Angular momentum (A-level only) — AQA A-Level Physics
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Angular momentum is the rotational analogue of linear momentum, defined as L = I-omega for a rigid body or L = m-v-r for a point mass, and is conserved when no external torque acts on a system. In AQA A-Level Physics, you must apply conservation of angular momentum to problems involving changing moment of inertia, such as a spinning ice skater pulling in their arms or a collapsing star.
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Angular momentum (A-level only) explained
Angular momentum (A-level only) covers 3 specification points, including Angular momentum = Iω Conservation of angular… and Angular impulse = change in angular momentum;….
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Open a point below for its explanation and revision guidance.
angular momentum = Iω Conservation of angular momentum.
Angular momentum is the rotational analogue of linear momentum. For a rigid body rotating about a fixed axis, it equals the product of moment of inertia I and angular velocity ω, so L = Iω. I depends on mass distribution about the axis; ω is the angular velocity in rad s⁻¹. Conservation of angular momentum states that if the resultant external torque on a system is zero, its total angular momentum remains constant. A skater pulling in their arms reduces I, so ω increases to keep Iω unchanged. Similarly, a planet in an elliptical orbit moves faster when closer to the Sun because its moment of inertia about the Sun is smaller. To solve problems, identify the axis, calculate I for the given shape, convert ω to rad s⁻¹, then apply L = Iω before and after any change.
Angular impulse = change in angular momentum; T ∆t = ∆ Iω where T is constant.
Angular impulse is the rotational analogue of linear impulse. When a constant torque T acts for a time ∆t, the angular impulse is T ∆t and it equals the change in angular momentum, ∆(Iω). This follows from Newton's second law in rotational form, T = ∆(Iω)/∆t, rearranged to T ∆t = ∆(Iω). If I is constant, the change is I∆ω; if I changes, the change is ∆(Iω) as a whole. For example, a flywheel with I = 0.50 kg m² accelerated from rest to 12 rad s⁻¹ by a constant torque of 3.0 N m takes ∆t = I∆ω/T = (0.50 × 12)/3.0 = 2.0 s. To solve problems, identify the torque, the time interval, and the initial and final angular momenta, then apply T ∆t = ∆(Iω) with consistent signs.
Applications may include examples from sport.
Angular momentum L = Iω is conserved when the net external torque is zero. In sport, this explains why a spinning skater or diver can change angular speed by altering body configuration. Pulling limbs in reduces the moment of inertia I, so ω must increase to keep L constant; extending limbs increases I and reduces ω. The rotational kinetic energy E = ½Iω² changes because internal work is done, even though L is conserved. Students should identify the system, confirm negligible external torque, compare I before and after, and then use conservation to find the new ω. They should also relate the effect to real sporting techniques, such as a gymnast tucking during a somersault or a diver tightening before entry, and explain the role of frictional torque in slowing rotation.
Your focus
- Define angular momentum as L = Iω and use correct units.
- Apply conservation of angular momentum to systems where the resultant external torque is zero.
- Calculate the new angular velocity when the moment of inertia of a rotating system changes.
Show all 9 objectives
- Define angular impulse as T ∆t and state its units.
- Apply T ∆t = ∆(Iω) to calculate torque, time, or change in angular momentum for constant torque.
- Explain how angular impulse relates to Newton's second law in rotational form.
- State the condition under which angular momentum is conserved.
- Apply conservation of angular momentum to calculate a changed angular velocity when the moment of inertia changes.
- Describe a sporting example that demonstrates conservation of angular momentum and explain the changes in I, ω and rotational kinetic energy.
Which command word is this?
You must use the given data and a correct formula to arrive at a numerical answer with appropriate units. Show all working; an answer alone may not gain full marks if the method is unclear.
Angular momentum (A-level only) exam tips
Topic Overview
Angular momentum is the rotational equivalent of linear momentum and is a conserved quantity in the absence of external torques. For a rigid body rotating about a fixed axis, it is defined as the product of moment of inertia and angular velocity, L = I-omega. For a point mass, it is L = m-v-r-sin(theta). This topic extends your understanding of circular motion and moments of inertia, and it appears in AQA A-Level Paper 3 (Turning Points in Physics) and in synoptic questions across all papers.
The principle of conservation of angular momentum explains a wide range of phenomena, from an ice skater spinning faster when pulling in their arms to the formation of neutron stars from collapsing stellar cores. It also underpins Kepler's second law of planetary motion. Mastering this topic requires confident use of radian measure, rotational kinetic energy, and the ability to link linear and angular quantities.
Key Concepts
- →Angular momentum L is a vector quantity, defined for a rigid body as L = I-omega, where I is the moment of inertia about the axis of rotation and omega is the angular velocity in rad per s.
- →For a point mass moving in a straight line, angular momentum about a point is L = m-v-r-sin(theta), where r is the distance from the point to the mass and theta is the angle between r and v.
