Rotational Kinetic Energy Equation: Formula, 7 Forms, and Worked Examples
The rotational kinetic energy equation is KE_rot = ½ I ω², where I is the moment of inertia and ω is the angular velocity. The result is expressed in joules (J). The 7 forms use angular velocity, angular momentum, frequency, revolutions per minute, linear speed, particle sums, or torque and angle. Rolling objects add a translational term (½ m v²).
For a reliable reference, see OpenStax derives rotational kinetic energy from moment of inertia and angular speed.
Rotational energy and angular momentum use the same rotational variables; see Angular Momentum and Conservation of Angular Momentum: Definition, Formulas, Conditions, 3 Worked Examples, and 8 Applications in Sports, Astronomy, and Quantum Physics.
What Is Rotational Kinetic Energy and What Are Examples of It?
Rotational kinetic energy is the energy an object has because it spins about an axis. According to OpenStax University Physics Volume 1, a rigid body rotating about a fixed axis has a rotational kinetic energy of one half the moment of inertia times the square of the angular velocity. It is a scalar quantity, so it has a magnitude and no direction. The SI unit is the joule (J).
Spinning objects have rotational kinetic energy. Examples include flywheels, wheels, turbines, spinning tops, planets, and molecules.
The table below defines each symbol used in rotational kinetic energy formulas and lists its SI unit.
| Symbol | Meaning | SI Unit |
|---|---|---|
| KE_rot | Rotational kinetic energy | joule (J) |
| I | Moment of inertia about the rotation axis | kg·m² |
| ω | Angular velocity | rad/s |
| L | Angular momentum (L = Iω) | kg·m²/s |
| f | Rotation frequency | hertz (Hz) |
| N | Rotation rate | revolutions per minute (rpm) |
| m | Mass | kilogram (kg) |
| r | Distance from the rotation axis | meter (m) |
| v | Linear speed at distance r (v = ωr) | m/s |
| τ | Net torque | N·m |
| Δθ | Angular displacement | rad |
What Are the 7 Forms of the Rotational Kinetic Energy Equation?
The 7 forms use angular velocity, angular momentum, frequency, revolutions per minute, linear speed, a sum over particles, or torque and angle. The table below lists the equation, the known quantities, and the condition for each form.
| No. | Form | Equation | Known Quantities | Applies When |
|---|---|---|---|---|
| 1 | Angular velocity | KE = ½ I ω² | I, ω | A rigid body rotates about a fixed axis |
| 2 | Angular momentum | KE = L² / (2I) | L, I | Angular momentum is known |
| 3 | Frequency | KE = 2π² I f² | I, f | Rotation frequency is in hertz |
| 4 | Revolutions per minute | KE = π² I N² / 1800 | I, N | Rotation rate is in rpm |
| 5 | Linear speed | KE = ½ I v² / r² | I, v, r | Speed at radius r is known |
| 6 | Particle sum | KE = Σ ½ mᵢ rᵢ² ω² | mᵢ, rᵢ, ω | The body is a set of discrete masses |
| 7 | Work and torque | KE = τ Δθ | τ, Δθ | A constant net torque acts on a body starting from rest |
The equation rearranges to solve for other quantities:
- Find the angular velocity with ω = √(2 KE / I).
- Find the moment of inertia with I = 2 KE / ω².
- Find the angular momentum with L = √(2 I KE).
Example for rearrangement: A wheel has KE = 500 J and I = 0.4 kg·m². The angular velocity is √(2 × 500 / 0.4) = 50 rad/s, which equals 477.5 rpm.
Example for form 2: A body with L = 20 kg·m²/s and I = 2 kg·m² has KE = 20² / (2 × 2) = 100 J.
Example for form 3: A rotor with I = 0.5 kg·m² turns at f = 5 Hz. The energy is 2π² × 0.5 × 25 = 246.7 J.
Example for form 5: A wheel with I = 0.8 kg·m² and a rim speed of 6 m/s at r = 0.4 m has ω = 15 rad/s. The energy is ½ × 0.8 × 15² = 90 J.
