Physics · System Of Particles And Rotational Motion · NEET
No. Linear momentum p = mv describes straight-line motion and points along the velocity. Angular momentum L = r × p is the moment of that momentum about a chosen point O. It is a cross product, so it points perpendicular to the plane containing r and p, and its value depends on how far the particle's line of motion is from O. A particle can have constant linear momentum yet different angular momentum about different points.
From L = r p sin θ, L becomes zero in three cases: (1) the linear momentum vanishes, p = 0 (particle at rest); (2) the particle is at the origin, r = 0; or (3) the line of motion of p passes through the origin, so θ = 0° or 180° and sin θ = 0. This last case is important: a particle moving straight toward or away from O has zero angular momentum about O.
Yes, as long as its line of motion does not pass through the reference point O. Even in pure straight-line motion, L = r⊥ p, where r⊥ is the perpendicular distance from O to the line of motion. This distance stays constant for uniform straight-line motion, so the angular momentum about O stays constant too. It is zero only when the line passes through O.
r⊥ = r sin θ is the perpendicular distance from the point O to the straight line along which the momentum p points (the line of action of p). Using it, L = r⊥ p. It is often easier than L = r p sin θ, because you just drop a perpendicular from O onto the velocity line and read off that distance. This mirrors the moment arm used in torque.
Because r is measured from a chosen origin O. Change O and you change r, so you change L = r × p. There is no single 'angular momentum' of a particle in isolation; you must always state 'about point O'. This is exactly why NEET questions ask for L about a specific point, and why answers differ from point to point unless the momentum passes through both.
L = r × p is the general definition for a single particle about a point. L = Iω is the special result for a rigid body rotating about a fixed axis, where I is the moment of inertia and ω the angular velocity. L = Iω is derived by summing r × p for every particle of the body about the axis. So L = r × p is the foundation; L = Iω is a convenient shortcut for fixed-axis rotation.
A thin horizontal disc rotates about a vertical axis passing through its fixed centre O. Its angular momentum is L_A and L_B when computed about points A and B respectively, where OB = 2 × OA. The value of L_A / L_B is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The SI unit is kg m^2 s^-1, which is the same as J s (joule second). This comes from L = r p, i.e. (metre)(kg m s^-1) = kg m^2 s^-1.
L = r × p, where r is the position vector from the reference point and p = mv is the linear momentum. Its magnitude is L = r p sin θ = r⊥ p, where θ is the angle between r and p.
Use the right-hand rule for the cross product r × p: point the fingers of your right hand along r, curl them toward p, and the thumb gives the direction of L. It is always perpendicular to the plane containing r and p.
Angular momentum is a vector. It has both magnitude (L = r p sin θ) and direction (perpendicular to the r-p plane, set by the right-hand rule), and it obeys vector addition.
It is the base for the relation torque = dL/dt and for conservation of angular momentum, both heavily tested. Understanding L = r × p for one particle lets you extend to L = Iω for rigid bodies and solve rotation problems correctly.
