Theme A — Space, Time and Motion
A.4.9 Rotational Kinetic Energy — IB Physics HL Revision Notes & Topical Questions
5 min read 8 parts Lesson 51 of 63 1 quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Lesson summary
A.4.9 Rotational Kinetic Energy — IB PHYSICS HL notes: Calculate rotational kinetic energy and apply it to rolling bodies and bodies on slopes.A rotating body has kinetic energy associated with its angular velocity.Ek = ½Iω2, the rotational analogue of ½mv2.Ek = L2/2I, the analog…
- Rolling motion
- Without slipping, each point has a different linear velocity
- Solid cylinder (I
- Solid sphere (I
- For a solid sphere mgh
- A rolling body reaches a lower speed than a sliding one
Part 1 of 6Rotational kinetic energy
A rotating body has kinetic energy associated with its angular velocity.
Ek
½Iω2, the rotational analogue of ½mv2.
Ek
L2/2I, the analogue of p2/2m.
I is in kg m2, ω in rad s-1 and L in kg m2 s-1.
Part 2 of 6Rolling without slipping
1Rolling motion
Rolling motion rotation combined with translation.
2
With enough friction the object rolls without slipping.
- The contact point is at rest.
- The centre of mass has speed v = ωr.
- The top point moves at 2v = 2ωr.
3
When a disc slips, every point has the same linear velocity and ω is zero.
4Without slipping, each point has a different linear velocity
- Without slipping, each point has a different linear velocity proportional to r from the centre.
Part 3 of 6Total kinetic energy of a rolling body
Ek total
½mv2 + ½Iω2.
Use v
ωr to write it in terms of one speed.
Solid cylinder (I
½mr2) — Ek total = ¾mv2.
Solid sphere (I
(2/5)mr2) — Ek total = (7/10)mv2.
Part 4 of 6Rolling down a slope
1
At the top the body has Ep = mgh and is at rest.
2
At the bottom the potential energy has become translational and rotational kinetic energy.
3
Equate mgh = ½mv2 + ½Iω2.
4
Substitute I and use v = ωr to find v or ω.
1For a solid sphere mgh
(7/10)mv2.
2A rolling body reaches a lower speed than a sliding one
- A rolling body reaches a lower speed than a sliding one because some energy goes into rotation.
Part 5 of 6Changes of rotational energy
Changes of rotational energy
Combining flywheels changes I and ω together.
- For a flywheel of ½MR2 with a second of ½M and ½R placed on top, the moment of inertia rises to (9/16)MR2.
- At 2/3 the angular velocity, Ek becomes ½ of the original.
Compute Ek before and after using ½Iω2
- Compute Ek before and after using ½Iω2 and track both I and ω.
Rotational energy can be converted to other forms, e.g. thermal energy through friction.
Part 6 of 6More from the specification
More from the specification
Rotational kinetic work in flywheel systems
- Rotational kinetic work in flywheel systems calculating half moment of inertia times angular velocity squared gives energy stored in a flywheel system.
Rotational kinetic energy
- E equals one half I omega squared, where I is moment of inertia and omega angular speed.
Torque and work
- Torque and work work done by a torque equals torque times angular displacement.
Angular acceleration
- Angular acceleration resultant torque divided by moment of inertia gives it.
Rolling
- Rolling a rolling object has both translational and rotational kinetic energy, so it accelerates down a slope more slowly than a sliding one.
Conservation of energy
- Conservation of energy on a slope, loss of gravitational energy equals the gain in kinetic energy of both types.
Equilibrium
- State with zero net force.
Perpendicular
- At right angles.
Impulse
- J = FΔt = Δp.What the symbols mean: F = force (N) · Δt = time the force acts (s) · Δp = change in momentum (kg m s−1); FΔt is the impulse (N s)
Radius
- Distance from the centre to the circular path.
Constant
- Not changing.
Rotational equilibrium
- Στ = 0; no angular acceleration.
Angular impulse
- Product of net torque and the time it acts, τΔt.
Perpendicular distance
- Distance from the axis to the line of motion, r⊥ = r sinθ.
Torque
- Rotational effect of a force, τ = rF sinθ.
Acceleration
- Rate of change of velocity.
Resultant
- Vector sum of two or more vectors.
Force
- A push or pull measured in newtons.
Distance
- Total length of path travelled, a scalar.
Displacement
- Change in position.
Couple
- A pair of equal opposite forces with different lines of action.
Clockwise
- Turning in the direction of a clock's hands.
Key terms
| Term | Meaning |
|---|---|
| Rotational kinetic energy | Energy of rotation, ½Iω^2. |
| Rolling without slipping | Rolling with the contact point at rest, v = ωr. |
| Moment of inertia | Resistance to change in rotational motion. |
| Angular velocity | Rate of change of angular displacement. |
| Translational kinetic energy | Energy of motion of the centre of mass, ½mv2. |
| Flywheel | Rotating disc used to store rotational energy. |
Avoid these mistakes
1
Using only ½mv2 for a rolling body — rotational energy is also present.
2
Forgetting that the top of a rolling wheel moves at 2v.
3
Using ω in rev/s or degrees/s in ½Iω^2.
4
Assuming the same Ek for all rolling shapes — it depends on I.
5
Treating a sliding and a rolling body as having equal speeds at the bottom of a slope.
Marked practiceQuizzes for this lesson
Last updated:
\n