- →Conservation of angular momentum: if the net external torque on a system is zero, the total angular momentum remains constant. This means I1-omega1 = I2-omega2 for a system with changing moment of inertia.
- →Rotational kinetic energy is E_k = (1/2)-I-omega^2. In problems where I changes, kinetic energy is not conserved because work is done by internal forces.
- →Angular impulse is the product of torque and time, and it equals the change in angular momentum: tau x delta t = delta L.
Marking Points
- State that angular momentum L is the product of moment of inertia I and angular velocity ω, with L = Iω and units kg m² rad s⁻¹ or kg m² s⁻¹.
- Explain that moment of inertia I depends on mass and how that mass is distributed about the chosen axis, so the same body can have different I values for different axes.
- Apply conservation of angular momentum when the resultant external torque is zero, setting I₁ω₁ = I₂ω₂ for a system whose mass distribution changes.
- Convert angular velocity between rev s⁻¹ and rad s⁻¹ using ω = 2πf before substituting into L = Iω.
- Interpret a decrease in I as causing an increase in ω when L is conserved, and link this to the skater or collapsing-star example.
- Use consistent sign conventions for clockwise and anticlockwise rotation when combining angular momenta of interacting bodies.
- State that angular impulse is the product of torque and the time for which it acts, with units N m s.
- Apply the relationship T ∆t = ∆(Iω) for a constant torque, recognising that ∆(Iω) may involve changes in I, ω, or both.
- Rearrange the equation to find torque, time interval, or change in angular momentum as required.
- Use consistent sign conventions for torque and angular momentum about the chosen axis.
- Link the equation to Newton's second law in rotational form, T = ∆(Iω)/∆t, to justify why angular impulse equals the change in angular momentum.
- Calculate ∆ω from initial and final angular velocities in rad s⁻¹, converting units where necessary.
- States that angular momentum is conserved when the net external torque on the system is zero.
- Uses L = Iω to compare the moment of inertia before and after a change in body shape.
- Explains that reducing the moment of inertia increases angular velocity, and increasing it decreases angular velocity.
- Applies conservation of angular momentum to a named sporting action, such as a skater pulling in their arms or a diver tucking.
- Recognises that rotational kinetic energy changes because internal work is done, even though angular momentum is conserved.
- Discusses how frictional torque from air resistance or contact surfaces reduces angular momentum over time.
Examiner Tips
- 💡Write the conservation equation explicitly as I₁ω₁ = I₂ω₂ before substituting values, so the examiner can see the physical principle being applied.
- 💡Check that ω is in rad s⁻¹; if a frequency in Hz or rev s⁻¹ is given, convert using ω = 2πf.
- 💡State the condition for conservation clearly: resultant external torque is zero, not merely that no forces act.
- 💡Show the substitution into T ∆t = ∆(Iω) with units, so the examiner can follow the calculation and check the final unit.
- 💡If the torque is not constant, do not use T ∆t directly; instead consider the area under a torque–time graph or use the average torque.
- 💡Keep angular velocities in rad s⁻¹ and torques in N m throughout to avoid unit conversion errors.
- 💡Begin by stating the condition for conservation: net external torque is zero.
- 💡Show the before-and-after equation I₁ω₁ = I₂ω₂ and substitute values carefully, keeping units consistent.
- 💡When explaining a sporting example, name the action and describe how the body configuration changes I and ω.
- 💡Always state the principle of conservation of angular momentum and identify the axis about which it is conserved. Marks are awarded for explicitly saying 'no external torque acts' or 'net external torque is zero'.
- 💡Convert all angular velocities to rad per s before substituting into equations. Periods given in days or revolutions per minute must be converted.
- 💡When calculating angular momentum of a point mass, use the perpendicular distance from the axis to the line of action of the velocity, or use L = m-v-r-sin(theta). Do not simply use r as the distance from the axis to the mass unless the velocity is perpendicular to r.
Common Mistakes
- Treating angular momentum as a scalar and ignoring direction: correct by assigning positive and negative signs to opposite senses of rotation about the chosen axis.
- Using linear momentum p = mv instead of L = Iω for a rotating rigid body: correct by identifying the rotational variables and the axis of rotation.
- Assuming I is always mr² for any body: correct by selecting the appropriate expression for the shape, such as ½mr² for a solid disc about its central axis.
- Confusing angular impulse with linear impulse: correct by using torque T and angular momentum Iω rather than force F and linear momentum mv.
- Assuming T ∆t = I∆ω always, even when I changes: correct by writing the change as ∆(Iω) and evaluating initial and final values separately.
- Forgetting that torque must be constant for the simple form T ∆t = ∆(Iω): correct by checking the question states or implies a constant torque.
- Treating angular momentum as conserved even when a significant external torque acts; the correction is to check that the net external torque is zero before applying conservation.
- Confusing moment of inertia with mass alone; the correction is to use I = Σmr² and recognise that distribution of mass affects I.
- Assuming rotational kinetic energy is conserved when a skater pulls their arms in; the correction is to state that internal work increases kinetic energy while angular momentum stays constant.
