This MCQ module is based on: Vector Product Angular Velocity
TOPIC 25 OF 33
Vector Product Angular Velocity
🎓 Class 11
Physics
CBSE
Theory
Ch 6 – System of Particles and Rotational Motion
⏱ ~14 min
🌐 Language: [gtranslate]
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Vector Product Angular Velocity
6.4 Vector Product (Cross Product) of Two Vectors
Just as the dot product gave us a scalar from two vectors, the vector (cross) product produces a vector. It is essential for torque, angular momentum, and the magnetic Lorentz force.
\[\vec{A} \times \vec{B} = (|A||B|\sin\theta)\,\hat{n}\]
where \(\hat{n}\) is a unit vector perpendicular to the plane of \(\vec{A}\) and \(\vec{B}\), determined by the right-hand rule: curl the fingers of the right hand from \(\vec{A}\) to \(\vec{B}\); the thumb points along \(\hat{n}\).
Properties of the Cross Product:
- Anti-commutative: \(\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}\)
- Distributive: \(\vec{A} \times (\vec{B} + \vec{C}) = \vec{A}\times\vec{B} + \vec{A}\times\vec{C}\)
- \(\vec{A} \times \vec{A} = \vec{0}\)
- If \(\vec{A} \parallel \vec{B}\), \(\vec{A} \times \vec{B} = \vec{0}\)
- Unit-vector cross products: \(\hat{i}\times\hat{j} = \hat{k}\), \(\hat{j}\times\hat{k} = \hat{i}\), \(\hat{k}\times\hat{i} = \hat{j}\)
In components:
\[\vec{A}\times\vec{B} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix} = (A_yB_z-A_zB_y)\hat{i} + (A_zB_x-A_xB_z)\hat{j} + (A_xB_y-A_yB_x)\hat{k}\]
6.5 Angular Velocity and its Relation with Linear Velocity
Angular velocity \(\vec{\omega}\) is a vector along the rotation axis (direction by right-hand rule of curl). Its magnitude is \(\omega = d\theta/dt\), measured in rad/s.
Consider a particle at position \(\vec{r}\) (from any point on the rotation axis) on a rigid body rotating with \(\vec{\omega}\). Its linear velocity is:
\[\vec{v} = \vec{\omega} \times \vec{r}\]
This is a beautiful application of the cross product: the linear speed v = ωr sin α (where α is the angle between ω and r), and the direction is tangential to the circular path. For a particle at perpendicular distance ρ from the axis, v = ωρ.
6.6 Angular Acceleration
The angular acceleration is the rate of change of angular velocity:
\[\vec{\alpha} = \frac{d\vec{\omega}}{dt}\]
For rotation about a fixed axis, \(\vec{\alpha}\) lies along the axis; its sign tells whether ω is increasing (same direction as ω) or decreasing.
| Quantity | Linear (translational) | Angular (rotational) |
|---|---|---|
| Position | x | θ |
| Velocity | v = dx/dt | ω = dθ/dt |
| Acceleration | a = dv/dt | α = dω/dt |
| Mass / Inertia | m | I |
| Force / Torque | F = ma | τ = Iα |
| Momentum | p = mv | L = Iω |
| KE | ½ mv² | ½ Iω² |
6.7 Torque (Moment of a Force)
Torque measures the tendency of a force to cause rotation about an axis. For a force \(\vec{F}\) acting at position \(\vec{r}\) from the axis (or from a chosen origin):
\[\vec{\tau} = \vec{r} \times \vec{F}\]
Magnitude: \(\tau = rF\sin\theta = F \cdot d_\perp\) where \(d_\perp\) is the perpendicular distance from the line of action of F to the axis. Torque is greatest when force is applied perpendicular to position (θ = 90°).
🎯 Interactive Simulation: Torque Calculator
A force F is applied to a wrench of length r at angle θ from the wrench. Adjust to see how torque varies. Maximum torque is at θ = 90° (perpendicular).
τ = r·F·sin θ
15.00 N·m
(Effective lever arm = r sin θ = 0.30 m)
Example 6.4: Cross Product in Components
Find \(\vec{A}\times\vec{B}\) for \(\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{B} = \hat{i} - \hat{j} + 2\hat{k}\).
\[\vec{A}\times\vec{B} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&-1\\1&-1&2\end{vmatrix}\]
\[= \hat{i}(3\cdot2 - (-1)(-1)) - \hat{j}(2\cdot2 - (-1)(1)) + \hat{k}(2\cdot(-1) - 3\cdot1)\]
\[= \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) = \boxed{5\hat{i} - 5\hat{j} - 5\hat{k}}\]
Example 6.5: Linear Speed of a Carousel Rider
A merry-go-round rotates at 0.5 rad/s. A child sits 3 m from the centre. Find her linear speed and the direction of her velocity at the instant she is at the easternmost point. (Assume CCW rotation viewed from above.)
