Relation Between Rotational Kinetic Energy and Angular Momentum
Physics · System Of Particles And Rotational Motion · NEET
For a body rotating about a fixed axis, the link between rotational kinetic energy (KE) and angular momentum (L) is KE = L^2 / (2I), or equivalently L = sqrt(2 I KE), where I is the moment of inertia. This comes from KE = (1/2) I w^2 and L = I w. Memory hook: it is the exact rotational twin of the linear formula KE = p^2 / (2m), just swap mass m for moment of inertia I and momentum p for angular momentum L.
A rigid body spinning about a fixed axis has L = I w (green, along the axis). The same rotational kinetic energy can be written as (1/2) I w^2 or, using w = L/I, as L^2/(2I). This gives L = sqrt(2 I KE).
Your doubts, answered
Is KE = L^2/2I the same thing as KE = (1/2) I w^2?
Yes, they are the same energy written two ways. Start with KE = (1/2) I w^2. Since L = I w, we get w = L/I. Substitute: KE = (1/2) I (L/I)^2 = L^2 / (2I). Use (1/2) I w^2 when you know I and w. Use L^2/(2I) when the angular momentum L is the quantity given or conserved.
Why is it L^2 over 2I, and not simply L times w?
L times w is not energy. L*w = (I w)(w) = I w^2, which is TWICE the kinetic energy. The correct rotational KE has a factor of one-half: KE = (1/2) I w^2. Written in terms of L, that half stays, giving KE = L^2/(2I). A quick check: KE = (1/2) L w is also correct because (1/2)(I w)(w) = (1/2) I w^2.
Two bodies have equal kinetic energy. Which one has larger angular momentum?
From L = sqrt(2 I KE), if KE is the same for both, then L is larger for the body with the larger moment of inertia I. So a fatter, more spread-out body (bigger I) carries more angular momentum at the same energy. This is exactly what NEET 2016 tested.
How does this connect to the linear (straight-line) formulas?
It is a one-to-one analogy. Linear: KE = p^2/(2m). Rotational: KE = L^2/(2I). Replace mass m by moment of inertia I, and linear momentum p by angular momentum L. Remembering one instantly gives you the other, which saves time in the NEET exam.
If angular momentum L is conserved and I increases, what happens to KE?
When L is fixed (no external torque, like a spinning skater pulling arms in or out), KE = L^2/(2I). If I increases, KE decreases; if I decreases, KE increases. So a skater who pulls the arms IN reduces I, which speeds up spin AND raises rotational KE (the extra energy comes from the muscular work of pulling in).
⚠️ The NEET trap ✗ Students plug into KE = (1/2) I w^2 but forget that when L is held constant, changing I changes both w and KE, so they wrongly assume KE stays the same when L is conserved. ✓ When L is conserved, use KE = L^2/(2I). Only L is constant; KE is NOT. If I doubles, w halves (from L = I w), and KE becomes L^2/(2*2I) = half of the original. Energy is not conserved here. 🧠 L constant does not mean KE constant. KE = L^2/2I still depends on I.
Real NEET questions
2016
Two rotating bodies A and B of masses m and 2m with moments of inertia I_A and I_B (I_B > I_A) have equal kinetic energy of rotation. If L_A and L_B are their angular momenta respectively, then:
A · L_A = L_B / 2
B · L_A = 2 L_B
C · L_B > L_A ✓
D · L_A > L_B
Solution: Use the relation KE = L^2 / (2I), which rearranges to L = sqrt(2 I KE). Both bodies have equal rotational kinetic energy, so KE is the same for A and B. Then L is proportional to sqrt(I). Since I_B > I_A, we get sqrt(I_B) > sqrt(I_A), so L_B > L_A. Mass values are extra data meant to distract; only I and KE matter here. Answer: C.
Solved System Of Particles And Rotational Motion NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula relating rotational kinetic energy and angular momentum?
KE = L^2 / (2I), where L is angular momentum and I is moment of inertia. Equivalently L = sqrt(2 I KE), or KE = (1/2) L w.
Does this formula work only for rotation about a fixed axis?
Yes. The clean relations L = I w and KE = L^2/(2I) apply to a rigid body rotating about a fixed axis that is an axis of symmetry, where L and w point along the same axis.
What is the linear analogue of KE = L^2/2I?
The straight-line version is KE = p^2/(2m), with p = linear momentum and m = mass. Swap m for I and p for L to move between them.
For equal angular momentum, which body has more kinetic energy?
From KE = L^2/(2I), for equal L the body with the SMALLER moment of inertia I has the larger kinetic energy.
Why does a spinning skater speed up when pulling arms in?
With no external torque, L = I w stays constant. Pulling arms in lowers I, so w rises. Also KE = L^2/(2I) rises because I falls; the added energy comes from the work done pulling the arms inward.