Example for form 6: A dumbbell has two 1 kg masses at 0.5 m from the axis and ω = 4 rad/s. Each mass moves at 2 m/s and has 2 J. The total is 4 J, which equals ½ × 0.5 × 4².
How Do You Calculate Rotational Kinetic Energy Step by Step?
Rotational kinetic energy is calculated by multiplying one half, the moment of inertia, and the square of the angular velocity. The 6 steps are:
- Identify the rotation axis.
- Find the moment of inertia (I) about that axis.
- Convert the rotation rate to radians per second.
- Square the angular velocity.
- Multiply by I and by ½.
- Express the result in joules.
Example: A solid disk has a mass of 20 kg and a radius of 0.5 m. It turns at 300 rpm.
- The moment of inertia is ½ × 20 × 0.5² = 2.5 kg·m².
- The angular velocity is 300 × 2π / 60 = 31.42 rad/s.
- The square of the angular velocity is 986.96 rad²/s².
- The energy is ½ × 2.5 × 986.96 = 1,233.7 J.
How Is the Rotational Kinetic Energy Equation Derived?
The equation is derived by summing the translational kinetic energy of every particle in a rotating rigid body. Every particle moves at v = ωr, where r is the distance from the axis. The 5 steps are:
- Write the kinetic energy of 1 particle: ½ mᵢ vᵢ².
- Substitute vᵢ = ω rᵢ to obtain ½ mᵢ rᵢ² ω².
- Sum over all particles: KE = Σ ½ mᵢ rᵢ² ω².
- Factor out the common term ω²: KE = ½ (Σ mᵢ rᵢ²) ω².
- Define I = Σ mᵢ rᵢ², which gives KE = ½ I ω².
A continuous body replaces the sum with an integral: I = ∫ r² dm. Example for a thin rod of length L and mass m about its center: I = ∫ x² (m / L) dx from −L/2 to +L/2, which equals mL² / 12.
Leonhard Euler introduced the term moment of inertia in Theoria motus corporum solidorum seu rigidorum in 1765.
What Are the Moment of Inertia Formulas for Common Shapes?
The moment of inertia is the sum of mass times squared distance from the axis (I = Σ m r²), and it depends on the mass, the mass distribution, and the axis. The table below lists 9 shapes with the axis and the formula for I. The symbol m is the mass, r is the radius, L is the length, and a and b are the plate side lengths.
| Shape | Axis | Moment of Inertia I |
|---|---|---|
| Point mass at radius r | Axis at distance r | m r² |
| Thin hoop or ring | Central axis, perpendicular to the plane | m r² |
| Solid disk or solid cylinder | Central axis | ½ m r² |
| Hollow cylinder with inner radius r₁ and outer radius r₂ | Central axis | ½ m (r₁² + r₂²) |
| Solid sphere | Any diameter | (2/5) m r² |
| Thin spherical shell | Any diameter | (2/3) m r² |
| Thin rod | Perpendicular axis through the center | (1/12) m L² |
| Thin rod | Perpendicular axis through 1 end | (1/3) m L² |
| Rectangular plate | Axis through the center, perpendicular to the plane | (1/12) m (a² + b²) |
Mass placed far from the axis raises I, so a hoop has twice the moment of inertia of a solid disk with the same mass and radius.
How Does the Parallel Axis Theorem Change Rotational Kinetic Energy?
The parallel axis theorem states that the moment of inertia about any axis equals the moment of inertia about a parallel axis through the center of mass plus m d². The formula is I = I_cm + m d², where d is the distance between the 2 axes. The theorem is also called the Huygens–Steiner theorem, after Christiaan Huygens and Jakob Steiner.
Example: A solid disk has m = 4 kg and r = 0.3 m.
- The moment of inertia about the center is ½ × 4 × 0.3² = 0.18 kg·m².
- The moment of inertia about an axis on the edge is 0.18 + 4 × 0.3² = 0.54 kg·m².
- The energy at ω = 10 rad/s is 9 J about the center and 27 J about the edge.
