pot-
ential energy U(r) is
L=μ/2(r'^2+r^2θ'^2)-U(r),
where μ is some constant.
(a)(5 points) Use Euler-Lagrange equation to show that
∂L/∂θ'=μr^2θ'
is a constant of the motion. In the following, we will denote this const-
ant by l, thus,
μr^2θ'=l=constant. (1)
(b)(5 points) Use Euler-Lagrange equation to derive
μ(r''-rθ'^2)=-∂U/∂r
(c)(5 points) Show that
E≡μ/2(r'^2+r^2θ'^2)+U(r) (2)
is also a constant of the motion.
(d)(10 points)Eqn.(1) means
1/dt=1/(μr^2)1/dθ.
For Kepler's planetary problem, U(r)=-k/r for some positive constant k.
Let us define
u≡1/r.
Please show that Eqn.(2) becomes
E=l^2/(2μ)((du/dθ)^2+u^2)-ku. (3)
(e)(5 points)Completing squares, we can recast Eqn.(3) into
l^2 du 2 μk 2 l^2 μk 2
E=──((─) +(u-──) )-──(──) ,
2μ dθ l^2 2μ l^2
which becomes
2μ μk 2 du 2 μk
(──E+(──) )=(─) +(u-──)^2. (4)
l^2 l^2 dθ l^2
Please verify that
μk 2μ μk 2
u-── = √(──E+(──) )cosθ (5)
l^2 l^2 l^2
is a solution to Eqn.(4).
(f)(10points) Explain why Eqn.5 tells us that the orbit is a conic section,
then find the eccentricity εof this conic section in terms of the const-
ants μ,k,l, and E.
2.(40pts) The inertia tensor I can be viewed as a real symmetric operator. Let
R be a rotation by some definite angle φ about some axis we will convenien-
tly call the z-axis, assuming that φ≠0 or π. Suppose I enjoys the rotati-
onal semmetry under R, that is, for any vector u we always have
R(I(u))=I(R(u)) (6)
(a)(5 points)Consider the action of I on the unit vector ez. Why is it that
I(ez)=I(R(ez))?
(b)(5 points)Use Eqn.(6) to argue that I(ez)正比於ez. Thus, ez is necessari-
ly an eigenvector of I. Let's denote its associated eigenvalue by λ3.
(c)(5 points)Since the x-y plane is orthogonal to the eigenvector ez, accor-
ding to the proof in class, we know the other two eigenvectors of I must
lie in the x-y plane! Let e1 be the second eigenvector in the x-y plane,
that is,
I(e1)=λ1e1
for some real number λ1. Use Eqn.(6) to show that R(e1) is also an eige-
nvector of I, with the same eigenvalue λ1.
(d)(5 points)Argue why R(e1) can not be proportional to e1, and thus conclu-
de that e1 and R(e1) are two independent eigenvectors, with the same eig-
envalue λ1.
(e)(5 points)Explain why the matrix representation of I is of form
λ1 0 0
( 0 λ1 0 )
0 0 λ3
in the chosen Euclidean basis.
(f)(15 points)(You can setup any coordinate system you like to solve this
problem. Alternatively, you are welcome to freely quote the results from
Parts (a)-(e) above to simplify the calculations for this part. Either
method is fine.) The molecule CH3Cl has a three-dimensional geometry sh-
own in Fig.1 below, with the three hydrogen atoms and the chlorine atom
together forming a tetrahedron which is elongated in the direction of the
carbon-chlorine bond. We will treat all the atoms as particles. The iner-
tia tensor is defined as
I=(Σm_jr_j^2)I-Σm_jr_j張量積r_j
j j
Taking the carbon atom (the only one lying on the plane) as the origin,
and assume that the mass of Cl and H are M and m, repectively. (Relevant
lengths a,b, and c are also shown in the figure for your convenience.)
1.(5 points) Find the principal axes.
2.(10 points) Calculate the principal moments of inertia for this molecu-
le.
https://imgur.com/AmP1TcF
Fig.1: 3-D structure of CH3Cl molecule.
3.(20 points) A rectangular coordinate system x-y-z is rotating about the lab
at some constant angular velocity ω=ωez. The time derivative of a vector
A≡Axex+Ayey+Azez can be written as
dA/dt=(Ax'ex+Ay'ey+Az'ez) + (Axex'+Ayey'+Azez')
=(Ax'ex+Ay'ey+Az'ez) + (Axω×ex+Ayω×ex+Azω×ex)
≡dbar A/dt + ω×A,
where dbarA/dt is to be identified with the time derivative of A as measured
by the rotating observer.
(a)(5 points)Show that the position vector r of a particle satisfies
d^2r/dt^2=dbar^2r/dt^2+2ω×dbar r/dt+ω×(ω×r)
(b)(5 points)With F=md^2r/dt^2 for a force F acting on the particle of mass
m, we may cast Eqn.(7) into
mdbar^2r/dt^2=F+(dbar r/dt×2mω)+(-mω×(ω×r)).
The third term can be written as
(-mω×(ω×r))=-▽V(r)
for some function V(r). Please find a suitable form for V(r).
(c)(10 points)A closed shallow cylindrical water tank is fast spinning about
the z- axis of the figure below. The water inside is co-rotatinh with the
tank so that it appears to be motionless to the rotating observer. A tiny
air bubble is created somewhere at half the radius of the tank, with no
initial velocity to the rotating observer. Ignoring local gravity of the
earth, please describe the motion of the bubble relative to the rotating
observer and also explain why you think it moves that way.
https://imgur.com/KRAArxc
Fig.2:Air bubble in a rotating tank filled with water.