Physics 5300, Theoretical Mechanics Spring 2015
Assignment 3 solutions
Given: Wed, Jan 28,
Due Tue Feb 3
The problems numbers below are from Classical Mechanics, John R. Taylor, University
Science Books (2005).
Problem 1
Solution:
Taylor 7.1
We have
L=T −V
(1)
1
T = m(ẋ2 + ẏ 2 + ż 2 )
2
V = mgz
Thus
1
L = m(ẋ2 + ẏ 2 + ż 2 ) − mgz
2
(2)
(3)
(4)
The Lagrange equations are
x:
d ∂L
∂L
( )−
= 0,
dt ∂ ẋ
∂x
mẍ = 0
(5)
y:
∂L
d ∂L
( )−
= 0,
dt ∂ ẏ
∂y
mÿ = 0
(6)
∂L
d ∂L
( )−
= 0, mz̈ + mg = 0
(7)
dt ∂ ż
∂z
These are as expected since the x and y directons feel no gorce, while the z moton gives
mz̈ = −mg so the projectile feels a downward force mg.
z:
Problem 2
Solution:
Taylor 7.2
The force F = −kx comes from a potential
1
V = kx2
2
(8)
d
V
dx
(9)
since we have
F =−
1
Then we get
1
1
L = T − V = mẋ2 − kx2
2
2
(10)
d ∂L
∂L
( )−
= 0,
dt ∂ ẋ
∂x
(11)
The Lagrange equation is
x:
The solution is
r
x = A cos(
Problem 3
Solution:
mẍ + kx = 0
k
t + φ)
m
(12)
Taylor 7.3
We have
1
1
L = m(ẋ2 + ẏ 2 ) − k(x2 + y 2 )
2
2
(13)
The Lagrange equations are
Problem 4
x:
∂L
d ∂L
( )−
= 0,
dt ∂ ẋ
∂x
mẍ + kx = 0
(14)
y:
d ∂L
∂L
( )−
= 0,
dt ∂ ẏ
∂y
mÿ + ky = 0
(15)
Taylor 7.10
Solution: The coordinates are ρ, φ. The half angle is α, and the cone points down.
Consider any z < 0. Then we have
tan α =
Thus
ρ
(−z)
(16)
ρ
tan α
(17)
x = ρ cos φ
(18)
y = ρ sin φ
(19)
z=−
We also have
Problem 5
Taylor 7.17
2
Solution:
We have
1
1
1
T = m1 ẏ12 + m2 ẏ22 + Iω 2
2
2
2
V = m1 gy1 + m2 gy2
(20)
(21)
We have the constraint
y1 + y2 = Y = constant
(22)
ẏ1 = ẏ2
(23)
We solve this constraint. Thus
We also have
v
ẏ1
=
R
R
where R is the radius of the pulley. Thus we have
ω=
I
1
T = (m1 + m2 + 2 )ẏ12
2
R
(25)
V = (m1 − m2 )gy1 + m2 gY
(26)
1
I
L = T − V = (m1 + m2 + 2 )ẏ12 − (m1 − m2 )gy1 − m2 gY
2
R
The Lagranges equation is
d ∂L
∂L
y1 :
(
)−
=0
dt ∂ ẏ1
∂y1
I
)ÿ1 − [−(m1 − m2 )g] = 0
R2
(m1 − m2 )g
ÿ1 = −
(m1 + m2 + RI2 )
(m1 + m2 +
The solution is
y1 = y10 + v10 t −
Problem 6
Solution:
1 (m1 − m2 )g 2
t
2 (m1 + m2 + RI2 )
(27)
(28)
(29)
(30)
(31)
Taylor 7.20
We have
ż = λφ̇
Thus
(24)
1
1
R2
1
R2
T = m(ż 2 + R2 φ̇2 ) = m(ż 2 + 2 ż 2 ) = m(1 + 2 )ż 2
2
2
λ
2
λ
V = mgz
3
(32)
(33)
(34)
Thus
1
R2
L = m(1 + 2 )ż 2 − mgz
2
λ
(35)
The Lagrange equation is
z : m(1 +
R2
)z̈ − (−mg) = 0
λ2
z̈ = −
g
(1 +
R2
)
λ2
(36)
(37)
In the limit R → 0 we get
z̈ = −g
(38)
which makes sense since in this case we just have a vertical wire, and the bead will slide
straight down with acceleration −g.
4