Give the most detailed explanation of how angular momentum can be obtained L=rp---Here is the complete and accurate extraction of the content from the image: **1. Basic Concepts: Linear Momentum and Torque** Before arriving at angular momentum (L), we need to understand its two foundational precursors: * **Linear Momentum (p):** A measure of the "quantity of motion" of an object in translational motion. * **Formula:** p = m ⋅ v * *m* is mass, *v* is velocity. * Linear momentum answers "how hard is it to stop this moving object?". * **Torque (τ):** A measure of the effectiveness of a force in causing rotation about an axis. * **Formula:** τ = r × F * *r* is the position vector from the axis of rotation to the point where the force is applied. * *F* is the applied force. * Torque answers "how effectively does this force cause a spin?". **2. The Analogy: Building Intuition for "r × p"** Angular momentum is the rotational analogue of linear momentum. If linear momentum (p) is concerned with translation, angular momentum (L) is concerned with rotation. Imagine a particle moving in a straight line. It has momentum p and is hard to stop. Now, imagine that same particle tethered to a string and whirled in a circle (like a rock on a string). It is now rotating. The "difficulty in stopping its rotation" is what angular momentum measures. How do we quantify this "rotational difficulty"? It intuitively depends on two things: 1. **The magnitude of its linear motion (p).** A larger, faster rock (m and v are large) is harder to stop spinning. 2. **How far the particle is from the center of rotation (r).** Spinning a rock on a a 2-meter string is much harder to stop than on a 0.5-meter string, even if p is the same. This is because a longer lever arm (r) provides a "mechanical advantage" for sustaining rotational motion. Therefore, L should be proportional to both r and p. A simple product (r ⋅ p) is insufficient because direction is crucial in rotation. This is why we use the cross product, which naturally yields a quantity perpendicular to the plane of rotation. Hence, L = r × p. **3. Formal Derivation from Newton's Laws** The most rigorous explanation comes from Newton's second law for rotation. * **Newton's II Law for Translation:** Force is the rate of change of linear momentum. * F = dp/dt * **Newton's II Law for Rotation:** Torque is the rate of change of angular momentum. * τ = dL/dt We can derive the formula L = r × p from this relationship. 1. Start with the definition of torque: τ = r × F 2. Substitute F using Newton's law F = dp/dt: τ = r × (dp/dt) 3. Now, let's find the time derivative of (r × p) using the product rule for cross products: d/dt (r × p) = (dr/dt × p) + (r × dp/dt) 4. Analyze the first term, (dr/dt × p): * dr/dt is the velocity, v. * p is momentum, m ⋅ v. * Therefore, dr/dt × p = v × (m ⋅ v). (Note: m is a scalar) * The cross product of any two parallel vectors (v and v) is zero. * Thus, (dr/dt × p) = 0. 5. The equation now simplifies to: d/dt (r × p) = 0 + (r × dp/dt) 6. Notice that (r × dp/dt) is exactly equal to τ from step 2. Therefore, d/dt (r × p) = τ 7. However, from Newton's law for rotation, we know τ = dL/dt. dL/dt = d/dt (r × p) 8. Integrating both sides with respect to time yields: L = r × p + constant 9. If we choose the initial condition where angular momentum is zero (no rotation), the integration constant becomes zero. We thus arrive at the fundamental definition: L = r × p **4. Details of the Cross Product and its Magnitude** The cross product r × p is not simple multiplication. Its result is a vector whose direction is given by the right-hand rule and whose magnitude is given by: |L| = |r × p| = |r| |p| sinθ Where: * |r| is the distance from the particle to the origin/axis of rotation (in meters, m). * |p| is the magnitude of linear momentum (in kg·m/s). * θ is the angle between the vector r and the vector p. Now, consider the sinθ component. If a particle moves in a perfect circle, the vector r (radial) and the vector p (tangential) are always perpendicular. The angle between two perpendicular vectors is 90°, and sin(90°) = 1. Therefore, for circular motion, the equation simplifies greatly: |L| = |r| |p| (1) = r ⋅ p And since p = m ⋅ v, we can write: |L| = r ⋅ m ⋅ v This is why you often see the formula L = mvr in introductory textbooks. It is critical to remember that L = mvr is a special case that ONLY applies to circular motion or when the angle θ=90°. The most general and always correct formula is L = r × p with a magnitude of r p sinθ. **5. Conclusion and Summary** In summary, angular momentum L is fundamentally defined as r × p because: 1. **Conceptually:** It is the precise rotational analogue of linear momentum, accounting for both the "inertia" of motion (p) and the "lever arm" of rotation (r). 2. **Mathematically:** A formal derivation from Newton's second law (τ = dL/dt) directly leads to the definition L = r × p. 3. **Geometrically:** The cross product naturally captures the geometry of rotation, producing a vector whose direction indicates the axis of spin (via the right-hand rule) and whose magnitude depends on the "effectiveness" of r and p in producing rotation (sinθ). The formula L = Iω (where I is the moment of inertia and ω is angular velocity) is another form of the same concept, specialized for a rigid body rotating about a fixed axis. For a single particle, I = mr², so L = (mr²)ω = mr(rω) = mr v = r (m v) = r p, demonstrating the consistency between the two expressions.