Physics · System Of Particles And Rotational Motion · NEET
They describe the same quantity but at different levels. L = r × p is the angular momentum of a single particle about a point. When you add up r × p for every particle of a rigid body rotating about a fixed axis, the total component of angular momentum ALONG the axis comes out to L = Iω. So L = Iω is what you get after summing L = r × p over the whole spinning body for the axis direction. For NEET, use L = r × p for a single particle, and L = Iω for a rotating rigid body.
For a general body the full angular momentum vector L is not always parallel to ω. But when the body spins about a fixed axis (like a wheel, disc, or the axle of a fan), the component of angular momentum along that axis is exactly L = Iω, and that is the part that matters. NEET problems almost always give a symmetric body (disc, ring, sphere, cylinder) spinning about its symmetry axis, so L points along ω and L = Iω is fully valid.
I is always the moment of inertia about the ACTUAL axis of rotation. If the axis passes through the centre of mass, use the standard formula (for example I = MR²/2 for a disc). If the axis is shifted, first apply the parallel axis theorem I = I_cm + Md² to get I about the real axis, then use L = Iω.
Not by itself. L changes only when an external torque acts (τ = dL/dt). If no external torque acts and you change I (say a skater pulls arms in), then L = Iω stays constant, so ω must increase to compensate. That is the seed of conservation of angular momentum, the next topic.
They are linked but not the same. KE_rot = (1/2)Iω². Using L = Iω you can write KE = L²/(2I) or KE = (1/2)Lω. So if L stays fixed and I increases, the kinetic energy DECREASES (KE = L²/2I). This is exactly why the spinning-sphere NEET question works.
A solid sphere is rotating freely about its symmetry axis in free space. Its radius is increased keeping the mass constant. Which quantity stays constant?
The Sun rotates once in about 27 days. If it expanded to twice its present radius (uniform-density sphere, no external influence), its new period would be about:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
L = Iω, where I is the moment of inertia about that axis and ω is the angular speed in rad/s. It is the rotational analogue of linear momentum p = mv.
kg·m²/s (kilogram metre squared per second). This is the same as J·s (joule second). It comes from I (kg·m²) times ω (1/s).
For a symmetric body spinning about its symmetry axis, L points along the axis in the same direction as ω, given by the right-hand rule: curl the fingers in the direction of spin and the thumb points along L.
By τ = dL/dt. For a fixed axis this becomes τ = I(dω/dt) = Iα, the rotational form of Newton's second law. If τ = 0, then L is constant.
It is a vector. But for rotation about a single fixed axis we usually work with its component along the axis, L = Iω, and treat it like a signed scalar (positive one way, negative the other).