The edge axis gives 3 times the energy of the center axis at the same angular velocity.
What Does Rotational Kinetic Energy Depend On?
Rotational kinetic energy depends on 2 factors: the moment of inertia and the square of the angular velocity. The moment of inertia depends on the mass, the shape, and the axis. The table below lists 7 changes and the effect of each on the energy.
| Change | Effect on Rotational Kinetic Energy | Example |
|---|---|---|
| Doubling ω | Multiplies by 4 | 1,233.7 J becomes 4,934.8 J for I = 2.5 kg·m² |
| Tripling ω | Multiplies by 9 | 10 J becomes 90 J |
| Doubling I at the same ω | Doubles | 9 J becomes 18 J |
| Doubling the mass of the same shape and radius | Doubles | I is proportional to m |
| Doubling the radius of the same shape and mass | Multiplies by 4 | I is proportional to r² |
| Moving the axis of a rod from the center to 1 end | Multiplies by 4 | 4.5 J becomes 18 J for m = 3 kg, L = 1.2 m, ω = 5 rad/s |
| Reversing the direction of rotation | No change | ω² is always positive |
Rotational kinetic energy is never negative, because I and ω² are never negative.
How Do Shape and Mass Distribution Change Rotational Kinetic Energy?
Bodies with the same mass, radius, and angular velocity have different energies, because the moment of inertia grows as mass moves away from the axis. The table below compares 4 shapes with m = 2 kg, r = 0.10 m, and ω = 10 rad/s about the central axis or a diameter.
| Shape | Moment of Inertia Formula | I (kg·m²) | Rotational Kinetic Energy (J) |
|---|---|---|---|
| Thin hoop | m r² | 0.0200 | 1.00 |
| Thin spherical shell | (2/3) m r² | 0.0133 | 0.67 |
| Solid disk | ½ m r² | 0.0100 | 0.50 |
| Solid sphere | (2/5) m r² | 0.0080 | 0.40 |
The hoop stores 2.5 times the energy of the solid sphere at the same angular velocity.
How Does Rotational Kinetic Energy Change With Revolutions Per Minute?
Rotational kinetic energy grows with the square of the rpm value, so doubling the rpm multiplies the energy by 4. The formula is KE = π² I N² / 1800. The table below lists 5 rotation rates for an object with I = 1 kg·m².
| Rotation Rate (rpm) | Angular Velocity (rad/s) | Rotational Kinetic Energy (J) |
|---|---|---|
| 60 | 6.28 | 19.7 |
| 300 | 31.42 | 493.5 |
| 600 | 62.83 | 1,973.9 |
| 3,000 | 314.16 | 49,348 |
| 6,000 | 628.32 | 197,392 |
The conversion from rpm to rad/s multiplies by 2π / 60, which equals 0.10472.
How Do You Find Rotational Kinetic Energy From Angular Momentum?
Rotational kinetic energy is found with KE = L² / (2I), where L = Iω is the angular momentum. Angular momentum stays constant in an isolated system, so the energy varies inversely with the moment of inertia.
Example: A figure skater spins with I₁ = 4.0 kg·m² at ω₁ = 2.0 rad/s.
- The angular momentum is 4.0 × 2.0 = 8.0 kg·m²/s.
- The initial energy is ½ × 4.0 × 2.0² = 8 J.
- Pulling in the arms reduces I to 1.6 kg·m².
- The new angular velocity is 8.0 / 1.6 = 5.0 rad/s.
- The new energy is 8.0² / (2 × 1.6) = 20 J.
The energy rises by 12 J. The muscles of the skater supply this work as the arms move inward against the required centripetal force.
How Do Work, Torque, and Power Relate to Rotational Kinetic Energy?
The net work done by torque equals the change in rotational kinetic energy (W_net = τ Δθ = ΔKE_rot). The relationship is the rotational form of the work-energy theorem. Power equals P = dKE/dt = τ ω.
Example: A constant torque of 10 N·m acts on a body with I = 2 kg·m² that starts from rest.
- The angular acceleration is α = τ / I = 5 rad/s².
