This MCQ module is based on: Angular Momentum Equilibrium
TOPIC 26 OF 33
Angular Momentum Equilibrium
🎓 Class 11
Physics
CBSE
Theory
Ch 6 – System of Particles and Rotational Motion
⏱ ~14 min
🌐 Language: [gtranslate]
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Angular Momentum Equilibrium
6.8 Angular Momentum of a Particle
The Angular momentum of a particle of momentum \(\vec{p} = m\vec{v}\) at position \(\vec{r}\) about a chosen origin is:
\[\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}\]
Magnitude: \(L = mvr\sin\theta = m v r_\perp\) where \(r_\perp\) is the perpendicular distance from origin to the line of motion. Direction: by right-hand rule, perpendicular to the plane of \(\vec{r}\) and \(\vec{p}\).
Differentiating L with respect to time:
\[\frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}\]
The cross-term \(\frac{d\vec{r}}{dt} \times \vec{p}\) vanishes because v is parallel to p. So:
Newton's 2nd law for rotation:
\[\boxed{\;\vec{\tau} = \frac{d\vec{L}}{dt}\;}\]
The net torque on a particle equals the rate of change of its angular momentum — the angular analogue of \(\vec{F} = d\vec{p}/dt\).
6.9 Angular Momentum of a System of Particles
For a system, the total angular momentum is \(\vec{L}_{\text{total}} = \sum \vec{L}_i\). The internal torques cancel pairwise (Newton's 3rd law for non-collinear forces), so:
\[\frac{d\vec{L}_{\text{total}}}{dt} = \vec{\tau}_{\text{external}}\]
Conservation of Angular Momentum: If the net external torque is zero, the total angular momentum of a system is conserved.
\[\boxed{\;\vec{\tau}_{\text{ext}} = 0 \;\Rightarrow\; \vec{L}_{\text{total}} = \text{constant}\;}\]
Beautiful demonstrations:
- An ice skater pulls in arms — moment of inertia I decreases — ω increases (since L = Iω is constant).
- A diver tucks during somersault to spin faster, then extends to slow down before entry.
- A spinning top precesses but maintains spin angular momentum.
- Planets sweep equal areas in equal times (Kepler's 2nd law) — a consequence of L conservation.
6.10 Equilibrium of a Rigid Body
A rigid body is in mechanical equilibrium when both translational and rotational accelerations are zero. This requires:
Two Conditions for Equilibrium:
- Translational equilibrium: \(\sum \vec{F}_{\text{ext}} = 0\)
- Rotational equilibrium: \(\sum \vec{\tau}_{\text{ext}} = 0\) (about ANY point — if it holds for one point, it holds for all under translational equilibrium)
6.10.1 Couple
A pair of equal, opposite, non-collinear forces (a "couple") produces no net force but a non-zero net torque. Examples: the two forces on a steering wheel; the magnetic force on a compass needle. Such a couple causes pure rotation without translation.
6.10.2 Principle of Moments — The Lever
For a lever in rotational equilibrium with effort F₁ at distance d₁ and load F₂ at distance d₂ on opposite sides of the fulcrum:
\[F_1 d_1 = F_2 d_2 \;\;\;(\text{Mechanical Advantage} = F_2/F_1 = d_1/d_2)\]
6.10.3 Centre of Gravity
For a body in a uniform gravitational field, the gravitational force on each particle produces a torque. The CG is the point where the resultant gravitational force can be considered to act. In a uniform g, CG coincides with CoM. (For very tall bodies in non-uniform g, they may differ slightly.)
🎯 Interactive Simulation: Lever Balance
Place two weights on opposite sides of a fulcrum. Adjust positions and weights to find balance. The system tilts until F₁d₁ = F₂d₂.
τ_left = 80 N·m | τ_right = 80 N·m
⚖ Balanced
Example 6.7: Skater Conservation of L
A skater spinning at 1 rev/s with arms extended (I = 5 kg·m²) pulls her arms in to bring I down to 1.25 kg·m². Find her new angular speed.
No external torque ⇒ L conserved: I₁ω₁ = I₂ω₂.
\[\omega_2 = \frac{I_1}{I_2}\omega_1 = \frac{5}{1.25} \times 1 \text{ rev/s} = 4\text{ rev/s} = 8\pi\text{ rad/s}\] KE increases (½Iω² goes from ½·5·4π² = 10π² to ½·1.25·64π² = 40π²) — the skater does muscular work pulling arms inward!
\[\omega_2 = \frac{I_1}{I_2}\omega_1 = \frac{5}{1.25} \times 1 \text{ rev/s} = 4\text{ rev/s} = 8\pi\text{ rad/s}\] KE increases (½Iω² goes from ½·5·4π² = 10π² to ½·1.25·64π² = 40π²) — the skater does muscular work pulling arms inward!
Example 6.8: Ladder Equilibrium
A 5 m, 20 kg uniform ladder leans against a frictionless wall at angle 60° to the floor. The floor has friction. Find the normal forces and the minimum coefficient of friction needed for equilibrium. (g = 10 m/s²)
Wall normal N_w (horizontal), Floor normal N_f (vertical), Floor friction f (horizontal). Weight W = 200 N at midpoint.
Translational: N_f = W = 200 N (vertical). f = N_w (horizontal).
Rotational about base: N_w × 5sin60° = W × 2.5cos60° ⇒ N_w × 4.33 = 200 × 1.25 = 250.
N_w = 250/4.33 ≈ 57.74 N. So f = 57.74 N.
Min μ: μ ≥ f/N_f = 57.74/200 = 0.289.
Translational: N_f = W = 200 N (vertical). f = N_w (horizontal).
