SME
Dynamics · Problem type
Energy stored in a flywheel
Use this for “A flywheel with I = 2 kg·m² spins at 3000 rpm. How much kinetic energy does it store?”
Diagram
ω
=
2πN/60
=
2π
×
3,000
60
ω = 314.2 rad/s
v = rω = 0.3 × 314.2 = 94.25 m/s
Every point turns at the same ω, but the rim moves fastest.
This is an example. Type your own values to start a new problem.
ω (rad/s)
=
2π
×
N (rpm)
÷
60
E (kJ)
=
1
2
×
I (kg·m²)
×
(
ω (rad/s)
)
2
r
Rim radius
example · m
v
Rim speed
worked out (not typed) · m/s
Formulas
ω
=
2π
×
N
60
314.1593
=
2π
×
3,000
60
E
=
0.5
×
I
×
ω
2
98,696.044
=
0.5
×
2
×
314.1593
2
v
=
r
×
ω
94.2478 = 0.3 × 314.1593
Formulas are worked in J. The step-by-step shows the conversions.
Assumptions
•
The flywheel turns about a fixed axle through its centre, so all its energy is spin: E = ½Iω².
•
ω must be in rad/s: one turn is 2π rad and a minute is 60 s, so ω = 2πN ÷ 60.
•
The rim moves at v = rω; the rim speed limits how fast a real wheel may spin.
Step-by-step
WE KNOW
I = 2 kg·m², N = 3,000 rpm, r = 0.3 m
FIND
angular speed (ω), kinetic energy (E), rim speed (v)
Step 1 · Find angular speed (ω)
Use ω = 2π × N ÷ 60
Each turn is 2π radians, and a minute is 60 seconds.
ω
=
2π
×
3,000
60
ω
=
314.159265
rad/s
Step 2 · Find kinetic energy (E)
Use E = 0.5 × I × ω²
Each
bit
of
mass
dm
at
radius
ρ
moves
at
ρω,
so
E
=
∫
1
2
(ρω)
2
dm
=
1
2
Iω².
E
=
0.5
×
2
×
314.159265
2
E
=
0.5
×
2
×
98,700
E
=
98,696.044
J
Step 3 · Find rim speed (v)
Use v = r × ω
Each radian turned moves the rim one radius along.
v
=
0.3
×
314.159265
v
=
94.2478
m/s
Step 4 · Convert the answers
Change each answer back to the unit you chose.
E = 98,696.044 J = 98.696 kJ (1 kJ = 1000 J)
Check
Put the numbers back in. Both sides should match.
314.159265
=
2π
×
3,000
60
✓
98,696.04
=
0.5
×
2
×
314.159265
2
✓
94.2478 = 0.3 × 314.159265 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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