- The angular velocity at t = 4 s is 5 × 4 = 20 rad/s.
- The angular displacement is ½ × 5 × 4² = 40 rad.
- The work is 10 × 40 = 400 J.
- The energy is ½ × 2 × 20² = 400 J, which matches the work.
The power at t = 4 s is 10 × 20 = 200 W. The average power over the 4 s is 400 / 4 = 100 W.
What Is the Total Kinetic Energy of a Rolling Object?
The total kinetic energy of an object that rolls without slipping is KE_total = ½ m v² + ½ I ω², which equals ½ m v² (1 + k). The constant k is I / (m r²), and the condition for rolling is v = ωr. The first term is translational energy of the center of mass, and the second term is rotational energy about the center of mass.
The table below lists 4 shapes with k, the total energy, the rotational share, and the speed at the bottom of an incline with a vertical drop of 1.0 m (g = 9.8 m/s²).
| Shape | k = I / (m r²) | Total Kinetic Energy | Rotational Share of Total | Speed After 1.0 m Drop (m/s) |
|---|---|---|---|---|
| Solid sphere | 2/5 | (7/10) m v² | 28.6% | 3.74 |
| Solid cylinder | 1/2 | (3/4) m v² | 33.3% | 3.61 |
| Thin spherical shell | 2/3 | (5/6) m v² | 40.0% | 3.43 |
| Thin hoop | 1 | m v² | 50.0% | 3.13 |
The rotational share equals k / (1 + k).
Example: A solid sphere with m = 5 kg rolls at v = 4 m/s. The translational energy is ½ × 5 × 4² = 40 J. The rotational energy is ½ × (2/5) × 5 × 4² = 16 J. The total is 56 J, and the rotational share is 16 / 56 = 28.6%.
Rolling without slipping is also pure rotation about the contact point. The energy is ½ I_contact ω², where I_contact = I_cm + m r². A solid cylinder gives ½ × (3/2) m r² ω² = (3/4) m v².
How Do You Use Rotational Kinetic Energy in Energy Conservation Problems?
Energy conservation problems set the lost potential energy equal to the gained translational and rotational kinetic energy. The equation is m g h = ½ m v² + ½ I ω². Friction that causes rolling without slipping does no work on the object, so mechanical energy stays constant.
How Fast Does a Rolling Object Move at the Bottom of an Incline?
The speed at the bottom is v = √(2 g h / (1 + k)), and it does not depend on the mass or the radius. A smaller k gives a higher speed. The ranking from fastest to slowest is the solid sphere, the solid cylinder, the thin spherical shell, and the thin hoop.
Example: Objects start from rest at a height of 1.0 m. The sphere reaches √(2 × 9.8 / 1.4) = 3.74 m/s, and the hoop reaches √(2 × 9.8 / 2) = 3.13 m/s.
How Do You Solve a Pulley Problem With Rotational Kinetic Energy?
A pulley problem is solved with m g h = ½ m v² + ½ I ω² and the condition ω = v / R. A pulley that is a solid disk has I = ½ M R², so the equation becomes m g h = ½ (m + M/2) v².
Example: A 2 kg mass hangs from a string on a 4 kg solid disk pulley and falls 1.5 m from rest.
- The potential energy released is 2 × 9.8 × 1.5 = 29.4 J.
- The effective mass is 2 + 4 / 2 = 4 kg.
- The speed is √(2 × 29.4 / 4) = 3.83 m/s.
- The hanging mass has ½ × 2 × 3.83² = 14.7 J.
- The pulley has ¼ × 4 × 3.83² = 14.7 J of rotational energy.
How Fast Does a Swinging Rod Rotate at the Bottom?
A rod pivoted at 1 end and released from horizontal has ω = √(3g / L) at the lowest point. The result follows from m g (L/2) = ½ × (1/3) m L² × ω².
Example: A rod with m = 2 kg and L = 1.0 m loses 2 × 9.8 × 0.5 = 9.8 J of potential energy. The angular velocity is √(3 × 9.8 / 1.0) = 5.42 rad/s, and the tip speed is 5.42 m/s.
