SME
Classical Mechanics · Problem type
A pendulum: swing or go over the top
Use this for “A 0.4 kg bob on a 0.8 m rod passes the bottom at 6 rad/s. Does it swing back or go over the top? If it swings, how far?”
Diagram
E = ½mL²ω₀² = ½ × 1 × 1² × 5² = 12.5 J; Eₛ = 2mgL = 2 × 1 × 9.81 × 1 = 19.62 J
E
<
Eₛ:
it
swings
(libration).
cos
θₘₐₓ
=
1
−
E
mgL
=
−0.2742,
θₘₐₓ
=
105.9°
The flow is (θ̇, ω̇) = (ω, −(g/L) sin θ): clockwise round each oval.
This is an example. Type your own values to start a new problem.
m
Mass
example
L
Length
example
ω₀
Angular speed at the bottom
example · rad/s
E
Energy
worked out (not typed) · J
Eₛ
Separatrix energy
worked out (not typed) · J
ωₜₒₚ
Angular speed to go over
worked out (not typed) · rad/s
θₘₐₓ
Largest angle
worked out (not typed) · °
Formulas
E
=
1
2
×
m
×
L
2
×
ω₀
2
12.5
=
1
2
×
1
×
1
2
×
5
2
Eₛ
=
2
×
m
×
9.81
×
L
19.62 = 2 × 1 × 9.81 × 1
ωₜₒₚ
=
√(4
×
9.81
L
)
6.2642
=
√(4
×
9.81
1
)
cos(θₘₐₓ°)
=
1
−
E
m
× 9.81 ×
L
cos(105.9149°)
=
1
−
12.5
1 × 9.81 × 1
Assumptions
•
A point mass on a light rigid rod, no friction; θ is measured from the bottom; g = 9.81 m/s².
•
H = pθ² ÷ 2mL² + mgL(1 − cos θ) is conserved: E is the energy above the bottom, all kinetic there.
•
The separatrix Eₛ = 2mgL just reaches the top: below it the curve is a closed oval (a swing), above it the pendulum goes round.
Step-by-step
WE KNOW
m = 1 kg, L = 1 m, ω₀ = 5 rad/s
FIND
energy (E), separatrix energy (Eₛ), angular speed to go over (ωₜₒₚ), largest angle (θₘₐₓ)
Step 1 · Find energy (E)
Use E = ½ × m × L² × ω₀²
At
the
bottom
all
the
energy
is
kinetic:
1
2
Iω₀²
with
I
=
mL².
E
=
1
2
×
1
×
1
2
×
5
2
E
=
1
2
×
25
E
=
12.5
J
Step 2 · Find separatrix energy (Eₛ)
Use Eₛ = 2 × m × 9.81 × L
The
energy
that
lifts
the
bob
2
L
,
from
the
bottom
to
the
top:
the
separatrix.
Eₛ = 2 × 1 × 9.81 × 1
Eₛ = 19.62 J
Step 3 · Find angular speed to go over (ωₜₒₚ)
Use ωₜₒₚ = √(4 × 9.81 ÷ L)
Set
1
2
mL²
ω₀
2
=
2mgL:
m
cancels,
and
any
faster
speed
at
the
bottom
goes
over
the
top.
ωₜₒₚ
=
√(4
×
9.81
1
)
ωₜₒₚ
=
√
39.24
ωₜₒₚ = 6.2642 rad/s
Step 4 · Find largest angle (θₘₐₓ)
Use cos(θₘₐₓ°) = 1 − E ÷ (m × 9.81 × L)
12.5
<
19.62:
E
is
below
Eₛ,
so
it
swings
back.
At
θₘₐₓ
the
bob
stops:
all
of
E
is
height,
mgL(1
−
cos
θₘₐₓ).
θₘₐₓ
=
cos⁻¹(1
−
E
m
× 9.81 ×
L
)
θₘₐₓ
=
cos⁻¹(1
−
12.5
1 × 9.81 × 1
)
θₘₐₓ = 105.9149°
Check
Put the numbers back in. Both sides should match.
12.5
=
1
2
×
1
×
1
2
×
5
2
✓
19.62 = 2 × 1 × 9.81 × 1 ✓
6.2642
=
√(4
×
9.81
1
)
✓
cos(105.9149°)
=
1
−
12.5
1 × 9.81 × 1
✓
Topic check
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