Phys. 35100CD Mechanics PS #5 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) From the tidal potential t =
Gm
2
D 3 (x
1
+ 2 y 2 ), we could derive the tidal accelerations
Gm
= t = 2D3 x
x
= yt = Gm y
D3
gx
gy
b) The total po
Phys. 35100CD Mechanics PS #4 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) We have a damped oscillator: m x + b x + k x = 0. Using the new set of parameters o =
= b/(2m)
2
x + 2 x + 0 x = 0
k /m and
2
2
For underdamped os
Phys. 35100CD Mechanics PS #3 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1
a) If the force elds are conservative, they must be expressible as F = U , where U is the scalar potential.
Applying the curl on these force elds, we h
Phys. 35100CD Mechanics PS #2 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1
(a) In the required coordinate system, the equation of motion will be
m
dv
= mg v 2
dt
The terminal velocity u will satisfy the condition: v =
mg u2 =
Phys. 35100CD Mechanics PS #6 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) The kinetic energy of the system is
T=
1
1
mx2 + M y 2
2
2
b) The gravitational potential energy is
Ug = mgx M gy
c) The elastic energy of the spri
Phys. 35100CD Mechanics PS #7 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) Due to the nite velocity and acceleration, the position and velocity are continouse in time. So right after
t = 0, we have
x(0) = x(0) = 0
b) The p
Phys. 35100CD Mechanics PS #10 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) The disk has a azimuthal symmetry through its z -axis passing through its center, meaning I11 = I22 . The
perpendicular axis theorem tell us that
Phys. 35100CD Mechanics PS #10 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) As derived in class, the velocities of the two balls after collision will be at a right angle to each other and will
add up vectorially to the ini
Phys. 35100CD Mechanics PS #9 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) The potential energy is
U = G
Mm
r
where M is the mass of the Earth. The acceleration on the surface of the Earth is g = GM/R2 , so
U = mg
R2
r
For
Phys. 35100CD Mechanics PS #8 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1:
a) We divide the chain into two parts: the right part with mass
moving at speed x. The total kinetic energy of the chain is
T=
x
lm
and the left part
Phys. 35100CD Mechanics PS #1 Solution
Instructor: Alexios Polychronakos
Grader: Weikang Chen
Problem 1. 1.10
(a)From r = 2b sin t i + b cos t j , we have
=
dr
= 2b cos t i b sin t j
dt
dv
a=
= 2b 2 sin t i b 2 cos t j = 2 r
dt
v=
And speed is
4b2 2 cos2