Q.Write the relation between angular momentum and rotational kinetic energy. Draw a graph for the same. For two objects of the same angular momentum, compare the moment of inertia using the graph.

Step 1. Derive the relation between angular momentum and rotational kinetic energy.

For a rigid body rotating with angular velocity $\omega$ and moment of inertia $I$ about the axis, the rotational kinetic energy is $KE=\tfrac12 I\omega^2$ and the angular momentum is $L=I\omega$. Multiplying the numerator and denominator of the KE expression by $I$:

So the two rotational quantities are linked by

This is the exact rotational counterpart of $KE=p^2/2m$ for translational motion, with $L\to p$ and $I\to m$.

Step 2. Fix $L$ and describe how KE depends on $I$ alone.

If a particular value of angular momentum $L$ is held fixed (for instance, because no external torque acts and $L$ is conserved), then $L^2$ in the formula above is just a constant, so

Rotational kinetic energy is INVERSELY proportional to the moment of inertia when angular momentum is held fixed: increasing $I$ decreases KE, and decreasing $I$ increases KE (this is exactly the physics behind an ice skater spinning faster, and gaining kinetic energy, when pulling their arms in to reduce $I$, at the cost of doing muscular work against the outward pseudo-force).

Step 3. Describe the graph precisely.

Plot moment of inertia $I$ along the horizontal (x) axis and rotational kinetic energy $KE$ along the vertical (y) axis, for one single fixed value of $L$. Because $KE=\dfrac{L^2/2}{I}$ is of the form $y=\text{constant}/x$, the curve is a rectangular hyperbola lying in the first quadrant: it starts very high (large KE) for small $I$ near the y-axis, falls steeply at first, and then curves over into a long, ever-flatter tail approaching (but never touching) the x-axis as $I$ grows large. The curve never crosses either axis and is smoothly, monotonically decreasing throughout — there is exactly one point on the curve for each possible value of $I$, and moving further right along the curve always means a strictly smaller KE.

Step 4. Use the graph to compare two objects with the SAME $L$. …

Algebraic derivation of $KE=L^2/2I$ plus graphical (hyperbo …

• Confusing this with the (different, non-fixed-$L$) relation $KE=\tfrac12 I\omega^2$, and wrongly concluding KE increases with $I$ — that direct proportionality only holds at fixed $\omega$, not at fixed $L$; the question specifically fixes $L$, which flips the dependence to inverse. …