Kinetic Energy Calculator
KE = ½mv² for any variable, spinning and rolling objects, and a relativity check.
Where the energy is
Kinetic energy against speed
How it was worked out
About the Kinetic Energy Calculator
Kinetic energy is the energy an object has because it moves. For something travelling in a straight line it is K = ½mv²; for something spinning it is K = ½Iω², where I is the moment of inertia. This calculator solves either formula for any one unknown — the energy, the mass (or moment of inertia) or the speed — and converts between joules, foot-pounds, kilocalories, kilowatt-hours and electronvolts, kilograms, pounds and grains, and m/s, km/h, mph and ft/s.
Every straight-line answer is also checked against Einstein’s relativistic formula K = (γ − 1)mc². At everyday speeds the two agree to many decimal places; once an object moves faster than about a tenth of the speed of light the classical formula is noticeably low, so the calculator switches to the relativistic value and tells you by how much ½mv² is off. The spinning mode works out I for ten standard shapes and can add the forward motion of a ball, wheel or pipe that rolls without slipping.
How to use it
- Choose Moving for something travelling in a line, or Spinning for a rotating object.
- Pick what to solve for, then type the other two values and choose their units. Speeds can be entered as a fraction of the speed of light with the unit c.
- For a spinning object, either type the moment of inertia or choose a shape and give its mass and size. Tick Rolls without slipping for a ball, wheel or pipe rolling along the ground.
- Read the answer in your units, the same value in other units, and the step-by-step working; copy the result if you need it elsewhere.
Examples
80 kg at 10 m/s
K = ½ × 80 × 10² = 4,000 J = 4.0 kJ
OpenStax University Physics Vol. 1, Example 7.6(a).
1,500 kg at 100 km/h (27.78 m/s)
K = 578.7 kJ — four times the 144.7 kJ it has at 50 km/h
400 grains at 280 ft/s
K = 94.39 J = 69.62 ft·lbf; momentum 0.4973 lbf·s
K = 4.2 × 10²³ J at 22 km/s
m = 2K ÷ v² = 1.7 × 10¹⁵ kg
OpenStax Example 7.6(b), the Chicxulub impactor.
Four 50 kg, 4.00 m blades (rods about one end) at 300 rpm
I = 1,067 kg·m², K = 5.26 × 10⁵ J
OpenStax Example 10.9 — enter one 200 kg rod, since I adds up.
Electron (0.511 MeV/c²) at 0.990c
K = (γ − 1)mc² = 3.11 MeV with γ = 7.089; ½mv² would give only 0.250 MeV
OpenStax University Physics Vol. 3, Example 5.12.
Common uses
- Physics homework on kinetic energy, the work–energy theorem, rotation and rolling, with the substitution shown.
- Archery: arrow energy in foot-pounds from the arrow weight in grains and the chronograph speed in ft/s.
- Comparing the energy of a vehicle at two speeds — doubling the speed means four times the energy to dissipate when braking.
- Particle physics: the speed of an electron or proton from its kinetic energy in eV, keV or MeV.
- Flywheels, wheels and rotors: the energy stored at a given rpm.
The formulas
- Moving object:
K = ½ m v², som = 2K ÷ v²andv = √(2K ÷ m). Energy grows with the square of the speed: double the speed and the energy goes up four times. - Spinning object:
K = ½ I ω²with ω in radians per second (1 rpm = 2π/60 rad/s). - Rolling without slipping: the centre moves at
v = ωR, soK = ½ I ω² + ½ M v². The split is fixed by the shape: a solid sphere keeps 2/7 of its energy in rotation, a solid cylinder 1/3 and a hoop 1/2. - Relativistic:
K = (γ − 1) m c²withγ = 1 ÷ √(1 − v²/c²)and c = 299,792,458 m/s. From an energy,γ = 1 + K ÷ (mc²)andv = c √(1 − 1/γ²).
When the classical formula stops being accurate
The relativistic energy is always the larger one. How far it exceeds ½mv²:
- 1 % of c (3,000 km/s): 0.0075 %
- 10 % of c: 0.76 %
- 50 % of c: the relativistic value is 23.8 % higher (γ = 1.155)
- 99 % of c: 12.4 times higher (γ = 7.089)
The calculator uses ½mv² while it is within 1 % (below about 11.5 % of c, where the difference passes 1 %) and mentions the correction from 0.01 % on. A mass can never reach c: the energy needed grows without limit, so speeds of c or more are rejected.