How Much Energy Does a Flywheel Store?
A flywheel stores ΔKE = ½ I (ω_max² − ω_min²) between its maximum and minimum angular velocities. The stored energy is proportional to I and to the square of the angular velocity.
Example: A flywheel with I = 2 kg·m² slows from 20,000 rpm to 10,000 rpm.
- The maximum angular velocity is 2,094.4 rad/s.
- The minimum angular velocity is 1,047.2 rad/s.
- The usable energy is ½ × 2 × (2,094.4² − 1,047.2²) = 3.29 × 10⁶ J.
- The result equals 3.29 MJ, or 0.91 kWh.
Material strength limits the rotation rate. A thin rim has a hoop stress of σ = ρ v², so the maximum energy per kilogram is σ / (2ρ). Example: A steel with a design stress of 250 MPa and a density of 7,850 kg/m³ stores 15.9 kJ/kg, which equals 4.4 Wh/kg.
What Is the Rotational Kinetic Energy of the Earth and of Molecules?
Rotation energy appears at planetary and molecular scales, and Earth has a rotational kinetic energy of about 2.14 × 10²⁹ J. The table below lists the inputs and the result for Earth.
| Quantity | Value |
|---|---|
| Moment of inertia of Earth | 8.04 × 10³⁷ kg·m² |
| Angular velocity of Earth | 7.292 × 10⁻⁵ rad/s |
| Rotational kinetic energy | ½ × 8.04 × 10³⁷ × (7.292 × 10⁻⁵)² ≈ 2.14 × 10²⁹ J |
The moment of inertia of Earth equals 0.3307 M R², which is less than the 0.4 M R² of a uniform sphere, because density increases toward the core. The orbital kinetic energy of Earth is about 2.65 × 10³³ J, which is roughly 12,000 times the rotational energy.
Molecules have rotational energy as well. The classical equipartition theorem assigns ½ k_B T to each rotational degree of freedom. The table below lists the average rotational energy at 300 K, with k_B = 1.380649 × 10⁻²³ J/K.
| Molecule Type | Rotational Degrees of Freedom | Average Rotational Energy | Example Molecules |
|---|---|---|---|
| Monatomic | 0 | 0 | Helium, argon |
| Linear | 2 | k_B T = 4.14 × 10⁻²¹ J (0.0259 eV) | Nitrogen, oxygen, carbon monoxide |
| Nonlinear | 3 | (3/2) k_B T = 6.21 × 10⁻²¹ J | Water, methane |
Quantum mechanics quantizes the energy of a rigid rotor: E_J = ħ² J (J + 1) / (2I), where J = 0, 1, 2, and so on. Example: Carbon monoxide has a bond length of 112.8 pm and a reduced mass of 1.139 × 10⁻²⁶ kg. The moment of inertia is 1.45 × 10⁻⁴⁶ kg·m². The constant B = ħ² / (2I) is 3.84 × 10⁻²³ J, which equals 1.93 cm⁻¹. The first excited level has E₁ = 2B = 7.68 × 10⁻²³ J.
What Is the Difference Between Rotational and Translational Kinetic Energy?
Translational kinetic energy belongs to motion of the center of mass along a path, and rotational kinetic energy belongs to motion about an axis. The table below compares 9 linear quantities with their rotational counterparts.
| Linear Quantity | Linear Equation | Rotational Quantity | Rotational Equation |
|---|---|---|---|
| Mass (m) | kg | Moment of inertia (I) | kg·m² |
| Velocity (v) | m/s | Angular velocity (ω) | rad/s |
| Kinetic energy | ½ m v² | Kinetic energy | ½ I ω² |
| Momentum | p = m v | Angular momentum | L = I ω |
| Kinetic energy from momentum | p² / (2m) | Kinetic energy from angular momentum | L² / (2I) |
| Force (F) | N | Torque (τ) | N·m |
| Newton’s second law | F = m a | Newton’s second law | τ = I α |
| Work | W = F d | Work | W = τ Δθ |
| Power | P = F v | Power | P = τ ω |
A body that rotates and translates has a total energy of ½ M v_cm² + ½ I_cm ω², where v_cm is the speed of the center of mass. This split is known as König’s theorem.
