Library / Mechanical / MEC-15

Mechanical

Rigid-body kinetic energy

Translational plus rotational kinetic energy of a moving rigid body.

Ref: MEC-15
Rev: v1
Field: Mechanical
Result: T [J]

T = \frac{1}{2} m {v_{G}}^{2} + \frac{1}{2} I_{G} ω^{2}

Planar motion of a rigid body, with I_G about the centre of mass. The work–energy statement is W_ext = T₂ − T₁.

A rolling body stores energy both in moving forward and in spinning.

Calculator

Solve for

TKinetic energyComputed

mMass

0.11,000

v_GCentre speed

m/s

0.1100

I_GMoment of inertia

kg·m²

0.001100

ωSpin rate

rad/s

0.1500

Translational energy18 J
Rotational energy9 J

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Kinetic energy (J) current value 27 J

ωSpin rateKey driver

1822,500

v_GCentre speed

9.0220,000

I_GMoment of inertia

18.15,020

mMass

9.454,510

A solid disc rolling at 3 m/s with mass 4 kg has 18 J translational and 9 J rotational energy, 27 J in total.
Rolling bodies need more energy than sliding ones to reach the same speed.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (J)	Local step	Change in result
ω Spin rate	0.1 to 500 rad/s	18	22,520	22,500	+10%	+7 %
v_G Centre speed	0.1 to 100 m/s	9.02	20,010	20,000	+10%	+14 %
I_G Moment of inertia	0.001 to 100 kg·m²	18.05	5,018	5,000	+10%	+3.33 %
m Mass	0.1 to 1,000 kg	9.45	4,509	4,500	+10%	+6.67 %

Sweep of Spin rate

Spin rate (rad/s)	Kinetic energy (J)
0.1	18
0.29	18.01
0.8409	18.06
2.438	18.54
7.071	22.5
20.5	55.84
59.46	336.2
172.4	2,694
500	22,520

Sweep one variable

Vary

Kinetic energy (J)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Kinetic energy

Result · J

What the equation gives you. Everything else feeds it.

Mass

kg · typical 0.1 to 1,000

Mass of the body. Scales the translational term.

v_G

Centre speed

m/s · typical 0.1 to 100

Speed of the centre of mass. Squared in the translational term.

I_G

Moment of inertia

kg·m² · typical 0.001 to 100

Mass moment of inertia about G. Scales the rotational term.

Spin rate

rad/s · typical 0.1 to 500

Angular speed. Squared in the rotational term.

Worked example with the values above

Translational: ½mv² = 0.5 × 4 × 3² = 18 J.
Rotational: ½Iω² = 0.5 × 0.18 × 10² = 9 J.
T = 27 J.

Assumptions and limits

Planar rigid-body motion.
I_G about the centre of mass.

Common mistakes

Leaving out the rotational term.
Using I about the wrong axis.

Statistics and uncertainty

Moment of inertia is often the least certain input, since it depends on the mass distribution.

Where it is used

Flywheel and wheel energy storage.
Rolling-vehicle energy budgets.

Maths you need

Kinetic energy: ½mv²
Moment of inertia: kg m²

Related formulas

Practise this formula