\(\vec{\omega} = 0.5\,\hat{k}\) (upward, for CCW from above). Position \(\vec{r} = 3\,\hat{i}\).
\[\vec{v} = \vec{\omega}\times\vec{r} = 0.5\hat{k}\times 3\hat{i} = 1.5(\hat{k}\times\hat{i}) = 1.5\hat{j}\text{ m/s}\] Speed: 1.5 m/s. Direction: due north (\(\hat{j}\)).
\[\vec{v} = \vec{\omega}\times\vec{r} = 0.5\hat{k}\times 3\hat{i} = 1.5(\hat{k}\times\hat{i}) = 1.5\hat{j}\text{ m/s}\] Speed: 1.5 m/s. Direction: due north (\(\hat{j}\)).
Example 6.6: Torque from a Force
A force \(\vec{F} = 4\hat{i} - 3\hat{j}\) N acts at point \(\vec{r} = 2\hat{i} + \hat{j}\) m relative to origin. Find the torque about the origin.
\[\vec{\tau} = \vec{r}\times\vec{F} = (2\hat{i}+\hat{j})\times(4\hat{i}-3\hat{j})\]
\[= 2\cdot4(\hat{i}\times\hat{i}) + 2(-3)(\hat{i}\times\hat{j}) + 1\cdot4(\hat{j}\times\hat{i}) + 1(-3)(\hat{j}\times\hat{j})\]
\[= 0 - 6\hat{k} - 4\hat{k} - 0 = -10\hat{k}\text{ N·m}\]
The torque points into the page (clockwise rotation tendency).
🔧 Activity 6.2 — Door Handle and Torque
L3 Apply
Materials: A door, your finger.
Procedure:
- Try to push a door open by pressing very near the hinges.
- Now push at the handle (far from hinges) with the same finger pressure.
- Try pulling the handle straight away from the door (parallel to the door surface).
- Try pulling at 30° to the door surface.
Predict: Where and at what angle is the door easiest to open?
Conclusion: τ = rF sin θ. Step 1 (small r) gives small torque — door barely moves. Step 2 (full r, θ = 90°) gives maximum torque. Step 3 (θ = 0°) gives zero torque — no rotation. Step 4 (θ = 30°) gives 50% of step-2 torque. This is exactly why door handles are placed FAR from hinges — to maximize the lever arm.
🎯 Competency-Based Questions
A student tightens a lug nut on a car wheel using a 0.40 m wrench. She applies a 200 N force at the end of the wrench at angle 60° from the wrench shaft.
Q1. Calculate the torque applied.L3 Apply
Answer: (b). τ = 0.40 × 200 × sin 60° = 80 × 0.866 ≈ 69.3 N·m.
Q2. To maximize the torque with the same force, at what angle should she pull? L3 Apply
Answer: 90° — perpendicular to the wrench. sin 90° = 1 ⇒ τ_max = 0.40 × 200 = 80 N·m.
Q3. Find \(\hat{j} \times \hat{k}\) and \(\hat{k} \times \hat{j}\). L1 Remember
Answer: \(\hat{j}\times\hat{k} = \hat{i}\) and \(\hat{k}\times\hat{j} = -\hat{i}\). Cross product is anti-commutative.
Q4. State whether TRUE or FALSE: "Torque is the angular analogue of force." Justify. L5 Evaluate
Answer: TRUE. Torque produces angular acceleration as force produces linear acceleration: τ = Iα is the rotational analogue of F = ma. Units (N·m) and dimensions ([ML²T⁻²]) match this analogy.
Q5. HOT: A turbine blade tip moves at 250 m/s for ω = 100 rad/s. (a) What is the blade tip's distance from axis? (b) What linear acceleration does the tip experience if α = 50 rad/s²? L6 Create
(a) v = ωr ⇒ r = 250/100 = 2.5 m.
(b) Tangential acceleration a_t = αr = 50 × 2.5 = 125 m/s². Centripetal a_c = ω²r = 10000 × 2.5 = 25 000 m/s². Total magnitude ≈ √(125² + 25000²) ≈ 25 000 m/s² (centripetal dominates).
(b) Tangential acceleration a_t = αr = 50 × 2.5 = 125 m/s². Centripetal a_c = ω²r = 10000 × 2.5 = 25 000 m/s². Total magnitude ≈ √(125² + 25000²) ≈ 25 000 m/s² (centripetal dominates).
🧠 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): The cross product of two parallel vectors is zero.
Reason (R): sin 0° = 0.
Answer: (A). Both true; R explains A. |A × B| = |A||B|sin θ vanishes for θ = 0 or 180°.
Assertion (A): Torque depends on the choice of axis.
Reason (R): Torque is defined as r × F, where r is from the chosen origin.
Answer: (A). Both true; R explains A. Different origins give different r vectors and hence different torques.
Assertion (A): A × B and B × A are equal in magnitude but opposite in direction.
Reason (R): Cross product is anti-commutative.
Answer: (A). Both true; R explains A.
Frequently Asked Questions - Vector Product Angular Velocity
What is the main concept covered in Vector Product Angular Velocity?
In NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion), "Vector Product Angular Velocity" 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 Vector Product Angular Velocity useful in real-life applications?
Real-life applications of Vector Product Angular Velocity 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 Vector Product Angular Velocity?
Key formulas in Vector Product Angular Velocity (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. Vector Product Angular Velocity 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 Vector Product Angular Velocity?
CBSE board questions from Vector Product Angular Velocity 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 Vector Product Angular Velocity 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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