What Are the Units of the Rotational Kinetic Energy Equation and How Are They Converted?
The SI unit of rotational kinetic energy is the joule (J), which equals 1 kg·m²/s² and 1 N·m. The radian is a dimensionless unit, so ω² contributes a factor of 1/s². The table below lists 6 quantities with the SI unit and the conversions.
| Quantity | Symbol | SI Unit | Conversions |
|---|---|---|---|
| Energy | KE | J | 1 kWh = 3.6 × 10⁶ J, 1 ft·lbf = 1.3558 J, 1 eV = 1.602 × 10⁻¹⁹ J |
| Moment of inertia | I | kg·m² | 1 lb·ft² = 0.04214 kg·m², 1 g·cm² = 10⁻⁷ kg·m² |
| Angular velocity | ω | rad/s | 1 rpm = 0.10472 rad/s, 1 Hz = 6.2832 rad/s, 1°/s = 0.017453 rad/s |
| Frequency | f | Hz | 1 Hz = 60 rpm |
| Angular momentum | L | kg·m²/s | 1 kg·m²/s = 1 J·s |
| Torque | τ | N·m | 1 lbf·ft = 1.3558 N·m |
Example conversion: 1,233.7 J equals 910 ft·lbf and 0.343 Wh.
Torque and energy share the unit N·m, but torque is a vector and energy is a scalar. Torque is written in N·m, and energy is written in J.
How Do You Measure Rotational Kinetic Energy Experimentally?
Rotational kinetic energy is measured by determining the moment of inertia and the angular velocity separately and then applying KE = ½ I ω². Instruments for angular velocity include tachometers, rotary encoders, photogates, stroboscopes, and smartphone gyroscopes, which report ω in rad/s.
Methods for the moment of inertia include calculation from geometry, a torsion pendulum, and a falling-weight test. The torsion pendulum gives I = κ T² / (4π²), where κ is the torsion constant and T is the oscillation period.
The falling-weight method follows 5 steps:
- Wrap a string around an axle of radius r and attach a mass m.
- Release the mass from rest and measure the fall height h.
- Record the fall time t with a stopwatch or photogate.
- Calculate I = m r² (g t² / (2h) − 1).
- Calculate the final angular velocity with ω = 2h / (t r).
Example: m = 0.5 kg, r = 0.02 m, h = 1.0 m, and t = 4.0 s.
- The moment of inertia is 0.5 × 0.02² × (9.8 × 16 / 2 − 1) = 0.0155 kg·m².
- The angular velocity at the bottom is 2 × 1.0 / (4.0 × 0.02) = 25 rad/s.
- The rotational energy is ½ × 0.0155 × 25² = 4.84 J.
The potential energy released is 0.5 × 9.8 × 1.0 = 4.9 J. The translational energy of the mass is 0.06 J, so the energies add to 4.9 J.
Repeating the measurement in 3 trials reduces random error.
What Are the Limits of the Rotational Kinetic Energy Equation?
The equation applies to rigid bodies that rotate about a fixed axis or about the center of mass. 6 conditions limit the equation:
- Rigid body: the distances between particles stay constant.
- Single axis: I must match the actual rotation axis.
- Deformable bodies: a changing shape changes I during the motion.
- Fluids: liquid inside a spinning container does not rotate as a rigid body.
- Non-relativistic speeds: the rim speed ωr stays far below the speed of light.
- Quantum scale: molecules require the quantized levels E_J = ħ² J (J + 1) / (2I).
What Are the Common Mistakes When Using the Rotational Kinetic Energy Equation?
8 common mistakes cause incorrect rotational kinetic energy results. Rimoldini and Singh published a study in Physical Review Special Topics: Physics Education Research in 2005 and found that introductory physics students have difficulty with rotational and rolling motion concepts.
