SME
Classical Mechanics · Problem type
A bead on a spinning hoop
Use this for “A bead slides on a 0.25 m hoop spun at 9 rad/s about its vertical diameter. At what angle does it rest, and how fast does it rock about that point?”
Diagram
ωc
=
√(
g
R
)
=
√(
9.81
0.2
)
=
7.004
rad/s
cos
θ₀
=
g
ω
2
R
=
9.81
10
2
× 0.2
=
0.4905
θ₀ = 60.63°, Ω = ω sin θ₀ = 8.714 rad/s
Ueff
mgR
=
1
−
cos
θ
−
1
2
(
ω
ωc
)
2
sin²
θ
ω > ωc, so the bead rides up to θ₀ and rocks about it at Ω.
The bead rests where Ueff is lowest.
This is an example. Type your own values to start a new problem.
R
Hoop radius
example
ω
Spin
example · rad/s
ωc
Critical spin
worked out (not typed) · rad/s
θ₀
Resting angle
worked out (not typed) · °
Ω
Rocking frequency
worked out (not typed) · rad/s
Formulas
ωc
=
√(
9.81
R
)
7.0036
=
√(
9.81
0.2
)
cos(
θ₀
°)
=
ωc
2
ω
2
cos(60.6265°)
=
7.0036
2
10
2
Ω
=
√(
ω
2
−
ωc
4
ω
2
)
8.7144
=
√(
10
2
−
7.0036
4
10
2
)
Assumptions
•
The hoop spins at a steady ω about its vertical diameter; the bead slides without friction, and θ is measured from the bottom; g = 9.81 m/s².
•
One coordinate θ: L = ½mR²(θ̇² + ω² sin² θ) − mgR(1 − cos θ), so Ueff = mgR(1 − cos θ) − ½mω²R² sin² θ.
•
The bead rests where Ueff is least: the bottom up to ωc, off it above ωc.
Step-by-step
WE KNOW
R = 0.2 m, ω = 10 rad/s
FIND
critical spin (ωc), resting angle (θ₀), rocking frequency (Ω)
Step 1 · Find critical spin (ωc)
Use ωc = √(9.81 ÷ R)
Below
this
spin
the
bottom
stays
the
lowest
point
of
Ueff:
there
d
2
Ueff/dθ²
=
mR²(
ωc
2
−
ω
2
).
ωc
=
√(
9.81
0.2
)
ωc
=
√
49.05
ωc = 7.0036 rad/s
Step 2 · Find resting angle (θ₀)
Use cos(θ₀°) = ωc² ÷ ω²
Above
ωc,
dUeff/dθ
=
mR²
sin
θ(
ωc
2
−
ω
2
cos
θ)
is
0
off
the
bottom
where
cos
θ₀
=
g
ω
2
R
=
ωc
2
ω
2
.
10 > 7.0036: above ωc, so the bead rides up the hoop
θ₀
=
cos⁻¹(
ωc
2
ω
2
)
θ₀
=
cos⁻¹(
7.0036
2
10
2
)
θ₀
=
60.6265°
Step 3 · Find rocking frequency (Ω)
Use Ω = √(ω² − ωc⁴ ÷ ω²)
Linearize
the
θ
equation
about
θ₀
:
the
bead
rocks
at
Ω
=
ω
sin
θ₀
,
and
sin²
θ₀
=
1
−
(
ωc
ω
)
4
.
10
>
7.0036:
above
ωc,
so
it
rocks
about
θ₀
Ω
=
√(
10
2
−
7.0036
4
10
2
)
Ω
=
√(100
−
2,406
100
)
Ω
=
√
100 − 24.06
Ω
=
√
75.94
Ω
=
8.7144
rad/s
Check
Put the numbers back in. Both sides should match.
7.0036
=
√(
9.81
0.2
)
✓
cos(60.6265°)
=
7.0036
2
10
2
✓
8.7144
=
√(
10
2
−
7.0036
4
10
2
)
✓
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