Rotational about base: N_w × 5sin60° = W × 2.5cos60° ⇒ N_w × 4.33 = 200 × 1.25 = 250.
N_w = 250/4.33 ≈ 57.74 N. So f = 57.74 N.
Min μ: μ ≥ f/N_f = 57.74/200 = 0.289.
Example 6.9: Angular Momentum of a Projectile
A 2 kg ball is thrown horizontally at 10 m/s from a 5 m high cliff. Calculate its angular momentum about the throwing point at the moment just before it lands. (g = 10 m/s²)
Time to fall: t = √(2h/g) = √1 = 1 s. Vertical speed at landing: v_y = gt = 10 m/s. Horizontal: v_x = 10 m/s.
Horizontal range: x = 10 m. So r = (10, -5, 0) m relative to throw point.
v = (10, -10, 0) m/s, p = (20, -20, 0) kg·m/s.
\[\vec{L} = \vec{r}\times\vec{p} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\10&-5&0\\20&-20&0\end{vmatrix} = \hat{k}[(10)(-20) - (-5)(20)] = \hat{k}[-200 + 100] = -100\hat{k}\] L = 100 kg·m²/s directed into the page.
Horizontal range: x = 10 m. So r = (10, -5, 0) m relative to throw point.
v = (10, -10, 0) m/s, p = (20, -20, 0) kg·m/s.
\[\vec{L} = \vec{r}\times\vec{p} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\10&-5&0\\20&-20&0\end{vmatrix} = \hat{k}[(10)(-20) - (-5)(20)] = \hat{k}[-200 + 100] = -100\hat{k}\] L = 100 kg·m²/s directed into the page.
🌀 Activity 6.3 — Spinning Stool
L4 Analyse
Materials: Rotating stool/chair, two heavy books or dumbbells.
Procedure:
- Sit on the stool holding the books out at arm's length. Have a partner spin you gently.
- Quickly pull the books in close to your chest. Observe the change in rotation rate.
- Extend the books outward again. Observe the speed change.
Predict: If you halve your moment of inertia (by pulling in), by what factor will your angular speed change?
Conclusion: L = Iω is conserved in absence of external torque (the bearing friction is small for short experiments). Halving I doubles ω. This same principle is used by gymnasts, divers, ice skaters, and helicopter rotors. The KE = ½Iω² actually INCREASES when you pull in — the extra energy comes from the work your muscles do pulling the books inward.
🎯 Competency-Based Questions
A 4 m uniform plank weighing 80 N rests on two supports. Support A is at the left end; support B is 1 m from the right end. A 200 N child stands at the midpoint of the plank.
Q1. Find the upward force exerted by support B.L3 Apply
Answer: (c). Take torques about A: N_B × 3 = 80 × 2 + 200 × 2 = 160 + 400 = 560 ⇒ N_B = 186.67 N ≈ 187 N.
Q2. Find the upward force exerted by support A.L3 Apply
Answer: Vertical equilibrium: N_A + N_B = 80 + 200 = 280 ⇒ N_A = 280 − 187 = 93.33 N.
Q3. State whether TRUE or FALSE: "If the child walks toward support A, support B's reading decreases." L4 Analyse
Answer: TRUE. As the child moves closer to A, her contribution to torque about A decreases, so N_B (the force needed to balance) decreases. This is the principle behind weighbridges with two scales.
Q4. A figure skater's rotational KE doubles when she pulls in arms. State whether energy is conserved. L5 Evaluate
Answer: Mechanical KE is NOT conserved, but TOTAL energy is. The extra rotational KE comes from internal muscular work the skater does pulling her arms in against the centripetal pull. Angular momentum (L = Iω) IS conserved because no external torque acts.
Q5. HOT: Design a balance for measuring small masses using only a metre rule and a known mass. Specify operation. L6 Create
Sample design: Pivot the metre rule at its centre. Place known mass M at distance d_known from pivot on one side. Place unknown mass m on the other side at distance d_unknown so the rule balances horizontally. Then m × d_unknown = M × d_known ⇒ m = M·d_known/d_unknown. Sensitivity grows with longer arms — this is the principle of the chemical balance.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): A figure skater spins faster when she pulls her arms in.
Reason (R): Angular momentum is conserved when no external torque acts.
Answer: (A). Both true; R explains A. Smaller I → larger ω with L = Iω fixed.
Assertion (A): A body in translational equilibrium is also in rotational equilibrium.
Reason (R): If net force is zero, net torque is also zero.
Answer: (D). Assertion FALSE — a couple has zero net force but non-zero torque (causes rotation). Reason FALSE in general; only true if forces are concurrent. So A is false, R is false. Closest to (D) — but technically both FALSE.
Assertion (A): The centre of gravity of a body coincides with its centre of mass in a uniform gravitational field.
Reason (R): In a uniform field, gravity acts equally on every particle of a body.
Answer: (A). Both true; R explains A.
Frequently Asked Questions - Angular Momentum Equilibrium
What is the main concept covered in Angular Momentum Equilibrium?
In NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion), "Angular Momentum Equilibrium" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Angular Momentum Equilibrium useful in real-life applications?
Real-life applications of Angular Momentum Equilibrium from NCERT Class 11 Physics Chapter 6 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Angular Momentum Equilibrium?
Key formulas in Angular Momentum Equilibrium (NCERT Class 11 Physics Chapter 6 System of Particles and Rotational Motion) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 6?
NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion) is structured so each part builds on the previous one. Angular Momentum Equilibrium connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Angular Momentum Equilibrium?
CBSE board questions from Angular Momentum Equilibrium typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Angular Momentum Equilibrium lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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