Physics · System Of Particles And Rotational Motion · NEET
No external torque acts about her vertical spin axis (friction from the ice is tiny), so her angular momentum L = Iω is conserved. When she pulls her arms in, her mass moves closer to the axis, so her moment of inertia I becomes smaller. To keep Iω constant, ω must rise. Numerically, if I becomes half, ω doubles: Iω = I'ω' gives I(ω) = (I/2)(ω'), so ω' = 2ω. This is the exact idea NEET tests with skaters, divers and acrobats from NCERT.
No. Angular momentum L = Iω is conserved, but rotational kinetic energy K = (1/2)Iω^2 = L^2/(2I) is NOT. When I decreases, K actually increases. The extra energy comes from the muscular work the skater does to pull her arms inward against the outward centrifugal tendency. NEET loves this trap: L stays the same, K goes up. Remember K = L^2/(2I), so smaller I means larger K for fixed L.
Only when the net external torque about the chosen axis is zero: τ_ext = dL/dt, so τ_ext = 0 gives L = constant. Internal forces (like the skater's muscles, or two discs pressing together) cannot change the total L. So collisions, coupling of rotating discs, and a person walking on a turntable are all conservation-of-L problems as long as no outside torque acts.
Linear momentum p = mv is conserved when the net external force is zero. Angular momentum L = Iω is conserved when the net external torque is zero. They are separate conditions. A body can conserve L (spinning freely) while its linear momentum changes, or vice versa. For NEET, match the trigger: 'no external force' → p conserved; 'no external torque' → L conserved.
Gravity acts through the centre of mass in many symmetric cases, giving zero torque about that point, so L is conserved. For a planet moving under the Sun's pull, the force is central (always along the line joining them), so torque about the Sun is zero and L is conserved. This is exactly why a planet moves fastest at perihelion and slowest at aphelion (Kepler's law of areas).
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.
If the net external torque on a system is zero, its total angular momentum stays constant in both magnitude and direction. For rotation about a fixed axis this reads L = Iω = constant, so I1·ω1 = I2·ω2.
Iω = I'ω' (initial Iω equals final I'ω'). Combine it with I = Σmr^2 for point masses or the standard I of the body. To check energy, use K = (1/2)Iω^2 = L^2/(2I).
NCERT lists: (1) an ice skater or ballet dancer speeding up a spin by pulling in the arms, (2) a diver or acrobat curling up to somersault faster in mid-air, and (3) a planet moving faster near the Sun (Kepler's law of areas from a central force).
Yes. L = Iω points along the axis of rotation, given by the right-hand rule. Conservation means both its size and direction are fixed when torque is zero, which is why a spinning top or gyroscope keeps its axis pointing the same way.
It appears almost every year as a one-step 'Iω = I'ω'' calculation or as a conceptual trap about kinetic energy. Mastering K = L^2/(2I) lets you answer both the 'ω doubles' numericals and the 'is energy conserved' concept questions quickly.
