SME
Mechanical Vibrations · Problem type
Rotating unbalance: the shaking of an off-balance rotor
Use this for “A 50 kg motor on 20,000 N/m mounts with ζ = 0.1 has a 0.01 kg·m unbalance and runs at 40 rad/s. Find the steady amplitude.”
Diagram
F₀ = mₑeω² = 0.01 × 40² = 16 N, growing with speed
r
=
ω
ωₙ
=
40
20
=
2
X
=
(
mₑe
M
)
×
r
2
÷
√(
(1 −
r
2
)
2
+
(2ζr)
2
)
X = 0.0002 × 1.32 = 0.000264 m
tan
φ
=
2ζ
r
1 −
r
2
:
φ
=
172°
Near
r
=
1
the
response
peaks;
far
above
it
MX
mₑe
tends
to
1:
X
→
mₑe
M
,
the
machine
circling
opposite
the
unbalance.
This is an example. Type your own values to start a new problem.
ωₙ (rad/s)
=
k (N/m)
÷
M (kg)
r
=
ω (rad/s)
÷
ωₙ (rad/s)
X (mm)
=
(
mₑe (kg·m)
÷
M (kg)
)
×
r
2
÷
(1
−
r
2
)
2
+
(2
×
×
)
2
F₀
Unbalance force
worked out (not typed)
Formulas
ωₙ
=
√(
k
M
)
20
=
√(
20,000
50
)
r
=
ω
ωₙ
2
=
40
20
F₀
=
mₑe
×
ω
2
16
=
0.01
×
40
2
X
=
(
mₑe
M
)
×
r
2
÷
√(
(1 −
r
2
)
2
+
(2 ×
ζ
×
r
)
2
)
0.0002643
=
(
0.01
50
)
×
2
2
÷
√(
(1 −
2
2
)
2
+
(2 × 0.1 × 2)
2
)
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
The machine of total mass M moves only up and down, in steady state.
•
The rotor’s unbalance mₑe spinning at ω pushes with F₀ = mₑeω², so the force grows with speed.
•
Far above resonance X tends to mₑe ÷ M: the machine circles opposite the unbalance.
Step-by-step
WE KNOW
M = 50 kg, k = 20,000 N/m, ω = 40 rad/s, ζ = 0.1, mₑe = 0.01 kg·m
FIND
natural frequency (ωₙ), frequency ratio (r), unbalance force (F₀), steady amplitude (X)
Step 1 · Find natural frequency (ωₙ)
Use ωₙ = √(k ÷ M)
How fast the mass rings on its own: the square root of stiffness over mass.
ωₙ
=
√(
20,000
50
)
ωₙ
=
√
400
ωₙ = 20 rad/s
Step 2 · Find frequency ratio (r)
Use r = ω ÷ ωₙ
The frequency ratio: how fast the force drives against how fast the mass rings.
r
=
40
20
r
=
2
Step 3 · Find unbalance force (F₀)
Use F₀ = mₑe × ω²
The
pull
of
the
off-center
mass
going
round:
mass
times
radius
times
ω
2
.
F₀
=
0.01
×
40
2
F₀
=
0.01
×
1,600
F₀
=
16
N
Step 4 · Find steady amplitude (X)
Use X = (mₑe ÷ M) × r² ÷ √((1 − r²)² + (2 × ζ × r)²)
F₀
k
is
mₑeω²
k
=
(
mₑe
M
)
r
2
,
then
the
same
magnification
as
any
forced
block.
X
=
(
0.01
50
)
×
2
2
÷
√(
(1 −
2
2
)
2
+
(2 × 0.1 × 2)
2
)
X
=
0.01
50
×
2
2
÷
√(
(1 − 4)
2
+
(0.2 × 2)
2
)
X
=
0.01
50
×
2
2
÷
√(
(−3)
2
+
0.4
2
)
X
=
0.01
50
×
2
2
÷
√
9 + 0.16
X
=
0.01
50
×
2
2
÷
√
9.16
X
=
0.01
50
×
4
3.027
X
=
0.0002
×
4
3.027
X
=
0.0008
3.027
X
=
0.0002643
m
Step 5 · Convert the answers
Change each answer back to the unit you chose.
X = 0.0002643 m = 0.2643 mm (1 m = 1000 mm)
Check
Put the numbers back in. Both sides should match.
20
=
√(
20,000
50
)
✓
2
=
40
20
✓
16
=
0.01
×
40
2
✓
0.0002643
=
(
0.01
50
)
×
2
2
÷
√(
(1 −
2
2
)
2
+
(2 × 0.1 × 2)
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.