The mistakes include:
- Using rpm or degrees per second instead of rad/s.
- Forgetting to square the angular velocity.
- Dropping the factor ½.
- Using diameter instead of radius in I.
- Applying m r² to every shape.
- Using the moment of inertia about the wrong axis.
- Leaving out the translational term for a rolling object.
- Mixing up torque (N·m) with energy (J).
Example of the first mistake: A disk with I = 2.5 kg·m² turns at 300 rpm. Substituting 300 directly gives ½ × 2.5 × 300² = 112,500 J. The correct value is 1,233.7 J, so the error is a factor of 91.2.
What Are Real-Life Applications of the Rotational Kinetic Energy Equation?
The rotational kinetic energy equation applies to 6 technologies and natural systems. Examples include:
- Flywheel energy storage systems, which store electrical energy as spin and release it on demand.
- Engine flywheels, which smooth the torque pulses of reciprocating engines.
- Turbine rotors, which transfer fluid energy to generators in wind, steam, and hydroelectric plants.
- Reaction wheels, which orient satellites by exchanging angular momentum and rotational energy.
- Rolling vehicles and wheels, which require extra energy for acceleration because of their moment of inertia.
- Figure skaters and divers, who change the moment of inertia to control the spin rate.
What Do People Also Ask About the Rotational Kinetic Energy Equation?
What is the rotational kinetic energy equation?
The equation is KE_rot = ½ I ω², where I is the moment of inertia and ω is the angular velocity.
What is the unit of rotational kinetic energy?
The SI unit of rotational kinetic energy is the joule (J).
What is the formula for rotational kinetic energy with rpm?
The formula is KE = π² I N² / 1800, where N is in revolutions per minute.
What is the formula for rotational kinetic energy with angular momentum?
The formula is KE = L² / (2I).
Is rotational kinetic energy a scalar or a vector?
Rotational kinetic energy is a scalar.
Can rotational kinetic energy be negative?
Rotational kinetic energy cannot be negative, because ω² and I are never negative.
What is moment of inertia?
Moment of inertia is the sum of mass times squared distance from the axis (I = Σ m r²), and it measures resistance to angular acceleration.
What is the rotational kinetic energy of a solid disk?
A solid disk has KE = ¼ m r² ω², because its moment of inertia is ½ m r².
What is the rotational kinetic energy of a solid sphere?
A solid sphere has KE = (1/5) m r² ω², because its moment of inertia is (2/5) m r².
What happens to rotational kinetic energy when angular velocity doubles?
The rotational kinetic energy increases by a factor of 4.
What is the difference between rotational and translational kinetic energy?
Translational energy (½ m v²) belongs to straight-line motion, and rotational energy (½ I ω²) belongs to spinning about an axis.
What is the total kinetic energy of a rolling object?
The total is ½ m v² + ½ I ω², which equals ½ m v² (1 + I / (m r²)) for rolling without slipping.
What fraction of the energy of a rolling sphere is rotational?
A solid sphere has 2/7, or 28.6%, of its total kinetic energy as rotational energy.
Why does a skater gain energy when pulling in the arms?
The moment of inertia drops at constant angular momentum, so KE = L² / (2I) rises, and the muscles supply the work.
How do you find rotational kinetic energy from work?
The net work of the torque equals the change in rotational kinetic energy (W = τ Δθ = ΔKE).
What Is the Rotational Kinetic Energy Equation and How Is It Applied?
The rotational kinetic energy equation is KE_rot = ½ I ω², which gives the energy of a spinning body in joules. The 7 forms use angular velocity, angular momentum, frequency, revolutions per minute, linear speed, particle sums, or torque and angle. The moment of inertia depends on the mass, the shape, and the axis, and the angular velocity must be in rad/s. Doubling the angular velocity multiplies the energy by 4. A rolling object adds ½ m v² for a total of ½ m v² (1 + k). Applications include flywheel storage, turbines, engine flywheels, reaction wheels, rolling vehicles, and spinning athletes.
Title: Difference Between Distance and Displacement: 10 Key Differences, Formulas, and Examples
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