- Students often think angular velocity remains constant when moment of inertia changes. Correction: angular momentum is conserved, so if I decreases, omega must increase proportionally.
- Students confuse angular momentum with linear momentum and use p = m-v instead of L = I-omega. Correction: angular momentum depends on the distribution of mass about the axis, not just the total mass and linear speed.
- Students assume rotational kinetic energy is conserved in a collision or when a skater pulls in their arms. Correction: kinetic energy changes because internal work is done; only angular momentum is conserved.
Revision Plan
- 1Day 1-2: Review the definitions of angular momentum for rigid bodies and point masses. Learn the formulas L = I-omega and L = m-v-r-sin(theta). Practice converting between linear and angular quantities.
- 2Day 3-4: Study the principle of conservation of angular momentum. Work through examples such as a skater pulling in arms, a collapsing star, and a planet in elliptical orbit. Focus on identifying when external torque is zero.
- 3Day 5-6: Practice calculations involving changes in moment of inertia and angular velocity. Include problems where rotational kinetic energy changes and you must explain why. Use past paper questions from AQA Paper 3.
- 4Day 7-8: Revise the relationship between angular impulse and change in angular momentum. Solve problems involving torques acting for a short time.
- 5Day 9-10: Complete a full topic test under timed conditions. Review mark schemes carefully, focusing on the wording required for conservation statements and explanations of energy changes.
Exam Question Types
- 📋Calculation of angular momentum for a rigid body or point mass: you will be given mass, radius, speed, or period and asked to find L. Ensure you use the correct formula and units.
- 📋Conservation of angular momentum problem: a system changes its moment of inertia (e.g., skater, star, or merging objects) and you must find the new angular velocity or period. State the conservation principle and show clear working.
- 📋Explanation question on why rotational kinetic energy changes when angular momentum is conserved: you must refer to work done by internal forces and the fact that kinetic energy is not conserved. Use the equations to support your answer.
- 📋Synoptic question linking angular momentum to Kepler's second law or to the formation of neutron stars: you may need to combine with gravitational field concepts.
Command Word Expectations (AQA)
You must use the given data and a correct formula to arrive at a numerical answer with appropriate units. Show all working; an answer alone may not gain full marks if the method is unclear.
You must provide a reasoned justification, often in terms of conservation laws or physical principles. Use precise terminology such as 'no external torque' or 'moment of inertia decreases'. A simple statement of the result is not sufficient.
You must demonstrate the steps leading to a given result. This is typically used when the final answer is provided; you must show the substitution and manipulation clearly to gain full marks.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A solid sphere of mass 2.0 kg and radius 0.15 m rolls without slipping along a horizontal surface at a linear speed of 3.0 m per s. Calculate its total kinetic energy and its angular momentum about its centre of mass.
- 1.Step 1: Identify given facts: mass m = 2.0 kg, radius r = 0.15 m, linear speed v = 3.0 m per s. For a solid sphere, moment of inertia I = (2/5)-m-r^2.
- 2.Step 2: Calculate I = (2/5) x 2.0 x (0.15)^2 = 0.018 kg m^2.
- 3.Step 3: Since it rolls without slipping, angular velocity omega = v / r = 3.0 / 0.15 = 20 rad per s.
- 4.Step 4: Total kinetic energy = translational KE + rotational KE = (1/2)-m-v^2 + (1/2)-I-omega^2 = (1/2)x2.0x(3.0)^2 + (1/2)x0.018x(20)^2 = 9.0 + 3.6 = 12.6 J.
- 5.Step 5: Angular momentum L = I-omega = 0.018 x 20 = 0.36 kg m^2 per s.
Question: A star of radius 7.0 x 10^5 km rotates with a period of 30 days. It collapses to form a neutron star of radius 10 km. Assuming the star is a uniform sphere and no mass is lost, calculate the new rotation period.
- 1.Step 1: Identify given facts: initial radius R1 = 7.0 x 10^5 km = 7.0 x 10^8 m, final radius R2 = 10 km = 1.0 x 10^4 m, initial period T1 = 30 days = 30 x 24 x 3600 = 2.592 x 10^6 s.
- 2.Step 2: Angular momentum is conserved: I1-omega1 = I2-omega2. For a uniform sphere, I = (2/5)-M-R^2, so (2/5)-M-R1^2-(2-pi/T1) = (2/5)-M-R2^2-(2-pi/T2).
- 3.Step 3: Cancel common factors: R1^2 / T1 = R2^2 / T2, so T2 = T1 x (R2/R1)^2.
- 4.Step 4: Substitute values: T2 = 2.592 x 10^6 x (1.0 x 10^4 / 7.0 x 10^8)^2 = 2.592 x 10^6 x (1.4286 x 10^-5)^2 = 2.592 x 10^6 x 2.041 x 10^-10 = 5.29 x 10^-4 s.
- 5.Step 5: Convert to milliseconds: T2 = 0.529 ms.