SME
Classical Mechanics · Problem type
A plate’s principal moments, and the axis that tumbles
Use this for “A 2 kg tray is 0.5 m by 0.2 m. Find its three principal moments through the center. About which axis will a spin tumble?”
Diagram
I₁
=
3
×
0.3
2
12
=
0.0225
kg·
m
2
(axis
along
a).
I₂
=
3
×
0.4
2
12
=
0.04
kg·
m
2
(axis
along
b).
I₃ = I₁ + I₂ = 0.0625 kg·m², square to the plate: the largest.
Spin about axis 3 or axis 1 is steady; about axis 2, the middle moment, the plate tumbles.
This is an example. Type your own values to start a new problem.
I₁
=
M (kg)
(
b (
)
)
2
12
=
I₁ (kg·m²)
I₂
=
(
a (
)
)
2
12
=
I₂ (kg·m²)
I₃
=
I₁
+
I₂
=
I₃ (kg·m²)
Tumbles
Axis whose spin tumbles
worked out (not typed)
Formulas
I₁
=
M
×
b
2
12
0.0225
=
3
×
0.3
2
12
I₂
=
M
×
a
2
12
0.04
=
3
×
0.4
2
12
I₃
=
I₁
+
I₂
0.0625 = 0.0225 + 0.04
Tumbles = the axis with the middle moment
0.04
>
0.0225,
so
axis
2
(along
b
)
Assumptions
•
A thin uniform plate: its thickness is too small to count, and the axes pass through its center.
•
About an axis along one side the plate acts as a rod across the other side; the perpendicular-axis theorem gives I₃ = I₁ + I₂.
•
Spin about the largest or smallest moment is steady; about the middle one it tumbles (the tennis-racket theorem). A square plate has no middle axis.
Step-by-step
WE KNOW
M = 3 kg, a = 0.4 m, b = 0.3 m
FIND
moment about the axis along a (I₁), moment about the axis along b (I₂), moment square to the plate (I₃), axis whose spin tumbles (Tumbles)
Step 1 · Find moment about the axis along a (I₁)
Use I₁ = M × b² ÷ 12
About
axis
1,
along
side
a
,
the
mass
spreads
across
side
b
:
the
plate
turns
like
one
rod
of
that
length.
I₁
=
3
×
0.3
2
12
I₁
=
3
×
0.09
12
I₁
=
0.27
12
I₁
=
0.0225
kg·
m
2
Step 2 · Find moment about the axis along b (I₂)
Use I₂ = M × a² ÷ 12
About
axis
2,
along
side
b
,
the
mass
spreads
across
side
a
:
the
plate
turns
like
one
rod
of
that
length.
I₂
=
3
×
0.4
2
12
I₂
=
3
×
0.16
12
I₂
=
0.48
12
I₂
=
0.04
kg·
m
2
Step 3 · Find moment square to the plate (I₃)
Use I₃ = I₁ + I₂
The perpendicular-axis theorem: for any flat body, add the two in-plane moments.
I₃
=
0.0225
+
0.04
I₃
=
0.0625
kg·
m
2
Step 4 · Find axis whose spin tumbles (Tumbles)
Use Tumbles = the axis with the middle moment
I₃
is
always
the
largest,
so
the
larger
of
I₁
and
I₂
is
the
middle
moment:
spin
about
that
axis
tumbles.
0.04
>
0.0225:
axis
2
(along
b
)
Tumbles
=
axis
2
(along
b
)
Check
Put the numbers back in. Both sides should match.
0.0225
=
3
×
0.3
2
12
✓
0.04
=
3
×
0.4
2
12
✓
0.0625 = 0.0225 + 0.04 ✓
0.04
>
0.0225,
so
axis
2
(along
b
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.