11.8
Angular Momentum: Single Particle
Angular momentum is directed perpendicular to the plane of the rotation, and its magnitude depends on the choice of the origin. The perpendicular vect…
Every rotating object, like a spinning top, possesses an angular momentum L, which is mathematically expressed as the cross-product of the position vector and the linear momentum, which is mass times velocity.
The magnitude of angular momentum equals to mvrsinθ, and its SI units are kilogram meter square per second.
For example, consider a particle moving in the xy-plane and having a position vector r. The right-hand thumb gives the direction of angular momentum if the fingers are curled in the direction of rotation.
The time derivative of angular momentum can be expressed using the vector rule for the derivatives. The first term contains the cross-product of the velocity vector with itself and hence equals zero.
In the second term, mass times acceleration is the net force acting on the particle; recall that the cross product of r and F is torque. Thus, the rate of change of angular momentum is equal to the torque exerted by the net force acting on the particle.
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Q1: How is angular momentum defined mathematically for a single particle?
Angular momentum L is the cross-product of the position vector and linear momentum (mass times velocity). Its magnitude equals mvrsinθ, where θ is the angle between position and velocity vectors. The SI units are kilogram meter squared per second. Direction is determined using the right-hand rule: curl fingers in the rotation direction, and the thumb points along the angular momentum vector.
Q2: What role does the lever arm play in calculating angular momentum?
The lever arm is the perpendicular vector from the origin to the linear momentum vector. If the lever arm and linear momentum are collinear, angular momentum equals zero. The magnitude of angular momentum depends directly on the lever arm length and the choice of origin, making origin selection critical for accurate calculations.
Q3: How does torque relate to changes in angular momentum?
The rate of change of angular momentum equals the net torque acting on the particle. This is the rotational analog of Newton's second law for linear momentum. Mathematically, dL/dt = τ, where τ is the torque from the net force. Both torque and angular momentum must be measured relative to the same origin fixed in an inertial frame.
Q4: Why does angular momentum depend on the choice of origin?
Angular momentum is directed perpendicular to the plane of rotation, and its magnitude depends on the perpendicular distance from the origin to the linear momentum vector. Different origins produce different position vectors r, which directly affect the cross-product calculation. Therefore, angular momentum values are only meaningful when referenced to a specific, fixed origin in an inertial frame.
Q5: What does the right-hand rule tell us about angular momentum direction?
The right-hand rule determines angular momentum direction: curl your fingers in the direction of rotation, and your thumb points along the angular momentum vector. This method applies because angular momentum is a cross-product, which always produces a vector perpendicular to both the position and momentum vectors involved in the calculation.
Q6: How does net torque cause changes in a rotating particle's motion?
Net torque acting on rotating bodies causes angular momentum to change over time. This relationship mirrors Newton's second law for rotational motion: the time derivative of angular momentum equals the applied torque. Without net torque, angular momentum remains constant, demonstrating the rotational analog of inertia in linear systems.
Q7: When is angular momentum zero for a particle in motion?
Angular momentum is zero when the lever arm and linear momentum vector are collinear, meaning they lie along the same line. In this case, the particle moves directly toward or away from the origin, producing no rotational motion about that point. The cross-product of parallel vectors always equals zero, resulting in zero angular momentum.