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PC2132 CLASSICAL MECHANICS
AY2004/2005 Semester 2 Part I
maximum velocity, vmax = v + v 0 minimum velocity, vmin = v − v 0
} vmax + vmin = 2v vmax − vmin = 2v 0
From conservation of angular momentum,
mrmaxvmin = mrminvmax ⇒ rmax rmin
= vmax vmin
From the geometrical property of ellipses,
rmin = a(1 − e) , rmax = a(1 + e) ⇒
rmax rmin^ =^
1 + e 1 − e.^ (2)
From (1) and (2),
vmax(1 − e) = vmin(1 + e) vmax − vmin = e(vmax − vmin) 2 v 0 = 2 ve e =
v 0 v
- x = 13 ωgt^3 sin θ, where θ = 90◦− latitude = 48◦. Here, t is the time to fall vertically through z = 27m, assuming starting from rest.
z =^1 2
gt^2 ⇒ t =
√ 2 z g
x = 1 3
ωg
√ 8 z^3 g^3
sin θ
= ω 3
√ 8 z^3 g
sin θ
2 π 24 × 60 × 60
√ 8 × 273
- 8 sin 48
◦
= 2. 28 mm, to the east.
r∗ 1 = r 1 − Rcm r∗ 2 = r 2 − Rcm M = m 1 + m 2
T =
2 m^1 r˙^1
2 +^1
2 m^2 r˙^2
2
= 1 2
m 1 ( r˙∗ 1 + R˙cm)^2 +^1 2
m 2 ( r˙∗ 2 + R˙cm)^2
= 1 2
M R˙^2 cm +^1 2
m 1 r˙∗ 12 +^1 2
m 2 r˙ 2 ∗^2 ︸ ︷︷ ︸ T ∗
- R ︸˙ cm(m 1 r︷︷˙∗ 1 + m 2 r˙∗ 2 )︸ =0 since ∑^ mi R˙cm= = 12 M R˙^2 cm + T ∗
Constraint equation: X 1 + X 2 = L, X^ ˙ 1 = − X˙ 2.
T 1 = 1
· 3 M X˙ 12 =^3
M X˙ 22.
V 1 = 0.
T 2 = 12 · 3 M X˙ 22 =^32 M X˙ 22.
v 2 = − 3 M gX 2 sin θ.
L = 3 M X˙^22 + 3M gX 2 sin θ. ∂L ∂x =^3 M g^ sin^ θ. ∂L ∂ x˙
= 6 M X˙ 2.
Ae−^1 = Ae−γnT^ , 1 = γnT, γ = 1 nT
= ω^1 2 πn
ω^21 = ω 02 − γ^2 , ω^20 = ω 12 + γ^2 = ω 12
( 1 + 1 4 π^2 n^2
) ,
ω 1 ω 0
( 1 + 1 4 π^2 n^2
)− (^12)
8 π^2 n^2.
1c.
x = A cos(ωt + θ 0 ) V = 12 kx^2 x ˙ = −ωA sin(ωt + θ 0 ) T = 12 m x˙^2
Over one period τ = (^2) ωπ ,
T¯ =
∫ (^) τ 0
1 2 m^ x˙^2 dt τ = 1 2 τ
mA^2 ω^2
∫ (^) τ
0
sin^2 (ωt + θ 0 )dt
= (^21) τ mA^2 ω
∫ (^2) π
0
sin^2 (ωt + θ 0 )d(ωt).
Since
∫ (^2) π 0 sin^2 xdx^ =^ π, T¯^ =^
mA^2 ωπ 2 τ.
V¯ =
∫ (^) τ 0
1 2 kx^2 dt τ = 1 τ
∫ (^) τ 0
k[A cos(ωt + θ 0 )]^2 dt
=
τ
kA^2 ω
∫ (^2) π 0
cos(ωt + θ 0 )d(ωt).
Since
∫ (^2) π 0 cos
(^2) xdx = π, V¯ = 1 τ
kA^2 π 2 ω.
ω^2 =
k m ⇒^
V¯ = 1
2 τ mωA
(^2) π = T .¯
2a. From the conservation of energy,
Ti + Ui = Tf + Uf = 0 ⇒ Ti = −Ui,
mv^2 esc = GM m R
vesc =
√ 2 GM R
=
√ 2 G × 7. 36 × 1022
- 74 × 106 = 2. 38 kms−^1.
On the surface, GM mr (^2) m = mg ⇒ g = GMr (^2) m = 1. 62 ms−^2.
2b.
1 2 mv esc^2 =^ GM m R ,
vesc =
√ 2 GM R
=
√ 2 3
Gρ 4 πR^2
=
√ 8 3
GρπR.
2c.
v = v 0 + u ln
M 0
M =^ u^ ln^
M 0
M (assuming^ v^0 = 0), p = mv = mu ln M M^0 , dp dM
= u
[ ln M^0 M
( ln M^0 M
)]
= u
( ln M^0 M
) .
When (^) dMdp = 0, ln M M^0 = 1 ⇒ (^) MM 0 = e−^1. This is a maximum as
d^2 p dM 2
∣∣ ∣∣ ∣M =M 0 e−^1
= − ue M 0
x = l sin θ. y = 1 2
at^2 − l cos θ. x ˙ = l θ˙ cos θ. y ˙ = at + l θ˙ sin θ.
T = 1
m( ˙x^2 + ˙y^2 )
= 1 2
m(l^2 θ˙^2 cos^2 θ + l^2 θ˙^2 sin^2 θ + a^2 t^2 + 2atl θ˙ sin θ)
