Physics · System Of Particles And Rotational Motion · NEET
Every small piece of a spinning body moves with a different linear speed v = rω (pieces far from the axis move faster). If you add up ½(Δm)v² for all pieces, v² = r²ω², so K = ½(Σ Δm r²)ω². The bracket Σ Δm r² is exactly the moment of inertia I. So K = ½Iω². Mass m alone cannot appear because different parts have different speeds — only I captures how the mass is spread out from the axis.
Use K = L²/(2I). This comes from L = Iω, so ω = L/I, and putting that into K = ½Iω² gives K = ½I(L/I)² = L²/(2I). This is the fastest form when L is fixed (like conservation of angular momentum problems) or when comparing two bodies with equal KE — a favourite NEET trick.
A rolling body both moves forward and spins, so its total KE has two parts: K_total = ½mv²(translation) + ½Iω²(rotation). For rolling without slipping, v = Rω, so you can combine them into K_total = ½mv²(1 + I/mR²). The extra I/mR² term is why a rolling ring is slower than a rolling sphere down the same incline — more of the energy goes into spinning.
Yes. I depends on the axis (through Σmr²), so the same body with the same ω has different rotational KE about different axes. A rod spun about its centre has less I (and less KE) than the same rod spun about its end at the same ω. Always identify the axis before using K = ½Iω².
Rearranged, L = √(2·I·K). If the KE is equal, the body with the larger moment of inertia I has the larger angular momentum L. This exact reasoning was asked in NEET 2016 — bigger I at equal KE means bigger L.
A solid sphere of mass m and radius R rotates about its diameter. A solid cylinder of the same mass and radius rotates about its geometric axis with twice the sphere's angular speed. The ratio of their rotational kinetic energies (sphere : cylinder) is:
Two rotating bodies A and B (masses m and 2m, moments of inertia I_A and I_B with I_B > I_A) have equal rotational kinetic energy. If L_A and L_B are their angular momenta, then:
A solid sphere rotates freely about its symmetry axis in free space. Its radius is increased keeping the mass constant. Which quantity stays constant?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
K = ½Iω², where I is the moment of inertia about the rotation axis and ω is the angular speed in rad/s. It is the rotational version of KE = ½mv², with I replacing m and ω replacing v.
The joule (J), the same as any energy. I is in kg·m² and ω is in rad/s, so ½Iω² gives kg·m²·(1/s²) = kg·m²/s² = J.
K = L²/(2I), where L = Iω is the angular momentum. This form is useful when L is fixed or when two bodies share equal KE or equal L.
A rolling ball translates and rotates at the same time, so K_total = ½mv² + ½Iω². For rolling without slipping v = Rω, letting you combine both into ½mv²(1 + I/mR²).
Yes, through I. For the same mass, spread out farther from the axis means larger I and therefore larger K at the same ω. That is why a ring stores more rotational energy than a disc of equal mass and radius spinning at the same speed.