Moments of inertia used for shapes
From OpenStax University Physics Vol. 1, Figure 10.20 (M = mass):
- Solid cylinder or disc about its axis: ½MR² (rolls)
- Hoop or thin ring about its axis: MR² (rolls)
- Thick ring or pipe about its axis: ½M(R₁² + R₂²) (rolls on the outer radius)
- Solid sphere about a diameter: ⅖MR² (rolls)
- Thin spherical shell: ⅔MR² (rolls)
- Thin rod about its centre: ML²/12; about one end: ML²/3
- Hoop about a diameter: ½MR²
- Solid cylinder about a central diameter: ¼MR² + ML²/12
- Rectangular slab about a perpendicular axis through its centre: M(a² + b²)/12
For anything else, type I directly. Moments of inertia add, so several identical parts can be entered as one part with their combined mass when they share the same axis and shape.
Units for archery and ballistics
Arrow and projectile weights are usually given in grains (1 gr = 1/7000 lb = 64.79891 mg exactly) and speeds in feet per second. With exact conversions, KE (ft·lbf) = grains × (ft/s)² ÷ 450,437. The widely quoted archery shortcut divides by 450,240 instead, which gives values 0.04 % higher — the difference is far smaller than chronograph error. Momentum p = mv is shown in kg·m/s and in lbf·s (the same as slug·ft/s), the unit archery charts use.
Sources
- OpenStax, University Physics Volume 1 (CC BY 4.0): §7.2 Kinetic Energy (Eq. 7.6, Examples 7.6–7.8) and §10.4 Moment of Inertia and Rotational Kinetic Energy (Figure 10.20, Examples 10.9 and 10.10); §11.1 Rolling Motion (v = Rω).
- OpenStax, University Physics Volume 3 (CC BY 4.0): §5.9 Relativistic Energy (Example 5.12).
- BIPM, The International System of Units (SI Brochure, 9th ed., 2019): exact c, h and e (1 eV = 1.602176634 × 10⁻¹⁹ J).
- NIST, CODATA 2022 recommended values: electron, proton, neutron and alpha-particle masses, atomic mass constant.
- NIST, SP 811 Appendix B.8: grain, foot-pound, calorie, Btu and other conversion factors.
Limitations
- Speeds are relative to whatever frame you measure them in; kinetic energy depends on that choice (OpenStax Example 7.7).
- The spinning mode assumes a rigid body turning about a fixed axis; for an axis that does not pass through the centre of mass, add Md² to I (parallel-axis theorem) and enter I directly.
- Rolling assumes no slipping and no deformation; a skidding or bouncing object does not obey v = ωR.
- The moments of inertia assume uniform density and ideal shapes (a thin rod has negligible thickness, a hoop negligible wall).
- Rest energy mc² and kinetic energy are different things: this tool reports only the kinetic part, (γ − 1)mc².
- Very close to the speed of light a typed speed pins the energy down poorly (each extra 9 in 0.999… c multiplies it by about 3.2), so for fast particles enter the energy and solve for the speed.
Privacy
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Frequently asked questions
What is the formula for kinetic energy?
K = ½mv² for an object moving in a straight line, with m in kilograms and v in metres per second giving joules. An 80 kg runner at 10 m/s has ½ × 80 × 10² = 4,000 J.
Why does doubling the speed quadruple the kinetic energy?
Because the speed is squared. A 1,500 kg car has 144.7 kJ at 50 km/h and 578.7 kJ at 100 km/h — four times as much, which is why braking distance grows so quickly with speed.
How do I find speed from kinetic energy?
Rearrange K = ½mv² to v = √(2K ÷ m). 500 kJ in a 1,500 kg car gives v = √(2 × 500,000 ÷ 1,500) = 25.82 m/s (92.95 km/h). At speeds near light the calculator uses v = c√(1 − 1/γ²) with γ = 1 + K ÷ (mc²) instead.
How do I calculate arrow kinetic energy in foot-pounds?
Multiply the arrow weight in grains by the speed in ft/s squared and divide by 450,437 (the archery shortcut uses 450,240). A 400-grain arrow at 280 ft/s carries 400 × 280² ÷ 450,437 = 69.6 ft·lbf.
When do I need the relativistic kinetic energy formula?
Above roughly a tenth of the speed of light. At 10 % of c the classical ½mv² is 0.75 % too low; at 50 % of c the true value is 23.8 % higher; for an electron at 0.99c it is 12.4 times higher (3.11 MeV instead of 0.25 MeV).
What is rotational kinetic energy?
The energy of spinning: K = ½Iω², where I (kg·m²) says how the mass is spread around the axis and ω is the angular speed in rad/s. Four 50 kg helicopter blades 4 m long turning at 300 rpm store about 526 kJ.
Can kinetic energy be negative?
No. Mass is positive and the speed is squared, so kinetic energy is zero for an object at rest and positive otherwise. It is a scalar: the direction of motion does not matter.